# 7.5: Solve Rational Equations

After defining the terms ‘expression’ and ‘equation’ earlier, we have used them throughout this book. Now we will solve a rational equation.

You must make sure to know the difference between rational expressions and rational equations. The equation contains an equal sign.

## Solve Rational Equations

We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.

We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then, we will have an equation that does not contain rational expressions and thus is much easier for us to solve. But because the original equation may have a variable in a denominator, we must be careful that we don’t end up with a solution that would make a denominator equal to zero.

So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard.

An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an extraneous solution to a rational equation.

Extraneous Solution to a Rational Equation

An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined.

We note any possible extraneous solutions, (c), by writing (x eq c) next to the equation.

Example (PageIndex{1}): How to Solve a Rational Equation

Solve: [dfrac{1}{x}+dfrac{1}{3}=dfrac{5}{6} onumber ]

Solution

Step 1. Note any value of the variable that would make any denominator zero.

If (x=0), then (dfrac{1}{x}) is undefined. So we'll write (x eq 0) next to the equation.

[dfrac{1}{x}+dfrac{1}{3}=dfrac{5}{6}, x eq 0 onumber ]

Step 2. Find the least common denominator of all denominators in the equation.

Find the LCD of (dfrac{1}{x}), (dfrac{1}{3}), and (dfrac{5}{6})

The LCD is (6x).

Step 3. Clear the fractions by multiplying both sides of the equation by the LCD.

Multiply both sides of the equation by the LCD, (6x).

[{color{red}6 x} cdotleft(dfrac{1}{x}+dfrac{1}{3} ight)={color{red}6 x} cdotleft(dfrac{5}{6} ight) onumber ]

Use the Distributive Property.

[{color{red}6 x} cdot dfrac{1}{x}+{color{red}6 x} cdot dfrac{1}{3}={color{red}6 x} cdotleft(dfrac{5}{6} ight) onumber ]

Simplify - and notice, no more fractions!

[6+2 x=5 x onumber ]

Step 4. Solve the resulting equation.

Simplify.

[egin{aligned} &6=3 x &2=x end{aligned} onumber ]

Step 5. Check.

If any values found in Step 1 are algebraic solutions, discard them. Check any remaining solutions in the original equation.

We did not get 0 as an algebraic solution.

[dfrac{1}{x}+dfrac{1}{3}=dfrac{5}{6} onumber ]

We substitute (x=2) into the original equation.

[egin{aligned} frac{1}{2}+frac{1}{3}&overset{?}{=}frac{5}{6} frac{3}{6}+frac{2}{6}&overset{?}{=}frac{5}{6} frac{5}{6}&=frac{5}{6} surd end{aligned} onumber ]

The solution is (x=2)

Exercise (PageIndex{1})

Solve: [dfrac{1}{y}+dfrac{2}{3}=dfrac{1}{5} onumber ]

(y=-dfrac{7}{15})

Exercise (PageIndex{2})

Solve: [dfrac{2}{3}+dfrac{1}{5}=dfrac{1}{x} onumber ]

(x=dfrac{13}{15})

The steps of this method are shown.

how to Solve equations with rational expressions.

• Step 1. Note any value of the variable that would make any denominator zero.
• Step 2. Find the least common denominator of all denominators in the equation.
• Step 3. Clear the fractions by multiplying both sides of the equation by the LCD.
• Step 4. Solve the resulting equation.
• Step 5. Check:
• If any values found in Step 1 are algebraic solutions, discard them.
• Check any remaining solutions in the original equation.

We always start by noting the values that would cause any denominators to be zero.

Example (PageIndex{2}): How to Solve a Rational Equation using the Zero Product Property

Solve: [1-dfrac{5}{y}=-dfrac{6}{y^{2}} onumber ]

Solution

Note any value of the variable that would make any denominator zero.

[1-dfrac{5}{y}=-dfrac{6}{y^{2}}, y eq 0 onumber ]

Find the least common denominator of all denominators in the equation. The LCD is (y^2).

Clear the fractions by multiplying both sides of the equation by the LCD.

[y^{2}left(1-dfrac{5}{y} ight)=y^{2}left(-dfrac{6}{y^{2}} ight) onumber ]

Distribute.

[y^{2} cdot 1-y^{2}left(dfrac{5}{y} ight)=y^{2}left(-dfrac{6}{y^{2}} ight) onumber ]

Multiply.

[y^{2}-5 y=-6 onumber ]

Solve the resulting equation. First write the quadratic equation in standard form.

[y^{2}-5 y+6=0 onumber ]

Factor.

[(y-2)(y-3)=0 onumber ]

Use the Zero Product Property.

[y-2=0 ext { or } y-3=0 onumber ]

Solve.

[y=2 ext { or } y=3 onumber ]

Check. We did not get (0) as an algebraic solution.

Check (y=2) and (y=3)in the original equation.

The solution is (y=2,y=3)

Exercise (PageIndex{3})

Solve: [1-dfrac{2}{x}=dfrac{15}{x^{2}} onumber ]

(x=-3, x=5)

Exercise (PageIndex{4})

Solve: [1-dfrac{4}{y}=dfrac{12}{y^{2}} onumber ]

(y=-2, y=6)

In the next example, the last denominators is a difference of squares. Remember to factor it first to find the LCD.

Example (PageIndex{3})

Solve: [dfrac{2}{x+2}+dfrac{4}{x-2}=dfrac{x-1}{x^{2}-4} onumber ]

Solution

Note any value of the variable that would make any denominator zero.

[dfrac{2}{x+2}+dfrac{4}{x-2}=dfrac{x-1}{(x+2)(x-2)}, x eq-2, x eq 2 onumber ]

Find the least common denominator of all denominators in the equation. The LCD is ((x+2)(x-2)).

Clear the fractions by multiplying both sides of the equation by the LCD.

[(x+2)(x-2)left(dfrac{2}{x+2}+dfrac{4}{x-2} ight)=(x+2)(x-2)left(dfrac{x-1}{x^{2}-4} ight) onumber ]

Distribute.

[(x+2)(x-2) dfrac{2}{x+2}+(x+2)(x-2) dfrac{4}{x-2}=(x+2)(x-2)left(dfrac{x-1}{x^{2}-4} ight) onumber ]

Remove common factors.

[cancel {(x+2)}(x-2) dfrac{2}{cancel {x+2}}+(x+2){cancel {(x-2)}} dfrac{4}{cancel {x-2}}=cancel {(x+2)(x-2)}left(dfrac{x-1}{cancel {x^{2}-4}} ight) onumber ]

Simplify.

[2(x-2)+4(x+2)=x-1 onumber ]

Distribute.

[2 x-4+4 x+8=x-1 onumber ]

Solve.

[egin{aligned} 6 x+4&=x-1 5 x&=-5 x&=-1 end{aligned}]

Check: We did not get 2 or −2 as algebraic solutions.

Check (x=-1) in the original equation.

[egin{aligned} dfrac{2}{x+2}+dfrac{4}{x-2} &=dfrac{x-1}{x^{2}-4} dfrac{2}{(-1)+2}+dfrac{4}{(-1)-2} &overset{?}{=} dfrac{(-1)-1}{(-1)^{2}-4} dfrac{2}{1}+dfrac{4}{-3} &overset{?}{=} dfrac{-2}{-3} dfrac{6}{3}-dfrac{4}{3} &overset{?}{=} dfrac{2}{3} dfrac{2}{3} &=dfrac{2}{3} surd end{aligned} onumber ]

The solution is (x=-1).

Exercise (PageIndex{5})

Solve: [dfrac{2}{x+1}+dfrac{1}{x-1}=dfrac{1}{x^{2}-1} onumber ]

(x=dfrac{2}{3})

Exercise (PageIndex{6})

Solve: [dfrac{5}{y+3}+dfrac{2}{y-3}=dfrac{5}{y^{2}-9} onumber ]

(y=2)

In the next example, the first denominator is a trinomial. Remember to factor it first to find the LCD.

