## College Algebra (10th Edition)

STEP 1: Restrictions. There must be no zeros in the denominator, so $x\neq\pm 2$ STEP 2: Fractions. The LCM of all denominators is $x^{2}-4=(x+2)(x-2).$ After multiplying both sides with the LCM, $2x=4-3(x-2)$ STEP 3: Remove all parentheses and simplify. $2x=4-3x+6$ $2x=10-3x$ STEP 4: Variables on one side, all other terms on the other... $2x=10-3x\qquad/+3x$ $5x=10$ STEP 5: Simplify and solve. $5x=10\qquad /\div 5$ $x=2$ In step 1, the number $2$ is excluded as a possible solution. Thus, the equation has no solutions.