Solve 1-2xleq7/9-2/3x

$$1 - 2 x \leq \frac { 7 } { 9 } - \frac { 2 } { 3 } x$$

Solve for x

$$x\geq \frac{1}{6}$$

Solution Steps

Add \frac{2}{3}x to both sides.

$$1-2x+\frac{2}{3}x\leq \frac{7}{9}$$

Combine -2x and \frac{2}{3}x to get -\frac{4}{3}x.

$$1-\frac{4}{3}x\leq \frac{7}{9}$$

Subtract 1 from both sides.

$$-\frac{4}{3}x\leq \frac{7}{9}-1$$

Convert 1 to fraction \frac{9}{9}.

$$-\frac{4}{3}x\leq \frac{7}{9}-\frac{9}{9}$$

Since \frac{7}{9} and \frac{9}{9} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{4}{3}x\leq \frac{7-9}{9}$$

Subtract 9 from 7 to get -2.

$$-\frac{4}{3}x\leq -\frac{2}{9}$$

Multiply both sides by -\frac{3}{4}, the reciprocal of -\frac{4}{3}. Since -\frac{4}{3} is negative, the inequality direction is changed.

$$x\geq -\frac{2}{9}\left(-\frac{3}{4}\right)$$

Multiply -\frac{2}{9} times -\frac{3}{4} by multiplying numerator times numerator and denominator times denominator.

$$x\geq \frac{-2\left(-3\right)}{9\times 4}$$

Do the multiplications in the fraction \frac{-2\left(-3\right)}{9\times 4}.

$$x\geq \frac{6}{36}$$

Reduce the fraction \frac{6}{36} to lowest terms by extracting and canceling out 6.

$$x\geq \frac{1}{6}$$