Solve 3x-4/5+1/8leq6x+7/8-x+1

$$\frac{ 3x-4 }{ 5 } + \frac{ 1 }{ 8 } \leq \frac{ 6x+7 }{ 8 } -x+1$$

Solve for x

$$x\leq 3$$

Solution Steps

Multiply both sides of the equation by 40, the least common multiple of 5,8. Since 40 is positive, the inequality direction remains the same.

$$8\left(3x-4\right)+5\leq 5\left(6x+7\right)-40x+40$$

Use the distributive property to multiply 8 by 3x-4.

$$24x-32+5\leq 5\left(6x+7\right)-40x+40$$

Add -32 and 5 to get -27.

$$24x-27\leq 5\left(6x+7\right)-40x+40$$

Use the distributive property to multiply 5 by 6x+7.

$$24x-27\leq 30x+35-40x+40$$

Combine 30x and -40x to get -10x.

$$24x-27\leq -10x+35+40$$

Add 35 and 40 to get 75.

$$24x-27\leq -10x+75$$

Add 10x to both sides.

$$24x-27+10x\leq 75$$

Combine 24x and 10x to get 34x.

$$34x-27\leq 75$$

Add 27 to both sides.

$$34x\leq 75+27$$

Add 75 and 27 to get 102.

$$34x\leq 102$$

Divide both sides by 34. Since 34 is positive, the inequality direction remains the same.

$$x\leq \frac{102}{34}$$

Divide 102 by 34 to get 3.

$$x\leq 3$$