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

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

Solve for x

$$x\geq -\frac{3}{4}$$

Solution Steps

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

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

Express \frac{5}{6}\left(-3\right) as a single fraction.

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

Multiply 5 and -3 to get -15.

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

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

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

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

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

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

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

Multiply -1 and -4 to get 4.

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

Divide 4 by 2 to get 2.

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

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

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

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

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

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

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

Add -5 and 4 to get -1.

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

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

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

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

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

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

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

Divide -3 by 3 to get -1.

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

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

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

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

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

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

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

Add \frac{1}{2} to both sides.

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

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

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

Since -\frac{2}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

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

Add -2 and 1 to get -1.

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

Multiply both sides by \frac{3}{2}, the reciprocal of \frac{2}{3}. Since \frac{2}{3} is positive, the inequality direction remains the same.

$$x\geq -\frac{1}{2}\times \frac{3}{2}$$

Multiply -\frac{1}{2} times \frac{3}{2} by multiplying numerator times numerator and denominator times denominator.

$$x\geq \frac{-3}{2\times 2}$$

Do the multiplications in the fraction \frac{-3}{2\times 2}.

$$x\geq \frac{-3}{4}$$

Fraction \frac{-3}{4} can be rewritten as -\frac{3}{4} by extracting the negative sign.

$$x\geq -\frac{3}{4}$$