Solve 8-4(x+3)leq5+4(6x+10)

$$8 - 4 ( x + 3 ) \leq 5 + 4 ( 6 x + 10 )$$

Solve for x

$$x\geq -\frac{7}{4}$$

Solution Steps

Use the distributive property to multiply -4 by x+3.

$$8-4x-12\leq 5+4\left(6x+10\right)$$

Subtract 12 from 8 to get -4.

$$-4-4x\leq 5+4\left(6x+10\right)$$

Use the distributive property to multiply 4 by 6x+10.

$$-4-4x\leq 5+24x+40$$

Add 5 and 40 to get 45.

$$-4-4x\leq 45+24x$$

Subtract 24x from both sides.

$$-4-4x-24x\leq 45$$

Combine -4x and -24x to get -28x.

$$-4-28x\leq 45$$

Add 4 to both sides.

$$-28x\leq 45+4$$

Add 45 and 4 to get 49.

$$-28x\leq 49$$

Divide both sides by -28. Since -28 is negative, the inequality direction is changed.

$$x\geq \frac{49}{-28}$$

Reduce the fraction \frac{49}{-28} to lowest terms by extracting and canceling out 7.

$$x\geq -\frac{7}{4}$$