Solve (4)1/3(x-1)leq1/2(2x+1)

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

Solve for x

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

Solution Steps

Multiply 4 and \frac{1}{3} to get \frac{4}{3}.

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

Use the distributive property to multiply \frac{4}{3} by x-1.

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

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

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

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

$$\frac{4}{3}x-\frac{4}{3}\leq \frac{1}{2}\times 2x+\frac{1}{2}$$

Cancel out 2 and 2.

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

Subtract x from both sides.

$$\frac{4}{3}x-\frac{4}{3}-x\leq \frac{1}{2}$$

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

$$\frac{1}{3}x-\frac{4}{3}\leq \frac{1}{2}$$

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

$$\frac{1}{3}x\leq \frac{1}{2}+\frac{4}{3}$$

Least common multiple of 2 and 3 is 6. Convert \frac{1}{2} and \frac{4}{3} to fractions with denominator 6.

$$\frac{1}{3}x\leq \frac{3}{6}+\frac{8}{6}$$

Since \frac{3}{6} and \frac{8}{6} have the same denominator, add them by adding their numerators.

$$\frac{1}{3}x\leq \frac{3+8}{6}$$

Add 3 and 8 to get 11.

$$\frac{1}{3}x\leq \frac{11}{6}$$

Multiply both sides by 3, the reciprocal of \frac{1}{3}. Since \frac{1}{3} is positive, the inequality direction remains the same.

$$x\leq \frac{11}{6}\times 3$$

Express \frac{11}{6}\times 3 as a single fraction.

$$x\leq \frac{11\times 3}{6}$$

Multiply 11 and 3 to get 33.

$$x\leq \frac{33}{6}$$

Reduce the fraction \frac{33}{6} to lowest terms by extracting and canceling out 3.

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