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All courses Algebra II · A-REI.2 20 of 55
Solve simple rational and radical equations and identify extraneous solutions.

Solve a rational equation with one denominator by clearing the denominator and checking restrictions

Problem
Solve \((x~+~3)/(x~-~1)~=~5\). State the solution and the original denominator restriction.
Your answer
Show answer choicesHide answer choices Work through the mathematics first, then compare your reasoning.Select the answer that matches your work.
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Hint

State the denominator restriction first, then clear the denominator by multiplying both sides by \(x-1\).

In a rational equation, denominator values that make the expression undefined must be excluded even after solving.

Solution walkthrough

01

Record the restriction

\[x-1~\ne~0~->~x~\ne~1\]

The original denominator cannot be zero, so x equals 1 is excluded before any algebra is done.

02

Clear the denominator

\[(x+3)/(x-1)=5~->~x+3=5(x-1)\]

For allowed values of x, multiply both sides by x minus 1 to obtain an equivalent linear equation.

03

Solve the linear equation

\[x+3=5x-5~->~8=4x~->~x=2\]

Distribute 5, move the x-terms to one side, and move the constants to the other.

04

Check and answer

\[2~\ne~1;~(2+3)/(2-1)=5~->~x=2,~\text{with}~x\ne~1\]

Two is in the original domain and substitution gives 5, so the solution is x equals 2 while the original restriction remains x not equal to 1, choice A.

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Another way

  1. Cross-multiply 1 times x plus 3 with 5 times x minus 1, then apply the same domain check.

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Common mistake

Clearing a denominator does not erase its restriction; x equals 1 remains forbidden even though the resulting linear equation has no denominator.

Solution walkthrough video