Solve (7x-2)-6*[4(x+3)]+180leq15/(3

$$( 7 x - 2 ) - 6 \cdot [ 4 ( x + 3 ) ] + 180 \leq 15 / (3$$

Solve for x

$$x\geq \frac{101}{17}$$

Solution Steps

Multiply 6 and 4 to get 24.

$$7x-2-24\left(x+3\right)+180\leq \frac{15}{3}$$

Use the distributive property to multiply -24 by x+3.

$$7x-2-24x-72+180\leq \frac{15}{3}$$

Combine 7x and -24x to get -17x.

$$-17x-2-72+180\leq \frac{15}{3}$$

Subtract 72 from -2 to get -74.

$$-17x-74+180\leq \frac{15}{3}$$

Add -74 and 180 to get 106.

$$-17x+106\leq \frac{15}{3}$$

Divide 15 by 3 to get 5.

$$-17x+106\leq 5$$

Subtract 106 from both sides.

$$-17x\leq 5-106$$

Subtract 106 from 5 to get -101.

$$-17x\leq -101$$

Divide both sides by -17. Since -17 is negative, the inequality direction is changed.

$$x\geq \frac{-101}{-17}$$

Fraction \frac{-101}{-17} can be simplified to \frac{101}{17} by removing the negative sign from both the numerator and the denominator.

$$x\geq \frac{101}{17}$$