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

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

Solve for x

$$x\leq -2$$

Solution Steps

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

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

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

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

Cancel out 2 and 2.

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

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

$$-\frac{1}{4}x+\frac{1}{4}-1\geq -\frac{1}{4}$$

Convert 1 to fraction \frac{4}{4}.

$$-\frac{1}{4}x+\frac{1}{4}-\frac{4}{4}\geq -\frac{1}{4}$$

Since \frac{1}{4} and \frac{4}{4} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{1}{4}x+\frac{1-4}{4}\geq -\frac{1}{4}$$

Subtract 4 from 1 to get -3.

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

Add \frac{3}{4} to both sides.

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

Since -\frac{1}{4} and \frac{3}{4} have the same denominator, add them by adding their numerators.

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

Add -1 and 3 to get 2.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

Multiply both sides by -4, the reciprocal of -\frac{1}{4}. Since -\frac{1}{4} is negative, the inequality direction is changed.

$$x\leq \frac{1}{2}\left(-4\right)$$

Multiply \frac{1}{2} and -4 to get \frac{-4}{2}.

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

Divide -4 by 2 to get -2.

$$x\leq -2$$