Solve 2x/3+x-8x/12+4x=1/2x-4/6x+2x^2

$$\frac { 2 x } { 3 + x } - \frac { 8 x } { 12 + 4 x } = \frac { 1 } { 2 x } - \frac { 4 } { 6 x + 2 x ^ { 2 } }$$

Solve for x

$$x=1$$

Steps for Solving Linear Equation

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 4x\left(x+3\right), the least common multiple of 3+x,12+4x,2x,6x+2x^{2}.

$$4x\times 2x-x\times 8x=2x+6-2\times 4$$

Multiply 4 and 2 to get 8.

$$8xx-x\times 8x=2x+6-2\times 4$$

Multiply x and x to get x^{2}.

$$8x^{2}-x\times 8x=2x+6-2\times 4$$

Multiply x and x to get x^{2}.

$$8x^{2}-x^{2}\times 8=2x+6-2\times 4$$

Multiply -2 and 4 to get -8.

$$8x^{2}-x^{2}\times 8=2x+6-8$$

Subtract 8 from 6 to get -2.

$$8x^{2}-x^{2}\times 8=2x-2$$

Subtract 2x from both sides.

$$8x^{2}-x^{2}\times 8-2x=-2$$

Multiply -1 and 8 to get -8.

$$8x^{2}-8x^{2}-2x=-2$$

Combine 8x^{2} and -8x^{2} to get 0.

$$-2x=-2$$

Divide both sides by -2.

$$x=\frac{-2}{-2}$$

Divide -2 by -2 to get 1.

$$x=1$$