Example (PageIndex{4})

Solve: [dfrac{m+11}{m^{2}-5 m+4}=dfrac{5}{m-4}-dfrac{3}{m-1} onumber ]

Solution

Note any value of the variable that would make any denominator zero. Use the factored form of the quadratic denominator.

[dfrac{m+11}{(m-4)(m-1)}=dfrac{5}{m-4}-dfrac{3}{m-1}, m eq 4, m eq 1 onumber ]

Find the least common denominator of all denominators in the equation. The LCD is ((m-4)(m-1))

Clear the fractions by multiplying both sides of the equation by the LCD.

[(m-4)(m-1)left(dfrac{m+11}{(m-4)(m-1)} ight)=(m-4)(m-1)left(dfrac{5}{m-4}-dfrac{3}{m-1} ight) onumber ]

Distribute.

[(m-4)(m-1)left(dfrac{m+11}{(m-4)(m-1)} ight)=(m-4)(m-1) dfrac{5}{m-4}-(m-4)(m-1) dfrac{3}{m-1} onumber ]

Remove common factors.

[cancel {(m-4)(m-1)}left(dfrac{m+11}{cancel {(m-4)(m-1)}} ight)=cancel {(m-4)}(m-1) dfrac{5}{cancel{m-4}}-(m-4)cancel {(m-1)} dfrac{3}{cancel {m-1}} onumber ]

Simplify.

[m+11=5(m-1)-3(m-4) onumber ]

Solve the resulting equation.

[egin{aligned} m+11&=5 m-5-3 m+12 4&=m end{aligned} onumber ]

Check. The only algebraic solution was 4, but we said that 4 would make a denominator equal to zero. The algebraic solution is an extraneous solution.

There is no solution to this equation.

Exercise (PageIndex{7})

Solve: [dfrac{x+13}{x^{2}-7 x+10}=dfrac{6}{x-5}-dfrac{4}{x-2} onumber ]

There is no solution.

Exercise (PageIndex{8})

Solve: [dfrac{y-6}{y^{2}+3 y-4}=dfrac{2}{y+4}+dfrac{7}{y-1} onumber ]

There is no solution.

The equation we solved in the previous example had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. In the next example we get two algebraic solutions. Here one or both could be extraneous solutions.

Example (PageIndex{5})

Solve: [dfrac{y}{y+6}=dfrac{72}{y^{2}-36}+4 onumber ]

Solution

Factor all the denominators, so we can note any value of the variable that would make any denominator zero.

[dfrac{y}{y+6}=dfrac{72}{(y-6)(y+6)}+4, y eq 6, y eq-6 onumber ]

Find the least common denominator. The LCD is ((y-6)(y+6))

Clear the fractions.

[(y-6)(y+6)left(dfrac{y}{y+6} ight)=(y-6)(y+6)left(dfrac{72}{(y-6)(y+6)}+4 ight) onumber ]

Simplify.

[(y-6) cdot y=72+(y-6)(y+6) cdot 4 onumber ]

Simplify.

[y(y-6)=72+4left(y^{2}-36 ight) onumber ]

Solve the resulting equation.

[egin{aligned} y^{2}-6 y&=72+4 y^{2}-144 0&=3 y^{2}+6 y-72 0&=3left(y^{2}+2 y-24 ight) 0&=3(y+6)(y-4) y&=-6, y=4 end{aligned} onumber ]

Check.

(y=-6) is an extraneous solution. Check (y=4) in the original equation.

[egin{aligned} dfrac{y}{y+6} &=dfrac{72}{y^{2}-36}+4 dfrac{4}{4+6} &overset{?}{=}dfrac{72}{4^{2}-36}+4 dfrac{4}{10} &overset{?}{=} dfrac{72}{-20}+4 dfrac{4}{10} &overset{?}{=} -dfrac{36}{10}+dfrac{40}{10} dfrac{4}{10} &=dfrac{4}{10} surd end{aligned} onumber ]

The solution is (y=4).

Exercise (PageIndex{9})

Solve: [dfrac{x}{x+4}=dfrac{32}{x^{2}-16}+5 onumber ]

(x=3)

Exercise (PageIndex{10})

Solve: [dfrac{y}{y+8}=dfrac{128}{y^{2}-64}+9 onumber ]

(y=7)

In some cases, all the algebraic solutions are extraneous.

Example (PageIndex{6})

Solve: [dfrac{x}{2 x-2}-dfrac{2}{3 x+3}=dfrac{5 x^{2}-2 x+9}{12 x^{2}-12} onumber ]

Solution

We will start by factoring all denominators, to make it easier to identify extraneous solutions and the LCD.

[dfrac{x}{2(x-1)}-dfrac{2}{3(x+1)}=dfrac{5 x^{2}-2 x+9}{12(x-1)(x+1)} onumber ]

Note any value of the variable that would make any denominator zero.

[dfrac{x}{2(x-1)}-dfrac{2}{3(x+1)}=dfrac{5 x^{2}-2 x+9}{12(x-1)(x+1)}, x eq 1, x eq-1 onumber ]

Find the least common denominator. The LCD is (12(x-1)(x+1)).

Clear the fractions.

[12(x-1)(x+1)left(dfrac{x}{2(x-1)}-dfrac{2}{3(x+1)} ight)=12(x-1)(x+1)left(dfrac{5 x^{2}-2 x+9}{12(x-1)(x+1)} ight) onumber ]

Simplify.

[6(x+1) cdot x-4(x-1) cdot 2=5 x^{2}-2 x+9 onumber ]

Simplify.

[6 x(x+1)-4 cdot 2(x-1)=5 x^{2}-2 x+9 onumber ]

Solve the resulting equation.

[egin{aligned} 6 x^{2}+6 x-8 x+8&=5 x^{2}-2 x+9 x^{2}-1&=0 (x-1)(x+1)&=0 x&=1 ext { or } x=-1 end{aligned} onumber ]

Check.

(x=1) and (x=-1) are extraneous solutions.

The equation has no solution.

Exercise (PageIndex{11})

Solve: [dfrac{y}{5 y-10}-dfrac{5}{3 y+6}=dfrac{2 y^{2}-19 y+54}{15 y^{2}-60} onumber ]

There is no solution.

Exercise (PageIndex{12})

Solve: [dfrac{z}{2 z+8}-dfrac{3}{4 z-8}=dfrac{3 z^{2}-16 z-16}{8 z^{2}+2 z-64} onumber ]

There is no solution.

Example (PageIndex{7})

Solve: [dfrac{4}{3 x^{2}-10 x+3}+dfrac{3}{3 x^{2}+2 x-1}=dfrac{2}{x^{2}-2 x-3} onumber ]

Solution

Factor all the denominators, so we can note any value of the variable that would make any denominator zero.

[dfrac{4}{(3 x-1)(x-3)}+dfrac{3}{(3 x-1)(x+1)}=dfrac{2}{(x-3)(x+1)}, x eq-1, x eq dfrac{1}{3}, x eq 3 onumber ]

Find the least common denominator. The LCD is ((3 x-1)(x+1)(x-3)).

Clear the fractions.

[(3 x-1)(x+1)(x-3)left(dfrac{4}{(3 x-1)(x-3)}+dfrac{3}{(3 x-1)(x+1)} ight)=(3 x-1)(x+1)(x-3)left(dfrac{2}{(x-3)(x+1)} ight) onumber ]

Simplify.

[4(x+1)+3(x-3)=2(3 x-1) onumber ]

Distribute.

[4 x+4+3 x-9=6 x-2 onumber ]

Simplify.

[7 x-5=6 x-2 onumber ]

[x=3 onumber ]

The only algebraic solution was (x=3)$,$ but we said that (x=3) would make a denominator equal to zero. The algebraic solution is an extraneous solution.

There is no solution to this equation.

