Solve 2*1x+24geq3*6x-0*6

$$2 \cdot 1 x + 24 \geq 3 \cdot 6 x - 0 \cdot 6$$

Solve for x

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

Solution Steps

Multiply 2 and 1 to get 2.

$$2x+24\geq 3\times 6x-0\times 6$$

Multiply 3 and 6 to get 18.

$$2x+24\geq 18x-0\times 6$$

Multiply 0 and 6 to get 0.

$$2x+24\geq 18x-0$$

Subtract 18x from both sides.

$$2x+24-18x\geq -0$$

Combine 2x and -18x to get -16x.

$$-16x+24\geq -0$$

Multiply -1 and 0 to get 0.

$$-16x+24\geq 0$$

Subtract 24 from both sides. Anything subtracted from zero gives its negation.

$$-16x\geq -24$$

Divide both sides by -16. Since -16 is negative, the inequality direction is changed.

$$x\leq \frac{-24}{-16}$$

Reduce the fraction \frac{-24}{-16} to lowest terms by extracting and canceling out -8.

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