Solve +3(10b-4)leq-54+14b

$$+ 3 ( 10 b - 4 ) \leq - 54 + 14 b$$

Solve for b

$$b\leq -\frac{21}{8}$$

Solution Steps

Use the distributive property to multiply 3 by 10b-4.

$$30b-12\leq -54+14b$$

Subtract 14b from both sides.

$$30b-12-14b\leq -54$$

Combine 30b and -14b to get 16b.

$$16b-12\leq -54$$

Add 12 to both sides.

$$16b\leq -54+12$$

Add -54 and 12 to get -42.

$$16b\leq -42$$

Divide both sides by 16. Since 16 is positive, the inequality direction remains the same.

$$b\leq \frac{-42}{16}$$

Reduce the fraction \frac{-42}{16} to lowest terms by extracting and canceling out 2.

$$b\leq -\frac{21}{8}$$