Solve 3/2x-3leq2x-9/2

$$\frac { 3 } { 2 } x - 3 \leq 2 x - \frac { 9 } { 2 }$$

Solve for x

$$x\geq 3$$

Solution Steps

Subtract 2x from both sides.

$$\frac{3}{2}x-3-2x\leq -\frac{9}{2}$$

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

$$-\frac{1}{2}x-3\leq -\frac{9}{2}$$

Add 3 to both sides.

$$-\frac{1}{2}x\leq -\frac{9}{2}+3$$

Convert 3 to fraction \frac{6}{2}.

$$-\frac{1}{2}x\leq -\frac{9}{2}+\frac{6}{2}$$

Since -\frac{9}{2} and \frac{6}{2} have the same denominator, add them by adding their numerators.

$$-\frac{1}{2}x\leq \frac{-9+6}{2}$$

Add -9 and 6 to get -3.

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

Multiply both sides by -2, the reciprocal of -\frac{1}{2}. Since -\frac{1}{2} is negative, the inequality direction is changed.

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

Express -\frac{3}{2}\left(-2\right) as a single fraction.

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

Multiply -3 and -2 to get 6.

$$x\geq \frac{6}{2}$$

Divide 6 by 2 to get 3.

$$x\geq 3$$