Solve -2+3/4x+5>-10x/12x+15

$$- 2 + \frac { 3 } { 4 x + 5 } > - \frac { 10 x } { 12 x + 15 }$$

Solve for x

$$x\in \left(-\frac{3}{2},-\frac{5}{4}\right)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply -2 times \frac{4x+5}{4x+5}.

$$-\frac{2\left(4x+5\right)}{4x+5}+\frac{3}{4x+5}>-\frac{10x}{12x+15}$$

Since -\frac{2\left(4x+5\right)}{4x+5} and \frac{3}{4x+5} have the same denominator, add them by adding their numerators.

$$\frac{-2\left(4x+5\right)+3}{4x+5}>-\frac{10x}{12x+15}$$

Do the multiplications in -2\left(4x+5\right)+3.

$$\frac{-8x-10+3}{4x+5}>-\frac{10x}{12x+15}$$

Combine like terms in -8x-10+3.

$$\frac{-8x-7}{4x+5}>-\frac{10x}{12x+15}$$

Add \frac{10x}{12x+15} to both sides.

$$\frac{-8x-7}{4x+5}+\frac{10x}{12x+15}>0$$

Factor 12x+15.

$$\frac{-8x-7}{4x+5}+\frac{10x}{3\left(4x+5\right)}>0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4x+5 and 3\left(4x+5\right) is 3\left(4x+5\right). Multiply \frac{-8x-7}{4x+5} times \frac{3}{3}.

$$\frac{3\left(-8x-7\right)}{3\left(4x+5\right)}+\frac{10x}{3\left(4x+5\right)}>0$$

Since \frac{3\left(-8x-7\right)}{3\left(4x+5\right)} and \frac{10x}{3\left(4x+5\right)} have the same denominator, add them by adding their numerators.

$$\frac{3\left(-8x-7\right)+10x}{3\left(4x+5\right)}>0$$

Do the multiplications in 3\left(-8x-7\right)+10x.

$$\frac{-24x-21+10x}{3\left(4x+5\right)}>0$$

Combine like terms in -24x-21+10x.

$$\frac{-14x-21}{3\left(4x+5\right)}>0$$

Use the distributive property to multiply 3 by 4x+5.

$$\frac{-14x-21}{12x+15}>0$$

For the quotient to be positive, -14x-21 and 12x+15 have to be both negative or both positive. Consider the case when -14x-21 and 12x+15 are both negative.

$$-14x-21<0 12x+15<0$$

The solution satisfying both inequalities is x\in \left(-\frac{3}{2},-\frac{5}{4}\right).

$$x\in \left(-\frac{3}{2},-\frac{5}{4}\right)$$

Consider the case when -14x-21 and 12x+15 are both positive.

$$12x+15>0 -14x-21>0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\frac{3}{2},-\frac{5}{4}\right)$$