Solve 3(-30+3z)-12geq-6-2(23-2z)

$$3 ( - 30 + 3 z ) - 12 \geq - 6 - 2 ( 23 - 2 z )$$

Solve for z

$$z\geq 10$$

Solution Steps

Use the distributive property to multiply 3 by -30+3z.

$$-90+9z-12\geq -6-2\left(23-2z\right)$$

Subtract 12 from -90 to get -102.

$$-102+9z\geq -6-2\left(23-2z\right)$$

Use the distributive property to multiply -2 by 23-2z.

$$-102+9z\geq -6-46+4z$$

Subtract 46 from -6 to get -52.

$$-102+9z\geq -52+4z$$

Subtract 4z from both sides.

$$-102+9z-4z\geq -52$$

Combine 9z and -4z to get 5z.

$$-102+5z\geq -52$$

Add 102 to both sides.

$$5z\geq -52+102$$

Add -52 and 102 to get 50.

$$5z\geq 50$$

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

$$z\geq \frac{50}{5}$$

Divide 50 by 5 to get 10.

$$z\geq 10$$