Solve 5/6(3-x)-1/2(x-4)geq1/3(2x-3)-x

$$\frac { 5 } { 6 } ( 3 - x ) - \frac { 1 } { 2 } ( x - 4 ) \geq \frac { 1 } { 3 } ( 2 x - 3 ) - x$$

Solve for x

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

Solution Steps

Use the distributive property to multiply \frac{5}{6} by 3-x.

$$\frac{5}{6}\times 3+\frac{5}{6}\left(-1\right)x-\frac{1}{2}\left(x-4\right)\geq \frac{1}{3}\left(2x-3\right)-x$$

Express \frac{5}{6}\times 3 as a single fraction.

$$\frac{5\times 3}{6}+\frac{5}{6}\left(-1\right)x-\frac{1}{2}\left(x-4\right)\geq \frac{1}{3}\left(2x-3\right)-x$$

Multiply 5 and 3 to get 15.

$$\frac{15}{6}+\frac{5}{6}\left(-1\right)x-\frac{1}{2}\left(x-4\right)\geq \frac{1}{3}\left(2x-3\right)-x$$

Reduce the fraction \frac{15}{6} to lowest terms by extracting and canceling out 3.

$$\frac{5}{2}+\frac{5}{6}\left(-1\right)x-\frac{1}{2}\left(x-4\right)\geq \frac{1}{3}\left(2x-3\right)-x$$

Multiply \frac{5}{6} and -1 to get -\frac{5}{6}.

$$\frac{5}{2}-\frac{5}{6}x-\frac{1}{2}\left(x-4\right)\geq \frac{1}{3}\left(2x-3\right)-x$$

Use the distributive property to multiply -\frac{1}{2} by x-4.

$$\frac{5}{2}-\frac{5}{6}x-\frac{1}{2}x-\frac{1}{2}\left(-4\right)\geq \frac{1}{3}\left(2x-3\right)-x$$

Express -\frac{1}{2}\left(-4\right) as a single fraction.

$$\frac{5}{2}-\frac{5}{6}x-\frac{1}{2}x+\frac{-\left(-4\right)}{2}\geq \frac{1}{3}\left(2x-3\right)-x$$

Multiply -1 and -4 to get 4.

$$\frac{5}{2}-\frac{5}{6}x-\frac{1}{2}x+\frac{4}{2}\geq \frac{1}{3}\left(2x-3\right)-x$$

Divide 4 by 2 to get 2.

$$\frac{5}{2}-\frac{5}{6}x-\frac{1}{2}x+2\geq \frac{1}{3}\left(2x-3\right)-x$$

Combine -\frac{5}{6}x and -\frac{1}{2}x to get -\frac{4}{3}x.

$$\frac{5}{2}-\frac{4}{3}x+2\geq \frac{1}{3}\left(2x-3\right)-x$$

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

$$\frac{5}{2}-\frac{4}{3}x+\frac{4}{2}\geq \frac{1}{3}\left(2x-3\right)-x$$

Since \frac{5}{2} and \frac{4}{2} have the same denominator, add them by adding their numerators.

$$\frac{5+4}{2}-\frac{4}{3}x\geq \frac{1}{3}\left(2x-3\right)-x$$

Add 5 and 4 to get 9.

$$\frac{9}{2}-\frac{4}{3}x\geq \frac{1}{3}\left(2x-3\right)-x$$

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

$$\frac{9}{2}-\frac{4}{3}x\geq \frac{1}{3}\times 2x+\frac{1}{3}\left(-3\right)-x$$

Multiply \frac{1}{3} and 2 to get \frac{2}{3}.

$$\frac{9}{2}-\frac{4}{3}x\geq \frac{2}{3}x+\frac{1}{3}\left(-3\right)-x$$

Multiply \frac{1}{3} and -3 to get \frac{-3}{3}.

$$\frac{9}{2}-\frac{4}{3}x\geq \frac{2}{3}x+\frac{-3}{3}-x$$

Divide -3 by 3 to get -1.

$$\frac{9}{2}-\frac{4}{3}x\geq \frac{2}{3}x-1-x$$

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

$$\frac{9}{2}-\frac{4}{3}x\geq -\frac{1}{3}x-1$$

Add \frac{1}{3}x to both sides.

$$\frac{9}{2}-\frac{4}{3}x+\frac{1}{3}x\geq -1$$

Combine -\frac{4}{3}x and \frac{1}{3}x to get -x.

$$\frac{9}{2}-x\geq -1$$

Subtract \frac{9}{2} from both sides.

$$-x\geq -1-\frac{9}{2}$$

Convert -1 to fraction -\frac{2}{2}.

$$-x\geq -\frac{2}{2}-\frac{9}{2}$$

Since -\frac{2}{2} and \frac{9}{2} have the same denominator, subtract them by subtracting their numerators.

$$-x\geq \frac{-2-9}{2}$$

Subtract 9 from -2 to get -11.

$$-x\geq -\frac{11}{2}$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x\leq \frac{-\frac{11}{2}}{-1}$$

Express \frac{-\frac{11}{2}}{-1} as a single fraction.

$$x\leq \frac{-11}{2\left(-1\right)}$$

Multiply 2 and -1 to get -2.

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

Fraction \frac{-11}{-2} can be simplified to \frac{11}{2} by removing the negative sign from both the numerator and the denominator.

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