Exercise (PageIndex{13})

Solve: [dfrac{15}{x^{2}+x-6}-dfrac{3}{x-2}=dfrac{2}{x+3} onumber ]

There is no solution.

Exercise (PageIndex{14})

Solve: [dfrac{5}{x^{2}+2 x-3}-dfrac{3}{x^{2}+x-2}=dfrac{1}{x^{2}+5 x+6} onumber ]

There is no solution.

## Use Rational Functions

Working with functions that are defined by rational expressions often lead to rational equations. Again, we use the same techniques to solve them.

Example (PageIndex{8})

For rational function, (f(x)=dfrac{2 x-6}{x^{2}-8 x+15}):

1. Find the domain of the function
2. Solve (f(x)=1)
3. Find the points on the graph at this function value.

Solution

1. The domain of a rational function is all real numbers except those that make the rational expression undefined. So to find them, we will set the denominator equal to zero and solve.

[egin{aligned} x^{2}-8 x+15&=0 (x-3)(x-5)&=0 quad ext{Factor the trinomial.} x-3&=0 quad ext {Use the Zero Product Property.} x-5&=0 quad ext {Use the Zero Product Property.} x=3 &; x=5 ext{ Solve.} end{aligned} onumber ]

The domain is all real numbers except (x eq 3, x eq 5)

1. [f(x)=1 onumber ]

Substitute in the rational expression.

[dfrac{2 x-6}{x^{2}-8 x+15}=1 onumber ]

Factor the denominator.

[dfrac{2 x-6}{(x-3)(x-5)}=1 onumber ]

Multiply both sides by the LCD, ((x-3)(x-5))

[(x-3)(x-5)left(dfrac{2 x-6}{(x-3)(x-5)} ight)=(x-3)(x-5)(1) onumber ]

Simplify.

[2 x-6=x^{2}-8 x+15 onumber ]

Solve.

[0=x^{2}-10 x+21 onumber ]

Factor.

[0=(x-7)(x-3) onumber ]

Use the Zero Product Property.

Solve.

1. The value of the function is 1 when (x=7, x=3)$.$ So the points on the graph of this function when (f(x)=1)$,$ will be ((7,1),(3,1)).

Exercise (PageIndex{15})

For rational function, (f(x)=dfrac{8-x}{x^{2}-7 x+12})

1. Find the domain of the function.
2. Solve (f(x)=3).
3. Find the points on the graph at this function value.
1. The domain is all real numbers except (x eq 3) and (x eq 4)
2. (x=2, x=dfrac{14}{3})
3. ((2,3),left(dfrac{14}{3}, 3 ight))

Exercise (PageIndex{16})

For rational function, (f(x)=dfrac{x-1}{x^{2}-6 x+5})

1. Solve (f(x)=4).
2. Find the points on the graph at this function value.
1. The domain is all real numbers except (x eq 1) and (x eq 5)
2. (x=dfrac{21}{4})
3. (left(dfrac{21}{4}, 4 ight))

## Solve a Rational Equation for a Specific Variable

When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable.

When we developed the point-slope formula from our slope formula, we cleared the fractions by multiplying by the LCD.

[egin{aligned} m &=frac{y-y_{1}}{x-x_{1}} mleft(x-x_{1} ight) &=left(frac{y-y_{1}}{x-x_{1}} ight)left(x-x_{1} ight) quad ext{Multiply both sides of the equation by } x-x_1. mleft(x-x_{1} ight) &=y-y_{1} quad ext {Simplify.} y-y_{1} &=mleft(x-x_{1} ight) quad ext {Rewrite the equation with the y terms on the left.} end{aligned} onumber ]

In the next example, we will use the same technique with the formula for slope that we used to get the point-slope form of an equation of a line through the point ((2,3)). We will add one more step to solve for (y).

Example (PageIndex{9})

Solve: (m=dfrac{y-2}{x-3}) for (y).

Solution

[m=dfrac{y-2}{x-3} onumber ]

Note any value of the variable that would make any denominator zero.

[m=dfrac{y-2}{x-3}, x eq 3 onumber ]

Clear the fractions by multiplying both sides of the equation by the LCD, (x-3).

[(x-3) m=(x-3)left(dfrac{y-2}{x-3} ight) onumber ]

Simplify.

[x m-3 m=y-2 onumber ]

Isolate the term with (y).

[x m-3 m+2=y onumber ]

Exercise (PageIndex{17})

Solve: (m=dfrac{y-5}{x-4}) for (y).

(y=m x-4 m+5)

Exercise (PageIndex{18})

Solve: (m=dfrac{y-1}{x+5}) for (y).

(y=m x+5 m+1)

Remember to multiply both sides by the LCD in the next example.

Example (PageIndex{10})

Solve: (dfrac{1}{c}+dfrac{1}{m}=1) for (c)

Solution

[dfrac{1}{c}+dfrac{1}{m}=1 ext { for } c onumber ]

Note any value of the variable that would make any denominator zero.

[dfrac{1}{c}+dfrac{1}{m}=1, c eq 0, m eq 0 onumber ]

Clear the fractions by multiplying both sides of the equations by the LCD, (cm).

[cmleft(dfrac{1}{c}+dfrac{1}{m} ight)=cm(1) onumber ]

Distribute.

[cmleft(frac{1}{c} ight)+cm frac{1}{m}=cm(1) onumber ]

Simplify.

[m+c=cm onumber ]

Collect the terms with (c) to the right.

[m=cm-c onumber ]

Factor the expression on the right.

[m=c(m-1) onumber ]

To isolate (c), divide both sides by (m-1).

[dfrac{m}{m-1}=dfrac{c(m-1)}{m-1} onumber ]

Simplify by removing common factors.

[dfrac{m}{m-1}=c onumber ]

Notice that even though we excluded (c=0) and (m=0) from the original equation, we must also now state that (m eq 1).

Exercise (PageIndex{19})

Solve: (dfrac{1}{a}+dfrac{1}{b}=c) for (a).

(a=dfrac{b}{c b-1})

Exercise (PageIndex{20})

Solve: (dfrac{2}{x}+dfrac{1}{3}=dfrac{1}{y}) for (y)

(y=dfrac{3 x}{x+6})

## 67 Solve Rational Equations

If you miss a problem, go back to the section listed and review the material.

1. Solve:
If you missed this problem, review (Figure).
2. Solve:
If you missed this problem, review (Figure).
3. Solve for in terms of : for
If you missed this problem, review (Figure).

After defining the terms expression and equation early in Foundations, we have used them throughout this book. We have simplified many kinds of expressions and solved many kinds of equations. We have simplified many rational expressions so far in this chapter. Now we will solve rational equations.

The definition of a rational equation is similar to the definition of equation we used in Foundations.

A rational equation is two rational expressions connected by an equal sign.

You must make sure to know the difference between rational expressions and rational equations. The equation contains an equal sign.

### Solve Rational Equations

We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.

Here is an example we did when we worked with linear equations:

 We multiplied both sides by the LCD. Then we distributed. We simplified—and then we had an equation with no fractions. Finally, we solved that equation.

We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then we will have an equation that does not contain rational expressions and thus is much easier for us to solve.

But because the original equation may have a variable in a denominator we must be careful that we don’t end up with a solution that would make a denominator equal to zero.

So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard.

An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an extraneous solution.

An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined.

We note any possible extraneous solutions, c, by writing next to the equation.

Solve:

Solve:

Solve:

The steps of this method are shown below.

1. Note any value of the variable that would make any denominator zero.
2. Find the least common denominator of all denominators in the equation.
3. Clear the fractions by multiplying both sides of the equation by the LCD.
4. Solve the resulting equation.
5. Check.
• If any values found in Step 1 are algebraic solutions, discard them.
• Check any remaining solutions in the original equation.

We always start by noting the values that would cause any denominators to be zero.

