Solve 3x-2/8+3x+3/5geq4-15x+2/4

$$\frac { 3 x - 2 } { 8 } + \frac { 3 x + 3 } { 5 } \geq 4 - \frac { 15 x + 2 } { 4 }$$

Solve for x

$$x\geq \frac{2}{3}$$

Solution Steps

Multiply both sides of the equation by 40, the least common multiple of 8,5,4. Since 40 is positive, the inequality direction remains the same.

$$5\left(3x-2\right)+8\left(3x+3\right)\geq 160-10\left(15x+2\right)$$

Use the distributive property to multiply 5 by 3x-2.

$$15x-10+8\left(3x+3\right)\geq 160-10\left(15x+2\right)$$

Use the distributive property to multiply 8 by 3x+3.

$$15x-10+24x+24\geq 160-10\left(15x+2\right)$$

Combine 15x and 24x to get 39x.

$$39x-10+24\geq 160-10\left(15x+2\right)$$

Add -10 and 24 to get 14.

$$39x+14\geq 160-10\left(15x+2\right)$$

Use the distributive property to multiply -10 by 15x+2.

$$39x+14\geq 160-150x-20$$

Subtract 20 from 160 to get 140.

$$39x+14\geq 140-150x$$

Add 150x to both sides.

$$39x+14+150x\geq 140$$

Combine 39x and 150x to get 189x.

$$189x+14\geq 140$$

Subtract 14 from both sides.

$$189x\geq 140-14$$

Subtract 14 from 140 to get 126.

$$189x\geq 126$$

Divide both sides by 189. Since 189 is positive, the inequality direction remains the same.

$$x\geq \frac{126}{189}$$

Reduce the fraction \frac{126}{189} to lowest terms by extracting and canceling out 63.

$$x\geq \frac{2}{3}$$