Solve 4a-6/5a+3geq3/2

$$\frac { 4 a - 6 } { 5 a + 3 } \geq \frac { 3 } { 2 }$$

Solve for a

$$a\in [-3,-\frac{3}{5})$$

Solution Steps

Denominator 5a+3 cannot be zero since division by zero is not defined. There are two cases.

$$5a+3>0 5a+3<0$$

Consider the case when 5a+3 is positive. Move 3 to the right hand side.

$$5a>-3$$

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

$$a>-\frac{3}{5}$$

The initial inequality does not change the direction when multiplied by 5a+3 for 5a+3>0.

$$4a-6\geq \frac{3}{2}\left(5a+3\right)$$

Multiply out the right hand side.

$$4a-6\geq \frac{15}{2}a+\frac{9}{2}$$

Move the terms containing a to the left hand side and all other terms to the right hand side.

$$4a-\frac{15}{2}a\geq 6+\frac{9}{2}$$

Combine like terms.

$$-\frac{7}{2}a\geq \frac{21}{2}$$

Divide both sides by -\frac{7}{2}. Since -\frac{7}{2} is negative, the inequality direction is changed.

$$a\leq -3$$

Consider condition a>-\frac{3}{5} specified above.

$$a\in \emptyset$$

Now consider the case when 5a+3 is negative. Move 3 to the right hand side.

$$5a<-3$$

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

$$a<-\frac{3}{5}$$

The initial inequality changes the direction when multiplied by 5a+3 for 5a+3<0.

$$4a-6\leq \frac{3}{2}\left(5a+3\right)$$

Multiply out the right hand side.

$$4a-6\leq \frac{15}{2}a+\frac{9}{2}$$

Move the terms containing a to the left hand side and all other terms to the right hand side.

$$4a-\frac{15}{2}a\leq 6+\frac{9}{2}$$

Combine like terms.

$$-\frac{7}{2}a\leq \frac{21}{2}$$

Divide both sides by -\frac{7}{2}. Since -\frac{7}{2} is negative, the inequality direction is changed.

$$a\geq -3$$

Consider condition a<-\frac{3}{5} specified above.

$$a\in [-3,-\frac{3}{5})$$

The final solution is the union of the obtained solutions.

$$a\in [-3,-\frac{3}{5})$$