Solve 7/6x+4leq3x+2

$$\frac{ 7 }{ 6 } x+4 \leq 3x+2$$

Solve for x

$$x\geq \frac{12}{11}$$

Solution Steps

Subtract 3x from both sides.

$$\frac{7}{6}x+4-3x\leq 2$$

Combine \frac{7}{6}x and -3x to get -\frac{11}{6}x.

$$-\frac{11}{6}x+4\leq 2$$

Subtract 4 from both sides.

$$-\frac{11}{6}x\leq 2-4$$

Subtract 4 from 2 to get -2.

$$-\frac{11}{6}x\leq -2$$

Multiply both sides by -\frac{6}{11}, the reciprocal of -\frac{11}{6}. Since -\frac{11}{6} is negative, the inequality direction is changed.

$$x\geq -2\left(-\frac{6}{11}\right)$$

Express -2\left(-\frac{6}{11}\right) as a single fraction.

$$x\geq \frac{-2\left(-6\right)}{11}$$

Multiply -2 and -6 to get 12.

$$x\geq \frac{12}{11}$$