Solve 1/9(3k-3/2)geq-2(k/2+3)

$$\frac { 1 } { 9 } ( 3 k - \frac { 3 } { 2 } ) \geq - 2 ( \frac { k } { 2 } + 3 )$$

Solve for k

$$k\geq -\frac{35}{8}$$

Solution Steps

Multiply both sides of the equation by 18, the least common multiple of 9,2. Since 18 is positive, the inequality direction remains the same.

$$2\left(3k-\frac{3}{2}\right)\geq -36\left(\frac{k}{2}+3\right)$$

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

$$6k+2\left(-\frac{3}{2}\right)\geq -36\left(\frac{k}{2}+3\right)$$

Cancel out 2 and 2.

$$6k-3\geq -36\left(\frac{k}{2}+3\right)$$

Use the distributive property to multiply -36 by \frac{k}{2}+3.

$$6k-3\geq -36\times \frac{k}{2}-108$$

Cancel out 2, the greatest common factor in 36 and 2.

$$6k-3\geq -18k-108$$

Add 18k to both sides.

$$6k-3+18k\geq -108$$

Combine 6k and 18k to get 24k.

$$24k-3\geq -108$$

Add 3 to both sides.

$$24k\geq -108+3$$

Add -108 and 3 to get -105.

$$24k\geq -105$$

Divide both sides by 24. Since 24 is positive, the inequality direction remains the same.

$$k\geq \frac{-105}{24}$$

Reduce the fraction \frac{-105}{24} to lowest terms by extracting and canceling out 3.

$$k\geq -\frac{35}{8}$$