Solve:

 Note any value of the variable that would make any denominator zero. Find the least common denominator of all denominators in the equation. The LCD is. Clear the fractions by multiplying both sides of the equation by the LCD. Distribute. Multiply. Solve the resulting equation. First write the quadratic equation in standard form. Factor. Use the Zero Product Property. Solve. Check. We did not get 0 as an algebraic solution.

Solve:

Solve:

Solve:

 Note any value of the variable that would make any denominator zero. Find the least common denominator of all denominators in the equation. The LCD is. Clear the fractions by multiplying both sides of the equation by the LCD. Remove common factors. Simplify. Multiply. Solve the resulting equation. We did not get 0 or as algebraic solutions.

Solve:

Solve:

When one of the denominators is a quadratic, remember to factor it first to find the LCD.

Solve:

 Note any value of the variable that would make any denominator zero. Find the least common denominator of all denominators in the equation. The LCD is. Clear the fractions by multiplying both sides of the equation by the LCD. Distribute. Remove common factors. Simplify. Distribute. Solve. We did not get as algebraic solutions.

Solve:

Solve:

Solve:

 Note any value of the variable that would make any denominator zero. Find the least common denominator of all denominators in the equation. The LCD is. Clear the fractions by multiplying both sides of the equation by the LCD. Distribute. Remove common factors. Simplify. Simplify. Combine like terms. Solve. First write in standard form. Factor. Use the Zero Product Property. We did not get 4 or 3 as algebraic solutions.

Solve:

Solve:

Solve:

 Factor all the denominators, so we can note any value of the variable the would make any denominator zero. Find the least common denominator of all denominators in the equation. The LCD is. Clear the fractions. Distribute. Remove common factors. Simplify. Solve the resulting equation. Check. The only algebraic solution was 4, but we said that 4 would make a denominator equal to zero. The algebraic solution is an extraneous solution. There is no solution to this equation.

Solve:

Solve:

The equation we solved in (Figure) had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. Some equations have no solution.

Solve:

 Note any value of the variable that would make any denominator zero. Find the least common denominator of all denominators in the equation. The LCD is. Clear the fractions by multiplying both sides of the equation by the LCD. Distribute. Remove common factors. Simplify. Solve the resulting equation. Check. is an extraneous solution.

Solve:

Solve:

Solve:

 Factor all the denominators, so we can note any value of the variable that would make any denominator zero. Find the least common denominator. The LCD is. Clear the fractions. Simplify. Simplify. Solve the resulting equation. Check. is an extraneous solution.

Solve:

Solve:

Solve:

 We will start by factoring all denominators, to make it easier to identify extraneous solutions and the LCD. Note any value of the variable that would make any denominator zero. Find the least common denominator.The LCD is Clear the fractions. Simplify. Simplify. Solve the resulting equation. Check. and are extraneous solutions. The equation has no solution.

Solve:

Solve:

### Solve a Rational Equation for a Specific Variable

When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable.

We’ll start with a formula relating distance, rate, and time. We have used it many times before, but not usually in this form.

Solve:

 Note any value of the variable that would make any denominator zero. Clear the fractions by multiplying both sides of the equations by the LCD, T. Simplify. Divide both sides by R to isolate T. Simplify.

Solve: for

Solve: for

(Figure) uses the formula for slope that we used to get the point-slope form of an equation of a line.

Solve:

 Note any value of the variable that would make any denominator zero. Clear the fractions by multiplying both sides of the equations by the LCD, . Simplify. Isolate the term with y. Divide both sides by m to isolate y. Simplify.

Solve: for

Solve: for

Be sure to follow all the steps in (Figure). It may look like a very simple formula, but we cannot solve it instantly for either denominator.

Solve

 Note any value of the variable that would make any denominator zero. Clear the fractions by multiplying both sides of the equations by the LCD, . Distribute. Simplify. Collect the terms with c to the right. Factor the expression on the right. To isolate c, divide both sides by . Simplify by removing common factors.

Notice that even though we excluded from the original equation, we must also now state that .

Solve: for

Solve: for

### Key Concepts

• Strategy to Solve Equations with Rational Expressions
1. Note any value of the variable that would make any denominator zero.
2. Find the least common denominator of all denominators in the equation.
3. Clear the fractions by multiplying both sides of the equation by the LCD.
4. Solve the resulting equation.
5. Check.
• If any values found in Step 1 are algebraic solutions, discard them.
• Check any remaining solutions in the original equation.

#### Practice Makes Perfect

Solve Rational Equations

In the following exercises, solve.

Solve a Rational Equation for a Specific Variable

In the following exercises, solve.

#### Everyday Math

House Painting Alain can paint a house in 4 days. Spiro would take 7 days to paint the same house. Solve the equation for t to find the number of days it would take them to paint the house if they worked together.

days

Boating Ari can drive his boat 18 miles with the current in the same amount of time it takes to drive 10 miles against the current. If the speed of the boat is 7 knots, solve the equation for c to find the speed of the current.

#### Writing Exercises

Why is there no solution to the equation ?

Pete thinks the equation has two solutions, . Explain why Pete is wrong.

#### Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

## 7.5: Solve Rational Equations

Applying Rational Equations

· Solve real world problems using rational functions.

Rational expressions and rational equations can be useful tools for representing real life situations and for finding answers to real problems. In particular, they are quite good for describing distance-speed-time questions, and modeling multi-person work problems.

Solving Work Problems

Work problems often ask us to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, W = rt. The amount of work done (W) is the product of the rate of work (r) and the time spent working (t). The work formula has 3 versions:

Some work problems have multiple machines or people working on a project together for the same amount of time but at different rates. In that case, we can add their individual work rates together to get a total work rate. Let’s look at an example:

Myra takes 2 hours to plant 500 flower bulbs. Francis takes 3 hours to plant 450 flower bulbs. Working together, how long should it take them to plant 1500 bulbs?

Myra : 500 bulbs/2 hours = 250 bulbs/hour

Think about how many bulbs each person can plant in one hour. This is their planting rate.

250 + 150 bulbs/hour = 400 bulbs/hour

Combine their hourly rates to determine the rate they work together.

Use one of the work formulas to write a rational equation, for example . We know r, the combined work rate, and we know W, the amount of work that must be done. What we don't know is how much time it will take to do the required work at the designated rate.

Solve the equation by multiplying both sides by the common denominator, then isolating t.

It takes 3 hours 45 minutes for Myra and Francis to plant 1500 bulbs together.

Other work problems go the other way. We'll calculate how long it will take one person to do a job alone when we know how long it takes a group to get it done:

Jamie, Pria and Paul can paint a room together in 2 hours. If Pria does the job alone she can paint the room in 5 hours. If Paul works alone, he can paint the room in 6 hours. If Jamie works alone, how long would it take her to paint the room?

Pria + Paul + Jamie = room/ hour

Determine the hourly rates for each person and for the whole group using the formula .

Work is painting 1 room, so W = 1.

We don’t know how long Jamie will take, so we need to keep the variable t.

Write an equation to show that the sum of their individual rates equals the group rate.

(Think of it this way: Pria works for one hour and paints of the room. Paul works for an hour and paints of the room. Jamie works for an hour and paints of the room. Together they have painted half the room in one hour.

Solve the rational equation.

First find the least common denominator of the individual rates. It is 5 • 6 • t = 30t.

Then multiply each term on the left by a fractional form of 1 so that all rates have the same denominator and can be added. (Note: we could also have found the common denominator of the entire equation, which is also 30t, and multiplied both sides by it.)

11t - 11t + 30 = 15t – 11t

Now multiply both sides of the equation by the common denominator, then simplify.

It would take Jamie 7.5 hours to paint the room by herself.

Tanya and Cam can each wash a car and vacuum its interior in 2 hours. Pat needs 3 hours to do this same job by himself. If Pat, Cam and Tanya work together, how long will it take them to clean a car?

A). Incorrect. If all three worked at the same rate, then the time for one car could be calculated by dividing the time it takes them working alone to clean 3 cars by 3. But that doesn't work in this case because Pat has a different cleaning rate than the others. The correct answer is 45 minutes.

