Solve 18(2z+4)+30zgeq699z+4

$$18 ( 2 z + 4 ) + 30 z \geq 699 z + 4$$

Solve for z

$$z\leq \frac{68}{633}$$

Solution Steps

Use the distributive property to multiply 18 by 2z+4.

$$36z+72+30z\geq 699z+4$$

Combine 36z and 30z to get 66z.

$$66z+72\geq 699z+4$$

Subtract 699z from both sides.

$$66z+72-699z\geq 4$$

Combine 66z and -699z to get -633z.

$$-633z+72\geq 4$$

Subtract 72 from both sides.

$$-633z\geq 4-72$$

Subtract 72 from 4 to get -68.

$$-633z\geq -68$$

Divide both sides by -633. Since -633 is negative, the inequality direction is changed.

$$z\leq \frac{-68}{-633}$$

Fraction \frac{-68}{-633} can be simplified to \frac{68}{633} by removing the negative sign from both the numerator and the denominator.

$$z\leq \frac{68}{633}$$