Solve 2/3(x+1)-5/6(x-7)leq2

$$\frac { 2 } { 3 } ( x + 1 ) - \frac { 5 } { 6 } ( x - 7 ) \leq 2$$

Solve for x

$$x\geq 27$$

Solution Steps

Use the distributive property to multiply \frac{2}{3} by x+1.

$$\frac{2}{3}x+\frac{2}{3}-\frac{5}{6}\left(x-7\right)\leq 2$$

Use the distributive property to multiply -\frac{5}{6} by x-7.

$$\frac{2}{3}x+\frac{2}{3}-\frac{5}{6}x-\frac{5}{6}\left(-7\right)\leq 2$$

Express -\frac{5}{6}\left(-7\right) as a single fraction.

$$\frac{2}{3}x+\frac{2}{3}-\frac{5}{6}x+\frac{-5\left(-7\right)}{6}\leq 2$$

Multiply -5 and -7 to get 35.

$$\frac{2}{3}x+\frac{2}{3}-\frac{5}{6}x+\frac{35}{6}\leq 2$$

Combine \frac{2}{3}x and -\frac{5}{6}x to get -\frac{1}{6}x.

$$-\frac{1}{6}x+\frac{2}{3}+\frac{35}{6}\leq 2$$

Least common multiple of 3 and 6 is 6. Convert \frac{2}{3} and \frac{35}{6} to fractions with denominator 6.

$$-\frac{1}{6}x+\frac{4}{6}+\frac{35}{6}\leq 2$$

Since \frac{4}{6} and \frac{35}{6} have the same denominator, add them by adding their numerators.

$$-\frac{1}{6}x+\frac{4+35}{6}\leq 2$$

Add 4 and 35 to get 39.

$$-\frac{1}{6}x+\frac{39}{6}\leq 2$$

Reduce the fraction \frac{39}{6} to lowest terms by extracting and canceling out 3.

$$-\frac{1}{6}x+\frac{13}{2}\leq 2$$

Subtract \frac{13}{2} from both sides.

$$-\frac{1}{6}x\leq 2-\frac{13}{2}$$

Convert 2 to fraction \frac{4}{2}.

$$-\frac{1}{6}x\leq \frac{4}{2}-\frac{13}{2}$$

Since \frac{4}{2} and \frac{13}{2} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{1}{6}x\leq \frac{4-13}{2}$$

Subtract 13 from 4 to get -9.

$$-\frac{1}{6}x\leq -\frac{9}{2}$$

Multiply both sides by -6, the reciprocal of -\frac{1}{6}. Since -\frac{1}{6} is negative, the inequality direction is changed.

$$x\geq -\frac{9}{2}\left(-6\right)$$

Express -\frac{9}{2}\left(-6\right) as a single fraction.

$$x\geq \frac{-9\left(-6\right)}{2}$$

Multiply -9 and -6 to get 54.

$$x\geq \frac{54}{2}$$

Divide 54 by 2 to get 27.

$$x\geq 27$$