Solve 5/2(2-x)-1/2(x-4)geq(1/3)(3x-3)+x

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

Solve for x

$$x\leq \frac{8}{5}$$

Solution Steps

Use the distributive property to multiply \frac{5}{2} by 2-x.

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

Cancel out 2 and 2.

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

Multiply \frac{5}{2} and -1 to get -\frac{5}{2}.

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

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

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

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

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

Multiply -1 and -4 to get 4.

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

Divide 4 by 2 to get 2.

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

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

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

Add 5 and 2 to get 7.

$$7-3x\geq \frac{1}{3}\left(3x-3\right)+x$$

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

$$7-3x\geq \frac{1}{3}\times 3x+\frac{1}{3}\left(-3\right)+x$$

Cancel out 3 and 3.

$$7-3x\geq x+\frac{1}{3}\left(-3\right)+x$$

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

$$7-3x\geq x+\frac{-3}{3}+x$$

Divide -3 by 3 to get -1.

$$7-3x\geq x-1+x$$

Combine x and x to get 2x.

$$7-3x\geq 2x-1$$

Subtract 2x from both sides.

$$7-3x-2x\geq -1$$

Combine -3x and -2x to get -5x.

$$7-5x\geq -1$$

Subtract 7 from both sides.

$$-5x\geq -1-7$$

Subtract 7 from -1 to get -8.

$$-5x\geq -8$$

Divide both sides by -5. Since -5 is negative, the inequality direction is changed.

$$x\leq \frac{-8}{-5}$$

Fraction \frac{-8}{-5} can be simplified to \frac{8}{5} by removing the negative sign from both the numerator and the denominator.

$$x\leq \frac{8}{5}$$