Solve 16-x/3geqx

$$\frac { 16 - x } { 3 } \geq x$$

Solve for x

$$x\leq 4$$

Solution Steps

Divide each term of 16-x by 3 to get \frac{16}{3}-\frac{1}{3}x.

$$\frac{16}{3}-\frac{1}{3}x\geq x$$

Subtract x from both sides.

$$\frac{16}{3}-\frac{1}{3}x-x\geq 0$$

Combine -\frac{1}{3}x and -x to get -\frac{4}{3}x.

$$\frac{16}{3}-\frac{4}{3}x\geq 0$$

Subtract \frac{16}{3} from both sides. Anything subtracted from zero gives its negation.

$$-\frac{4}{3}x\geq -\frac{16}{3}$$

Multiply both sides by -\frac{3}{4}, the reciprocal of -\frac{4}{3}. Since -\frac{4}{3} is negative, the inequality direction is changed.

$$x\leq -\frac{16}{3}\left(-\frac{3}{4}\right)$$

Multiply -\frac{16}{3} times -\frac{3}{4} by multiplying numerator times numerator and denominator times denominator.

$$x\leq \frac{-16\left(-3\right)}{3\times 4}$$

Do the multiplications in the fraction \frac{-16\left(-3\right)}{3\times 4}.

$$x\leq \frac{48}{12}$$

Divide 48 by 12 to get 4.

$$x\leq 4$$