Solve 3-11xleq1/4x+6

$$3 - 11 x \leq \frac { 1 } { 4 } x + 6$$

Solve for x

$$x\geq -\frac{4}{15}$$

Solution Steps

Subtract \frac{1}{4}x from both sides.

$$3-11x-\frac{1}{4}x\leq 6$$

Combine -11x and -\frac{1}{4}x to get -\frac{45}{4}x.

$$3-\frac{45}{4}x\leq 6$$

Subtract 3 from both sides.

$$-\frac{45}{4}x\leq 6-3$$

Subtract 3 from 6 to get 3.

$$-\frac{45}{4}x\leq 3$$

Multiply both sides by -\frac{4}{45}, the reciprocal of -\frac{45}{4}. Since -\frac{45}{4} is negative, the inequality direction is changed.

$$x\geq 3\left(-\frac{4}{45}\right)$$

Express 3\left(-\frac{4}{45}\right) as a single fraction.

$$x\geq \frac{3\left(-4\right)}{45}$$

Multiply 3 and -4 to get -12.

$$x\geq \frac{-12}{45}$$

Reduce the fraction \frac{-12}{45} to lowest terms by extracting and canceling out 3.

$$x\geq -\frac{4}{15}$$