B) Correct. According to the formula, . Tanya and Cam each have a rate of car in one hour, and Pat’s rate is car in one hour. Working together, they have a rate of , or . W is one car, so the formula becomes = , and so t = . It takes three-quarters of an hour, or 45 minutes, to clean one car.

C) Incorrect. Tanya and Cam clean a car in two hours each, not as a team. They each have a rate of car in one hour, and Pat’s rate is car in one hour. Working together, they have a rate of . The correct answer is 45 minutes.

D) Incorrect. This is the time it would take for Cam and Tanya to clean one car together. Since Pat is helping, it will take less time than that. The correct answer is 45 minutes.

Sometimes work problems describe rates in a relative way: someone works 3 times as fast as someone else or a machine takes 2 fewer hours to finish a job than another model of machine. In these instances, we express one rate using information about another rate. Let’s look at an example:

One pipe can fill a pool 1.5 times faster than a second pipe. If both pipes are open, the pool can be filled in 6 hours. If only the slower pipe is open, how long would it take to fill the pool?

Find the rates of each pipe alone and the two working together.

Hours needed for fast pipe to fill pool: p

Hours needed for slow pipe to fill pool alone: 1.5 p

Hours needed for both pipes together: 6

Write an equation that shows that the amount of work completed by both pipes in one hour is equal to the sum of the work of each pipe.

Solve for p. One way to do this is to rewrite the rational expressions using a common denominator.

Common denominator of p, 1.5p and 6 is 6p.

Slow pipe takes 1.5p hours to do the work alone.

The slower pipe will take 15 hours to fill the pool alone.

Rational equations can be used to solve a variety of problems that involve rates, times and work. Using rational expressions and equations can help us answer questions about how to combine workers or machines to complete a job on schedule.

## Open Resources for Community College Algebra

We open this section looking back on Example 12.3.2. Julia is taking her family on a boat trip (12) miles down the river and back. The river flows at a speed of (2) miles per hour and she wants to drive the boat at a constant speed, (v) miles per hour downstream and back upstream. Due to the current of the river, the actual speed of travel is (v+2) miles per hour going downstream, and (v-2) miles per hour going upstream. If Julia plans to spend (8) hours for the whole trip, how fast should she drive the boat?

The time it takes Julia to drive the boat downstream is (frac<12>) hours, and upstream is (frac<12>) hours. The function to model the whole trip's time is

where (t) stands for time in hours. The trip will take (8) hours, so we want (t(v)) to equal (8 ext<,>) and we have:

Instead of using the function's graph, we will solve this equation algebraically. You may wish to review the technique of eliminating denominators discussed in Subsection 2.3.2. We can use the same technique with variable expressions in the denominators. To remove the fractions in this equation, we will multiply both sides of the equation by the least common denominator ((v-2)(v+2) ext<,>) and we have:

###### Remark 12.5.2 .

At this point, logically all that we know is that the only possible solutions are (-1) and (4 ext<.>) Because of the step where factors were canceled, it's possible that these might not actually be solutions to the original equation. They each might be what is called an . An extraneous solution is a number that would appear to be a solution based on the solving process, but actually does not make the original equation true. Because of this, it is important that these proposed solutions be checked. Note that we're not checking to see if we made a calculation error, but are instead checking to see if the proposed solutions actually solve the original equation.

Algebraically, both values do check out to be solutions. In the context of this scenario, the boat's speed can't be negative, so we only take the solution (4 ext<.>) If Julia drives at (4) miles per hour, the whole trip would take (8) hours. This result matches the solution in Example 12.3.2.

###### Definition 12.5.3 . Rational Equation.

A rational equation is an equation involving one or more rational expressions. Usually, we consider these to be equations that have the variable in the denominator of at least one term.

Let's look at another application problem.

###### Example 12.5.4 .

It takes Ku (3) hours to paint a room and it takes Jacob (6) hours to paint the same room. If they work together, how long would it take them to paint the room?

Since it takes Ku (3) hours to paint the room, he paints (frac<1><3>) of the room each hour. Similarly, Jacob paints (frac<1><6>) of the room each hour. If they work together, they paint (frac<1><3>+frac<1><6>) of the room each hour.

Assume it takes (x) hours to paint the room if Ku and Jacob work together. This implies they paint (frac<1>) of the room together each hour. Now we can write this equation:

To clear away denominators, we multiply both sides of the equation by the common denominator of (3 ext<,>) (6) and (x ext<,>) which is (6x ext<:>)

Does the possible solution (x=2) check as an actual solution?

It does, so it is a solution. If Ku and Jacob work together, it would take them (2) hours to paint the room.

We are ready to outline a general process for solving a rational equation.

###### Process 12.5.5 . Solving Rational Equations.

To solve a rational equation,

Find the least common denominator for all terms in the equation.

Multiply every term in the equation by the least common denominator

Every denominator should cancel leaving a simpler kind of equation to solve. Use previous method to solve that equation.

Let's look at a few more examples of solving rational equations.

###### Example 12.5.6 .

The common denominator is (y(y+1) ext<.>) We will multiply both sides of the equation by (y(y+1) ext<:>)

Does the possible solution (y=-3) check as an actual solution?

It checks, so (-3) is a solution. We write the solution set as (<-3> ext<.>)

###### Example 12.5.7 .

The common denominator is (z-4 ext<.>) We will multiply both sides of the equation by (z-4 ext<:>)

Do the possible solutions (z=1) and (z=4) check as actual solutions?

The possible solution (z=4) does not actually work, since it leads to division by (0) in the equation. It is an extraneous solution. However, (z=1) is a valid solution. The only solution to the equation is (1 ext<,>) and thus we can write the solution set as (<1> ext<.>)

###### Example 12.5.8 .

To find the common denominator, we need to factor all denominators if possible:

Now we can see the common denominator is ((p+2)(p-2) ext<.>) We will multiply both sides of the equation by ((p+2)(p-2) ext<:>)

Does the possible solution (p=2) check as an actual solution?

The possible solution (p=2) does not actually work, since it leads to division by (0) in the equation. So this is an extraneous solution, and the equation actually has no solution. We could say that its solution set is the empty set, (emptyset ext<.>)

###### Example 12.5.9 .

Solve (C(t)=0.35 ext<,>) where (C(t)=frac<3t>) gives a drug's concentration in milligrams per liter (t) hours since an injection. (This function was explored in the introduction of Section 12.1.)

To solve (C(t)=0.35 ext<,>) we'll begin by setting up (frac<3t>=0.35 ext<.>) We'll begin by identifying that the LCD is (t^2+8 ext<,>) and multiply each side of the equation by this:

This results in a quadratic equation so we will put it in standard form and use the quadratic formula:

Each of these answers should be checked in the original equation they both work. In context, this means that the drug concentration will reach (0.35) milligrams per liter about (1.066) hours after the injection was given, and again (7.506) hours after the injection was given.

### Subsection 12.5.2 Solving Rational Equations for a Specific Variable

Rational equations can contain many variables and constants and we can solve for any one of them. The process for solving still involves multiplying each side of the equation by the LCD. Instead of having a numerical answer though, our final result will contain other variables and constants.

###### Example 12.5.10 .

In physics, when two resistances, (R_1) and (R_2 ext<,>) are connected in a parallel circuit, the combined resistance, (R ext<,>) can be calculated by the formula

Solve for (R) in this formula.

The common denominator is (R R_1 R_2 ext<.>) We will multiply both sides of the equation by (R R_1 R_2 ext<:>)

###### Example 12.5.11 .

Here is the slope formula

Solve for (x_1) in this formula.

The common denominator is (x_2-x_1 ext<.>) We will multiply both sides of the equation by (x_2-x_1 ext<:>)

###### Example 12.5.12 .

Solve the rational equation (x=frac<4y-1><2y-3>) for (y ext<.>)

Our first step will be to multiply each side by the LCD, which is simply (2y-3 ext<.>) After that, we'll isolate all terms containing (y ext<,>) factor out (y ext<,>) and then finish solving for that variable.

