Rational, radical, and absolute-value relationships — Topic practice

$\frac{\text{1}}{\text{2}}$

Require x ≠ −1. Multiply 3/(x + 1) = 2 by the nonzero denominator:

3 = 2(x + 1), 1 = 2x, x = $\frac{\text{1}}{\text{2}}$.

Check in the original: 3/(3/2) = 2. The candidate is not excluded, so it is valid. Review 4.04.

7

Because $\sqrt{\text{x + 2}}$ is nonnegative, the right side x − 4 must be nonnegative; therefore x ≥ 4. Squaring gives

x + 2 = (x − 4)² x² − 9x + 14 = (x − 7)(x − 2) = 0.

The candidate 2 fails x ≥ 4. The candidate 7 works: $\sqrt{\text{9}}$ = 3 = 7 − 4. The only solution is 7. Review 4.08.

No solution

The original denominator excludes x = 3. Clearing that denominator gives 2x − 1 = 5, hence x = 3. This is the only candidate, and it is forbidden. It cannot become a solution merely because the transformed linear equation accepts it. Review 4.05.