Solve x+1=2x-7/x+5-5x+8/x+5

$$x + 1 = \frac { 2 x - 7 } { x + 5 } - \frac { 5 x + 8 } { x + 5 }$$

Solve for x

$$x=-4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -5 since division by zero is not defined. Multiply both sides of the equation by x+5.

$$\left(x+5\right)x+x+5=2x-7-\left(5x+8\right)$$

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

$$x^{2}+5x+x+5=2x-7-\left(5x+8\right)$$

Combine 5x and x to get 6x.

$$x^{2}+6x+5=2x-7-\left(5x+8\right)$$

To find the opposite of 5x+8, find the opposite of each term.

$$x^{2}+6x+5=2x-7-5x-8$$

Combine 2x and -5x to get -3x.

$$x^{2}+6x+5=-3x-7-8$$

Subtract 8 from -7 to get -15.

$$x^{2}+6x+5=-3x-15$$

Add 3x to both sides.

$$x^{2}+6x+5+3x=-15$$

Combine 6x and 3x to get 9x.

$$x^{2}+9x+5=-15$$

Add 15 to both sides.

$$x^{2}+9x+5+15=0$$

Add 5 and 15 to get 20.

$$x^{2}+9x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 9 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 20}}{2}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 20}}{2}$$

Multiply -4 times 20.

$$x=\frac{-9±\sqrt{81-80}}{2}$$

Add 81 to -80.

$$x=\frac{-9±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-9±1}{2}$$

Now solve the equation x=\frac{-9±1}{2} when ± is plus. Add -9 to 1.

$$x=-\frac{8}{2}$$

Divide -8 by 2.

$$x=-4$$

Now solve the equation x=\frac{-9±1}{2} when ± is minus. Subtract 1 from -9.

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=-4 x=-5$$

Variable x cannot be equal to -5.

$$x=-4$$

Steps for Completing the Square

Variable x cannot be equal to -5 since division by zero is not defined. Multiply both sides of the equation by x+5.

$$\left(x+5\right)x+x+5=2x-7-\left(5x+8\right)$$

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

$$x^{2}+5x+x+5=2x-7-\left(5x+8\right)$$

Combine 5x and x to get 6x.

$$x^{2}+6x+5=2x-7-\left(5x+8\right)$$

To find the opposite of 5x+8, find the opposite of each term.

$$x^{2}+6x+5=2x-7-5x-8$$

Combine 2x and -5x to get -3x.

$$x^{2}+6x+5=-3x-7-8$$

Subtract 8 from -7 to get -15.

$$x^{2}+6x+5=-3x-15$$

Add 3x to both sides.

$$x^{2}+6x+5+3x=-15$$

Combine 6x and 3x to get 9x.

$$x^{2}+9x+5=-15$$

Subtract 5 from both sides.

$$x^{2}+9x=-15-5$$

Subtract 5 from -15 to get -20.

$$x^{2}+9x=-20$$

Divide 9, the coefficient of the x term, by 2 to get \frac{9}{2}. Then add the square of \frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+9x+\left(\frac{9}{2}\right)^{2}=-20+\left(\frac{9}{2}\right)^{2}$$

Square \frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+9x+\frac{81}{4}=-20+\frac{81}{4}$$

Add -20 to \frac{81}{4}.

$$x^{2}+9x+\frac{81}{4}=\frac{1}{4}$$

Factor x^{2}+9x+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x+\frac{9}{2}=\frac{1}{2} x+\frac{9}{2}=-\frac{1}{2}$$

Subtract \frac{9}{2} from both sides of the equation.

$$x=-4 x=-5$$

Variable x cannot be equal to -5.

$$x=-4$$