### Subsection 12.5.3 Solving Rational Equations Using Technology

In some instances, it may be difficult to solve rational equations algebraically. We can instead use graphing technology to obtain approximate solutions. Let's look at one such example.

###### Example 12.5.13 .

Solve the equation (frac<2>=frac<8>) using graphing technology.

We will define (f(x)=frac<2>) and (g(x)=frac<8> ext<,>) and then look for the points of intersection.

Since the two functions intersect at approximately ((-1.524,-0.442)) and ((3.405,4.936) ext<,>) the solutions to (frac<2>=frac<8>) are approximately (-1.524) and (3.405 ext<.>) We can write the solution set as (<-1.524ldots, 3.405ldots>) or in several other forms. It may be important to do something to communicate that these solutions are approximations. Here we used “(ldots)”, but you could also just say in words that the solutions are approximate.

Describe what an “extraneous solution” to a rational equation is.

In general, when solving a rational equation, multiplying through by the will leave you with a simpler equation to solve.

When you believe you have the solutions to a rational equation, what is more important than usual (compared to other kinds of equations) for you to do?

### Exercises 12.5.5 Exercises

###### Review and Warmup

Recall the time that Filip traveled with his kids to kick a soccer ball on Mars? We should examine one more angle to our soccer kick question. The formula (H(t)=-6.07t^2+27.1t) finds the height of the soccer ball in feet above the ground at a time (t) seconds after being kicked.

Using technology, find out what the maximum height of the ball was and when it reached that height.

Using technology, solve for when (H(t)=20) and interpret the meaning of this in a complete sentence.

Using technology, solve for when (H(t)=0) and interpret the meaning of this in a complete sentence.

###### Solving Rational Equations for a Specific Variable

Solve this equation for (a ext<:>)

Solve this equation for (m ext<:>)

Solve this equation for (x ext<:>)

Solve this equation for (C ext<:>)

Solve this equation for (a ext<:>)

Solve this equation for (c ext<:>)

Solve this equation for (B ext<:>)

Solve this equation for (a ext<:>)

###### Solving Rational Equations Using Technology

Use technology to solve the equation.

Use technology to solve the equation.

Use technology to solve the equation.

Use technology to solve the equation.

Use technology to solve the equation.

Use technology to solve the equation.

###### Application Problems

Scot and Jay are working together to paint a room. If Scot paints the room alone, it would take him (18) hours to complete the job. If Jay paints the room alone, it would take him (12) hours to complete the job. Answer the following question:

If they work together, it would take them hours to complete the job. Use a decimal in your answer if needed.

There are three pipes at a tank. To fill the tank, it would take Pipe A (3) hours, Pipe B (12) hours, and Pipe C (4) hours. Answer the following question:

If all three pipes are turned on, it would take hours to fill the tank.

Casandra and Tien are working together to paint a room. Casandra works (1.5) times as fast as Tien does. If they work together, it took them (9) hours to complete the job. Answer the following questions:

If Casandra paints the room alone, it would take her hours to complete the job.

If Tien paints the room alone, it would take him hours to complete the job.

Two pipes are being used to fill a tank. Pipe A can fill the tank (4.5) times as fast as Pipe B does. When both pipes are turned on, it takes (18) hours to fill the tank. Answer the following questions:

If only Pipe A is turned on, it would take hours to fill the tank.

If only Pipe B is turned on, it would take hours to fill the tank.

Kandace and Jenny worked together to paint a room, and it took them (2) hours to complete the job. If they work alone, it would take Jenny (3) more hours than Kandace to complete the job. Answer the following questions:

If Kandace paints the room alone, it would take her hours to complete the job.

If Jenny paints the room alone, it would take her hours to complete the job.

If both Pipe A and Pipe B are turned on, it would take (2) hours to fill a tank. If each pipe is turned on alone, it takes Pipe B (3) fewer hours than Pipe A to fill the tank. Answer the following questions:

If only Pipe A is turned on, it would take hours to fill the tank.

If only Pipe B is turned on, it would take hours to fill the tank.

Town A and Town B are (570) miles apart. A boat traveled from Town A to Town B, and then back to Town A. Since the river flows from Town B to Town A, the boat’s speed was (25) miles per hour faster when it traveled from Town B to Town A. The whole trip took (19) hours. Answer the following questions:

The boat traveled from Town A to Town B at the speed of miles per hour.

The boat traveled from Town B back to Town A at the speed of miles per hour.

A river flows at (7) miles per hour. A boat traveled with the current from Town A to Town B, which are (260) miles apart. Then, the boat turned around, and traveled against the current to reach Town C, which is (160) miles away from Town B. The second leg of the trip (Town B to Town C) took the same time as the first leg (Town A to Town B). During this whole trip, the boat was driving at a constant still-water speed. Answer the following question:

During this trip, the boat’s speed on still water was miles per hour.

A river flows at (5) miles per hour. A boat traveled with the current from Town A to Town B, which are (100) miles apart. The boat stayed overnight at Town B. The next day, the water’s current stopped, and boat traveled on still water to reach Town C, which is (190) miles away from Town B. The second leg of the trip (Town B to Town C) took (9) hours longer than the first leg (Town A to Town B). During this whole trip, the boat was driving at a constant still-water speed. Find this speed.

Note that you should not consider the unreasonable answer.

During this trip, the boat’s speed on still water was miles per hour.

Town A and Town B are (600) miles apart. With a constant still-water speed, a boat traveled from Town A to Town B, and then back to Town A. During this whole trip, the river flew from Town A to Town B at (20) miles per hour. The whole trip took (16) hours. Answer the following question:

During this trip, the boat’s speed on still water was miles per hour.

Town A and Town B are (350) miles apart. With a constant still-water speed of (24) miles per hour, a boat traveled from Town A to Town B, and then back to Town A. During this whole trip, the river flew from Town B to Town A at a constant speed. The whole trip took (30) hours. Answer the following question:

During this trip, the river’s speed was miles per hour.

Suppose that a large pump can empty a swimming pool in (43 < m hr>) and that a small pump can empty the same pool in (53 < m hr> ext<.>) If both pumps are used at the same time, how long will it take to empty the pool?

If both pumps are used at the same time, it will take to empty the pool.

The winner of a (9 < m mi>) race finishes (14.73 < m min>) ahead of the second-place runner. On average, the winner ran (0.6 < extstylefrac< mmathstrut mi>< mmathstrut hr>>) faster than the second place runner. Find the average running speed for each runner.

The winner's average speed was and the second-place runner's average speed was .

In still water a tugboat can travel (15 < extstylefrac< mmathstrut mi>< mmathstrut hr>> ext<.>) It travels (42 < m mi>) upstream and then (42 < m mi>) downstream in a total time of (5.96 < m hr> ext<.>) Find the speed of the current.

Without any wind an airplane flies at (300 < extstylefrac< mmathstrut mi>< mmathstrut hr>> ext<.>) The plane travels (600 < m mi>) into the wind and then returns with the wind in a total time of (4.04 < m hr> ext<.>) Find the average speed of the wind.

When there is a (11.8 < extstylefrac< mmathstrut mi>< mmathstrut hr>>) wind, an airplane can fly (770 < m mi>) with the wind in the same time that it can fly (702 < m mi>) against the wind. Find the speed of the plane when there is no wind.

It takes one employee (2.5 < m hr>) longer to mow a football field than it does a more experienced employee. Together they can mow the grass in (1.9 < m hr> ext<.>) How long does it take each person to mow the football field working alone?

The more experienced worker takes to mow the field alone, and the less experienced worker takes .

It takes one painter (13 < m hr>) longer to paint a house than it does a more experienced painter. Together they can paint the house in (30 < m hr> ext<.>) How long does it take for each painter to paint the house working alone?

The more experienced painter takes to paint the house alone, and the less experienced painter takes .

