### 11.4 - Fractional equations

Before reading this section you may want to review the following topics:

A fractional equation is one that contains fraction terms. In section 4.2 we saw how to solve a linear equation that contains fractions. The steps for solving any fractional equation are exactly the same:
• Look at the denominators of all the fraction terms and find their lowest common multiple (LCM) (this is also called the lowest common denominator (LCD) of the fractions).

• Multiply both sides of the equation by the LCM.

• Distribute the LCM over both sides of the equation.

• The equation no longer contains fraction terms and you can continue solving it by using the basic procedures for solving equations.

• Check the solution. This is especially important with fractional equations. There are two possible problems:

• If the denominator of any fraction term contains x, then the LCM will also contain x, and multiplying both sides of the equation by the LCM will increase the degree of x in the equation. This often leads to extraneous solutions.

• When substituting the solutions back into the original equation to check them, any solution that causes any fraction term to have a denominator of zero must be dropped because division by zero is forbidden in mathematics.

Example 1: Solve this fractional equation for x: Solution: The fraction terms have denominators of 3, 2 and 6. The LCM of these numbers is 6. Multiply both sides of the equation by 6. (Don’t forget to put brackets around both sides of the equation.) Distribute on both sides of the equation:
x − 3 = 6 x + 7.
The fractions are now cleared so this is no longer a fractional equation. Finish solving the equation by collecting linear terms on the left-hand-side and constant terms on the right-hand-side. This gives:
−2 x = 10.
Divide both sides by −2. This gives the solution:
x = −5.
Check it by substituting it back into the original equation. This gives −23 / 6 = −23 / 6, so the solution checks out.

Example 2: Solve this fractional equation for x: Solution: The fraction terms have denominators of x 2 + x − 2,   x + 2,   and   x − 1. It might appear that the LCM is just the product of all three, but because x 2 + x − 2 can be factored as (x + 2)(x − 1), the LCM is actually just (x + 2)(x − 1). Multiply both sides of the equation by it. (Don’t forget to put brackets around both sides of the equation.) Distribute on both sides of the equation:
9 = 3 (x − 1) + 7 (x + 2).
The fractions are now cleared so this is no longer a fractional equation; it is a linear equation. Solve it using the usual techniques. Distribute once more on the right-hand-side:
9 = 10 x + 11.
Collect constant terms on the left-hand-side:
−2 = 10 x.
Divide both sides by 10. This gives the solution:
x = −1/5.
Check it by substituting it back into the original equation. This gives −25 / 6 = −25 / 6, so the solution checks out.

Example 3: The purpose of this example is to illustrate a solution that must be rejected because it causes a division by zero. The equation is identical to the one in the previous example except that it differs in the sign of one term. Solve this fractional equation for x: Solution: Compare each step here with the corresponding step in the example above. Multiply both sides of the equation by the LCM, which again is (x + 2)(x − 1): Distribute on both sides of the equation:
9 = −3 (x − 1) + 7 (x + 2).
Distribute once more on the right-hand-side:
9 = 4 x + 17.
This time the solution is x = −2. If we try to substitute it back into the original equation we get divisions by zero in two of the fractions. Therefore we must reject this solution and state that the equation has no solution.

 Algebra Coach Exercises

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