Solve 2-8/9xgeqx+7/6

$$2 - \frac { 8 } { 9 } x \geq x + \frac { 7 } { 6 }$$

Solve for x

$$x\leq \frac{15}{34}$$

Solution Steps

Subtract x from both sides.

$$2-\frac{8}{9}x-x\geq \frac{7}{6}$$

Combine -\frac{8}{9}x and -x to get -\frac{17}{9}x.

$$2-\frac{17}{9}x\geq \frac{7}{6}$$

Subtract 2 from both sides.

$$-\frac{17}{9}x\geq \frac{7}{6}-2$$

Convert 2 to fraction \frac{12}{6}.

$$-\frac{17}{9}x\geq \frac{7}{6}-\frac{12}{6}$$

Since \frac{7}{6} and \frac{12}{6} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{17}{9}x\geq \frac{7-12}{6}$$

Subtract 12 from 7 to get -5.

$$-\frac{17}{9}x\geq -\frac{5}{6}$$

Multiply both sides by -\frac{9}{17}, the reciprocal of -\frac{17}{9}. Since -\frac{17}{9} is negative, the inequality direction is changed.

$$x\leq -\frac{5}{6}\left(-\frac{9}{17}\right)$$

Multiply -\frac{5}{6} times -\frac{9}{17} by multiplying numerator times numerator and denominator times denominator.

$$x\leq \frac{-5\left(-9\right)}{6\times 17}$$

Do the multiplications in the fraction \frac{-5\left(-9\right)}{6\times 17}.

$$x\leq \frac{45}{102}$$

Reduce the fraction \frac{45}{102} to lowest terms by extracting and canceling out 3.

$$x\leq \frac{15}{34}$$