## 7.5: Solve Rational Equations

Solving Rational Equations

· Solve rational equations using the techniques for simplifying and manipulating rational expressions.

Equations that contain rational expressions are called rational equations. We can solve these equations using the techniques for performing operations with rational expressions and for solving algebraic equations.

Solving Rational Equations Using Common Denominators

One method for solving rational equations is to rewrite the rational expressions in terms of a common denominator. Then, since we know the numerators are equal, we can solve for the variable. To illustrate this, let’s look at a very simple equation:

Since the denominator of each expression is the same, the numerators must be equivalent as well. This means that x = 3.

This is true for rational equations with polynomials too:

Again, since the denominators are the same, we know the numerators must also be equal. So we can set them equal to one another and solve for x.

We should check our solution in the original rational expression:

The solution checks, and since x = 8 does not result in division by 0, the solution is valid.

When the terms in a rational equation have unlike denominators, solving the equation will involve some extra work. Here’s an example:

There are no excluded values because the denominators are both constants.

Find a common denominator and rewrite each expression with that denominator.

The common denominator is 8.

Since the denominators are the same, the numerators must be equal for the equation to be true. Solve for x.

Check the solution by substituting 4 for x in the original equation.

Another way of solving rational equations is to multiply both sides of the equation by the common denominator. This eliminates the denominators and turns the rational equation into a polynomial equation. Here is the same equation we just solved:

There are no excluded values because the denominators are both constants.

Multiply both sides by the least common denominator

Now that we understand the techniques, let’s look at an example that has variables in the denominator too. Remember that whenever there are variables in the denominator, we need to find any values that are excluded from the domain because they'd make the denominator zero.

To solve this equation, we can multiply both sides by the least common denominator:

(x + 2)(x – 2) = 0

First determine the excluded values. These are the values of x that result in a 0 denominator.

x 2 – 4 = (x – 2)(x + 2)

Find the common denominator of x – 2, x + 2, and x 2 – 4

Since (x – 2) and (x + 2) are both factors of x 2 – 4, the least common denominator is (x – 2)(x + 2) or x 2 – 4

Multiply both sides of the equation by the common denominator.

7x – 14 + 5x + 10 =10x – 2

12x – 4 =10x – 2

12x – 10x – 4 = 10x – 10x – 2

Check to be sure that the solution is not an excluded value. (It is not.)

Check the solution in the original equation.

Solve the equation , m 0 or 2

A) Incorrect. You probably found the common denominator correctly, but forgot to distribute when you were simplifying. You also forgot to check your solution or note the excluded values m ≠ 2 because it makes the expression on the right side undefined. Multiplying both sides by the common denominator gives , so . The correct answer is m = 8.

B) Incorrect. , so . The solution, 8, is not an excluded value. The correct answer is m = 8.

C) Correct. Multiplying both sides of the equation by the common denominator gives , so . . The correct answer is m = 8.

We've seen that there is more than one way to solve rational equations. Because both of these techniques manipulate and rewrite terms, sometimes they can produce solutions that don't work in the original form of the equation. These types of answers are called extraneous solutions. These solutions are artifacts of the solving process and not real answers at all. That's why we should always check solutions in the original equations—we may find that they yield untrue statements or produce undefined expressions.

A) Correct. x – 2 + x 2 – 6x = 4 (x – 6 )(x + 1) = 0. Since 6 is an excluded value, it is an extraneous solution. Only -1 is a real solution.

B) Incorrect. 6 is an excluded value because it makes the denominator of the first rational expression equal to 0. Since 6 is an extraneous solution, it can't be included in the solution. The correct answer is -1.

C) Incorrect. The common denominator is (x – 6 )(x -2). Each term on the left side must be multiplied by a fraction equivalent to 1 that will produce that denominator: = . The correct answer is -1.

D) Incorrect. When the equation is solved by finding the common denominator, the answers are -1 and 6. It is true that 6 is an excluded value and thus an extraneous solution that must be discarded. But -1 works in the original equation and it is a valid solution. The correct answer is -1.

We solve rational equations by finding a common denominator. We can then follow either of two methods. We can rewrite the equation so that all terms have the common denominator and we can solve for the variable with just the numerators. Or we can multiply both sides of the equation by the common denominator so that all terms become polynomials instead of rational expressions.

An important step in solving rational equations is to reject any extraneous solutions from the final answer. Extraneous solutions are solutions that don't satisfy the original form of the equation because they produce untrue statements or are excluded values that make a denominator equal to 0.

## University of Transnational Business Law

This is “Solving Rational Equations”, section 7.5 from the book Beginning Algebra (v. 1.0). For details on it (including licensing), click here.

Has this book helped you? Consider passing it on:

Creative Commons supports free culture from music to education. Their licenses helped make this book available to you.

DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators.

7.5 Solving Rational Equations

Learning Objectives
1.Solve rational equations.
2.Solve literal equations, or formulas, involving rational expressions.

Solving Rational Equations

A rational equationAn equation containing at least one rational expression. is an equation containing at least one rational expression. Rational expressions typically contain a variable in the denominator. For this reason, we will take care to ensure that the denominator is not 0 by making note of restrictions and checking our solutions.

Solve rational equations by clearing the fractions by multiplying both sides of the equation by the least common denominator (LCD).

Solution: We first make a note that x≠0 and then multiply both sides by the LCD, 3x:

Check your answer by substituting 12 for x to see if you obtain a true statement.

After multiplying both sides of the previous example by the LCD, we were left with a linear equation to solve. This is not always the case sometimes we will be left with a quadratic equation.

Solution: In this example, there are two restrictions, x≠0 and x≠−1. Begin by multiplying both sides by the LCD, x(x+1).

After distributing and dividing out the common factors, a quadratic equation remains. To solve it, rewrite it in standard form, factor, and then set each factor equal to 0.

Check to see if these values solve the original equation.

Answer: The solutions are −1/2 and 1.

Up to this point, all of the possible solutions have solved the original equation. However, this may not always be the case. Multiplying both sides of an equation by variable factors may lead to extraneous solutionsA solution that does not solve the original equation., which are solutions that do not solve the original equation. A complete list of steps for solving a rational equation is outlined in the following example.

Example 3: Solve: xx+2+2x2+5x+6=5x+3.

Step 1: Factor all denominators and determine the LCD.

Step 2: Identify the restrictions. In this case, they are x≠−2 and x≠−3.

Step 3: Multiply both sides of the equation by the LCD. Distribute carefully and then simplify.

Step 4: Solve the resulting equation. Here the result is a quadratic equation. Rewrite it in standard form, factor, and then set each factor equal to 0.

Step 5: Check for extraneous solutions. Always substitute into the original equation, or the factored equivalent. In this case, choose the factored equivalent to check:

Here −2 is an extraneous solution and is not included in the solution set. It is important to note that −2 is a restriction.

If this process produces a solution that happens to be a restriction, then disregard it as an extraneous solution.

Try this! Solve: xx−5+3x+2=7xx2−3x−10.

Video Solution
(click to see video)
Sometimes all potential solutions are extraneous, in which case we say that there is no solution to the original equation. In the next two examples, we demonstrate two ways in which a rational equation can have no solutions.

Example 4: Solve: 3xx2−4−2x+2=1x+2.

Solution: To identify the LCD, first factor the denominators.

Multiply both sides by the least common denonominator (LCD), (x+2)(x−2), distributing carefully.

The equation is a contradiction and thus has no solution.

Example 5: Solve: xx−4−4x+5=36x2+x−20.

Solution: First, factor the denominators.

Take note that the restrictions are x≠4 and x≠−5. To clear the fractions, multiply by the LCD, (x−4)(x+5).

Both of these values are restrictions of the original equation hence both are extraneous.

Try this! Solve: 1x+1+xx−3=4xx2−2x−3.

