Solve x+6/x-5+x-5/x+6=2x^2+23x+4/x^2+x-30

$$\frac { x + 6 } { x - 5 } + \frac { x - 5 } { x + 6 } = \frac { 2 x ^ { 2 } + 23 x + 4 } { x ^ { 2 } + x - 30 }$$

Solve for x

$$x = \frac{19}{7} = 2\frac{5}{7} \approx 2.714285714$$

Steps for Solving Linear Equation

Variable x cannot be equal to any of the values -6,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+6\right), the least common multiple of x-5,x+6,x^{2}+x-30.

$$\left(x+6\right)\left(x+6\right)+\left(x-5\right)\left(x-5\right)=2x^{2}+23x+4$$

Multiply x+6 and x+6 to get \left(x+6\right)^{2}.

$$\left(x+6\right)^{2}+\left(x-5\right)\left(x-5\right)=2x^{2}+23x+4$$

Multiply x-5 and x-5 to get \left(x-5\right)^{2}.

$$\left(x+6\right)^{2}+\left(x-5\right)^{2}=2x^{2}+23x+4$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+6\right)^{2}.

$$x^{2}+12x+36+\left(x-5\right)^{2}=2x^{2}+23x+4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}+12x+36+x^{2}-10x+25=2x^{2}+23x+4$$

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

$$2x^{2}+12x+36-10x+25=2x^{2}+23x+4$$

Combine 12x and -10x to get 2x.

$$2x^{2}+2x+36+25=2x^{2}+23x+4$$

Add 36 and 25 to get 61.

$$2x^{2}+2x+61=2x^{2}+23x+4$$

Subtract 2x^{2} from both sides.

$$2x^{2}+2x+61-2x^{2}=23x+4$$

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

$$2x+61=23x+4$$

Subtract 23x from both sides.

$$2x+61-23x=4$$

Combine 2x and -23x to get -21x.

$$-21x+61=4$$

Subtract 61 from both sides.

$$-21x=4-61$$

Subtract 61 from 4 to get -57.

$$-21x=-57$$

Divide both sides by -21.

$$x=\frac{-57}{-21}$$

Reduce the fraction \frac{-57}{-21} to lowest terms by extracting and canceling out -3.

$$x=\frac{19}{7}$$