Rational Equations w/Extraneous Solutions

​Extraneous solutions are solutions to an equation that are NOT valid answers.

1. Factor all denominators.

2. Determine the LCD.

3. Multiply each part of the equation by the LCD to eliminate all of the denominators.

4. Solve the remaining equation.

5. Check for extraneous solutions.

   • Substitute the solutions into the equations to see if you get a true statement.

​Solve the following rational equation.

​*LCD = (x+2)(x-2)

​When you multiply everything by the LCD you end up with...

​x - 2 + x + 2 = 4
2x = 4
x = 2

​This gives you...

​Check for extraneous solutions by substituting 2 everywhere you see an x in the original problem.

Since we can NEVER have 0 on the bottom of a fraction the answer to this problem is...
NO SOLUTION

When we simplify everything we get...

​Solve the following rational equation.

​*LCD = (x-2)

​When you multiply everything by the LCD you end up with...

5x = 3x + 4
2x = 4
x = 2

​This gives you...

​Check for extraneous solutions by substituting 2 everywhere you see an x in the original problem.

Since we can NEVER have 0 on the bottom of a fraction the answer to this problem is...
NO SOLUTION

When we simplify everything we get...

​Solve the following rational equation.

​*LCD = (x + 3)(x - 3)

​When you multiply everything by the LCD you end up with...

6(x + 3) = 8x² - 4x(x - 3)
6x + 18 = 8x² - 4x² + 12x
0 = 4x² + 6x - 18
0 = 2x² + 3x - 9
x = -3, 3/2

​This gives you...

​Check for extraneous solutions by substituting -3 and 3/2 into the denominator where you have x in the original problem.

When we use -3 we end up with 0 in the denominator for two of the fractions. Therefore, -3 is NOT a solution.

​There should be an x here!

​Use the X to factor. Remember...
a = 2 so don't forget to divide your sides.

​Divide by the GCF of 2

When we use 3/2 we DON'T end up with 0 in the denominator at all. Therefore...
X = 3/2