Solve 4x-3/5-5x+2/3geqx+3

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

Solve for x

$$x\leq -\frac{16}{7}$$

Solution Steps

Multiply both sides of the equation by 15, the least common multiple of 5,3. Since 15 is positive, the inequality direction remains the same.

$$3\left(4x-3\right)-5\left(5x+2\right)\geq 15x+45$$

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

$$12x-9-5\left(5x+2\right)\geq 15x+45$$

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

$$12x-9-25x-10\geq 15x+45$$

Combine 12x and -25x to get -13x.

$$-13x-9-10\geq 15x+45$$

Subtract 10 from -9 to get -19.

$$-13x-19\geq 15x+45$$

Subtract 15x from both sides.

$$-13x-19-15x\geq 45$$

Combine -13x and -15x to get -28x.

$$-28x-19\geq 45$$

Add 19 to both sides.

$$-28x\geq 45+19$$

Add 45 and 19 to get 64.

$$-28x\geq 64$$

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

$$x\leq \frac{64}{-28}$$

Reduce the fraction \frac{64}{-28} to lowest terms by extracting and canceling out 4.

$$x\leq -\frac{16}{7}$$