Solve 6x+3/5-1leq2-2x-3/10

$$\frac { 6 x + 3 } { 5 } - 1 \leq 2 - \frac { 2 x - 3 } { 10 }$$

Solve for x

$$x\leq \frac{27}{14}$$

Solution Steps

Multiply both sides of the equation by 10, the least common multiple of 5,10. Since 10 is positive, the inequality direction remains the same.

$$2\left(6x+3\right)-10\leq 20-\left(2x-3\right)$$

Use the distributive property to multiply 2 by 6x+3.

$$12x+6-10\leq 20-\left(2x-3\right)$$

Subtract 10 from 6 to get -4.

$$12x-4\leq 20-\left(2x-3\right)$$

To find the opposite of 2x-3, find the opposite of each term.

$$12x-4\leq 20-2x-\left(-3\right)$$

The opposite of -3 is 3.

$$12x-4\leq 20-2x+3$$

Add 20 and 3 to get 23.

$$12x-4\leq 23-2x$$

Add 2x to both sides.

$$12x-4+2x\leq 23$$

Combine 12x and 2x to get 14x.

$$14x-4\leq 23$$

Add 4 to both sides.

$$14x\leq 23+4$$

Add 23 and 4 to get 27.

$$14x\leq 27$$

Divide both sides by 14. Since 14 is positive, the inequality direction remains the same.

$$x\leq \frac{27}{14}$$