Solve 15-3a-4(a+1)-11geq3(a-9)+2(5a+11)

$$15 - 3 a - 4 ( a + 1 ) - 11 \geq 3 ( a - 9 ) + 2 ( 5 a + 11 )$$

Solve for a

$$a\leq \frac{1}{4}$$

Solution Steps

Use the distributive property to multiply -4 by a+1.

$$15-3a-4a-4-11\geq 3\left(a-9\right)+2\left(5a+11\right)$$

Combine -3a and -4a to get -7a.

$$15-7a-4-11\geq 3\left(a-9\right)+2\left(5a+11\right)$$

Subtract 4 from 15 to get 11.

$$11-7a-11\geq 3\left(a-9\right)+2\left(5a+11\right)$$

Subtract 11 from 11 to get 0.

$$-7a\geq 3\left(a-9\right)+2\left(5a+11\right)$$

Use the distributive property to multiply 3 by a-9.

$$-7a\geq 3a-27+2\left(5a+11\right)$$

Use the distributive property to multiply 2 by 5a+11.

$$-7a\geq 3a-27+10a+22$$

Combine 3a and 10a to get 13a.

$$-7a\geq 13a-27+22$$

Add -27 and 22 to get -5.

$$-7a\geq 13a-5$$

Subtract 13a from both sides.

$$-7a-13a\geq -5$$

Combine -7a and -13a to get -20a.

$$-20a\geq -5$$

Divide both sides by -20. Since -20 is negative, the inequality direction is changed.

$$a\leq \frac{-5}{-20}$$

Reduce the fraction \frac{-5}{-20} to lowest terms by extracting and canceling out -5.

$$a\leq \frac{1}{4}$$