Solve 1/4x+7leq1/3x-2

$$\frac { 1 } { 4 } x + 7 \leq \frac { 1 } { 3 } x - 2$$

Solve for x

$$x\geq 108$$

Solution Steps

Subtract \frac{1}{3}x from both sides.

$$\frac{1}{4}x+7-\frac{1}{3}x\leq -2$$

Combine \frac{1}{4}x and -\frac{1}{3}x to get -\frac{1}{12}x.

$$-\frac{1}{12}x+7\leq -2$$

Subtract 7 from both sides.

$$-\frac{1}{12}x\leq -2-7$$

Subtract 7 from -2 to get -9.

$$-\frac{1}{12}x\leq -9$$

Multiply both sides by -12, the reciprocal of -\frac{1}{12}. Since -\frac{1}{12} is negative, the inequality direction is changed.

$$x\geq -9\left(-12\right)$$

Multiply -9 and -12 to get 108.

$$x\geq 108$$