Video Solution
(click to see video)
It is important to point out that this technique for clearing algebraic fractions only works for equations. Do not try to clear algebraic fractions when simplifying expressions. As a reminder, we have

Expressions are to be simplified and equations are to be solved. If we multiply the expression by the LCD, x(2x+1), we obtain another expression that is not equivalent.

Literal equations, or formulas, are often rational equations. Hence the techniques described in this section can be used to solve for particular variables. Assume that all variable expressions in the denominator are nonzero.

Example 6: Solve for x: z=x−5y.

Solution: The goal is to isolate x. Assuming that y is nonzero, multiply both sides by y and then add 5 to both sides.

Example 7: Solve for c: 1c=1a+1b.

Solution: In this example, the goal is to isolate c. We begin by multiplying both sides by the LCD, a⋅b⋅c, distributing carefully.

On the right side of the equation, factor out c.

Next, divide both sides of the equation by the quantity (b+a).

Try this! Solve for y: x=y+1y−1.

Video Solution
(click to see video)

Key Takeaways
•Begin solving rational equations by multiplying both sides by the LCD. The resulting equivalent equation can be solved using the techniques learned up to this point.
•Multiplying both sides of a rational equation by a variable expression introduces the possibility of extraneous solutions. Therefore, we must check the solutions against the set of restrictions. If a solution is a restriction, then it is not part of the domain and is extraneous.
•When multiplying both sides of an equation by an expression, distribute carefully and multiply each term by that expression.
•If all of the resulting solutions are extraneous, then the original equation has no solutions.

## Solving Literal Equations and Applications Involving Reciprocals

Literal equations, or formulas, are often rational equations. Hence the techniques described in this section can be used to solve for particular variables. Assume that all variable expressions in the denominator are nonzero.

### Example 9

The reciprocal of the combined resistance R of two resistors R 1 and R 2 in parallel is given by the formula 1 R = 1 R 1 + 1 R 2 . Solve for R in terms of R 1 and R 2 .

The goal is to isolate R on one side of the equation. Begin by multiplying both sides of the equation by the LCD, R R 1 R 2 .

R R 1 R 2 ⋅ 1 R = R R 1 R 2 ⋅ 1 R 1 + R R 1 R 2 ⋅ 1 R 2 R 1 R 2 = R R 2 + R R 1 R 1 R 2 = R ( R 2 + R 1 ) R 1 R 2 R 2 + R 1 = R

Answer: R = R 1 R 2 R 1 + R 2

Try this! Solve for y: x = 2 y + 5 y − 3 .

Recall that the reciprocal of a nonzero number n is 1 n . For example, the reciprocal of 5 is 1 5 and 5 ⋅ 1 5 = 1 . In this section, the applications will often involve the key word “reciprocal.” When this is the case, we will see that the algebraic setup results in a rational equation.

### Example 10

A positive integer is 3 less than another. If the reciprocal of the smaller integer is subtracted from twice the reciprocal of the larger, then the result is 1 20 . Find the two integers.

Let n represent the larger positive integer.

Let n − 3 represent the smaller positive integer.

Set up an algebraic equation.

Solve this rational expression by multiplying both sides by the LCD. The LCD is 20 n ( n − 3 ) .

2 n − 1 n − 3 = 1 20 20 n ( n − 3 ) ⋅ ( 2 n − 1 n − 3 ) = 20 n ( n − 3 ) ⋅ ( 1 20 ) 20 n ( n − 3 ) ⋅ 2 n − 20 n ( n − 3 ) ⋅ 1 n − 3 = 20 n ( n − 3 ) ⋅ ( 1 20 )

40 ( n − 3 ) − 20 n = n ( n − 3 ) 40 n − 120 − 20 n = n 2 − 3 n 20 n − 120 = n 2 − 3 n 0 = n 2 − 23 n + 120 0 = ( n − 8 ) ( n − 15 ) n − 8 = 0 or n − 15 = 0 n = 8 n = 15

Here we have two viable possibilities for the larger integer n. For this reason, we will we have two solutions to this problem.

If n = 8 , then n − 3 = 8 − 3 = 5 .

If n = 15 , then n − 3 = 15 − 3 = 12 .

As a check, perform the operations indicated in the problem.

2 ( 1 8 ) − 1 5 = 1 4 − 1 5 = 5 20 − 4 20 = 1 20 ✓

2 ( 1 15 ) − 1 12 = 2 15 − 1 12 = 8 60 − 5 60 = 3 60 = 1 20 ✓

Answer: Two sets of positive integers solve this problem: <5, 8>and <12, 15>.

Try this! When the reciprocal of the larger of two consecutive even integers is subtracted from 4 times the reciprocal of the smaller, the result is 5 6 . Find the integers.

## Illustrative Mathematics Grade 7, Unit 5, Lesson 15: Solving Equations with Rational Numbers

The following diagram shows how to solve equations that include rational numbers and have rational solutions.

#### Lesson 15.1 Number Talk: Opposites and Reciprocals

The variables a through h all represent different numbers. Mentally find numbers that make each equation true.
3/5 · 5/3 = a
7 · b = 1
c · d = 1
-6 + 6 = e
11 + f = 0
g + h = 0

#### Lesson 15.2 Match Solutions

Match each equation to a value that makes it true by dragging the answer to the corresponding equation. Be prepared to explain your reasoning.
Open Applet

#### Lesson 15.3 Trip to the Mountains

The Hiking Club is on a trip to hike up a mountain.

1. The members increased their elevation 290 feet during their hike this morning. Now they are at an elevation of 450 feet.
a. Explain how to find their elevation before the hike.
b. Han says the equation describes the situation. What does the variable represent?
c. Han says that he can rewrite his equation as to solve for . Compare Han&rsquos strategy to your strategy for finding the beginning elevation.
2. The temperature fell 4 degrees in the last hour. Now it is 21 degrees. Write and solve an equation to find the temperature it was 1 hour ago.
3. There are 3 times as many students participating in the hiking trip this year than last year. There are 42 students on the trip this year.
a. Explain how to find the number of students that came on the hiking trip last year.
b. Mai says the equation 3s = 42 describes the situation. What does the variable represent?
c. Mai says that she can rewrite her equation as to solve for s = 1/3 · 42. Compare Mai&rsquos strategy to your strategy for finding the number of students on last year’s trip.
4. The cost of the hiking trip this year is 2/3 of the cost of last year&rsquos trip. This year&rsquos trip cost \$32. Write and solve an equation to find the cost of last year&rsquos trip.

#### Lesson 15.4 Card Sort: Matching Inverses

Your teacher will give you a set of cards with numbers on them.

1. Match numbers with their additive inverses.
2. Next, match numbers with their multiplicative inverses.
3. What do you notice about the numbers and their inverses?

#### Lesson 15 Practice Problems

1. Solve.
a. 2/5 t = 6
b. -4.5 = a - 8
c. 1/2 + p = -3
d. 12 = x · 3
e. -12 = -3y
2. Evaluate each expression if x is 2/5, y is -4, and z is -0.2.
a. x + y
b. 2x - z
c. x + y + z
d. y · x
3. Match each equation to a step that will help solve the equation.
4. a. Write an equation where a number is added to a variable, and a solution is -8.
b. Write an equation where a number is multiplied by a variable, and a solution is
5. The markings on the number line are evenly spaced. Label the other markings on the number line.
6. In 2012, James Cameron descended to the bottom of Challenger Deep in the Marianas Trench the deepest point in the ocean. The vessel he rode in was called DeepSea Challenger.
Challenger Deep is 35,814 feet deep at its lowest point
a. DeepSea Challenger’s descent was a change in depth of (-4) feet per second. We can use the equation y = -4x to model this relationship, where y is the depth and x is the time in seconds that have passed. How many seconds does this model suggest it would take for DeepSea Challenger to reach the bottom?
b. To end the mission DeepSea Challenger made a one-hour ascent to the surface. How many seconds is this?
c. The ascent can be modeled by a different proportional relationship y = kx. What is the value of k in this case?

The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.