Solve a^2-4/4a^2+12a+9-a/4a+6=1/2a+3

$$\frac { a ^ { 2 } - 4 } { 4 a ^ { 2 } + 12 a + 9 } - \frac { a } { 4 a + 6 } = \frac { 1 } { 2 a + 3 }$$

Solve for a

$$a=-2$$

Steps for Solving Linear Equation

Variable a cannot be equal to -\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by 2\left(2a+3\right)^{2}, the least common multiple of 4a^{2}+12a+9,4a+6,2a+3.

$$2\left(a^{2}-4\right)-\left(2a+3\right)a=4a+6$$

Use the distributive property to multiply 2 by a^{2}-4.

$$2a^{2}-8-\left(2a+3\right)a=4a+6$$

Use the distributive property to multiply 2a+3 by a.

$$2a^{2}-8-\left(2a^{2}+3a\right)=4a+6$$

To find the opposite of 2a^{2}+3a, find the opposite of each term.

$$2a^{2}-8-2a^{2}-3a=4a+6$$

Combine 2a^{2} and -2a^{2} to get 0.

$$-8-3a=4a+6$$

Subtract 4a from both sides.

$$-8-3a-4a=6$$

Combine -3a and -4a to get -7a.

$$-8-7a=6$$

Add 8 to both sides.

$$-7a=6+8$$

Add 6 and 8 to get 14.

$$-7a=14$$

Divide both sides by -7.

$$a=\frac{14}{-7}$$

Divide 14 by -7 to get -2.

$$a=-2$$