Solve x+1/x^2+2x-15=2x-6/x+5-1/x-3

$$\frac { x + 1 } { x ^ { 2 } + 2 x - 15 } = \frac { 2 x - 6 } { x + 5 } - \frac { 1 } { x - 3 }$$

Solve for x

$$x=1$$

$$x=6$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -5,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+5\right), the least common multiple of x^{2}+2x-15,x+5,x-3.

$$x+1=\left(x-3\right)\left(2x-6\right)-\left(x+5\right)$$

Use the distributive property to multiply x-3 by 2x-6 and combine like terms.

$$x+1=2x^{2}-12x+18-\left(x+5\right)$$

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

$$x+1=2x^{2}-12x+18-x-5$$

Combine -12x and -x to get -13x.

$$x+1=2x^{2}-13x+18-5$$

Subtract 5 from 18 to get 13.

$$x+1=2x^{2}-13x+13$$

Subtract 2x^{2} from both sides.

$$x+1-2x^{2}=-13x+13$$

Add 13x to both sides.

$$x+1-2x^{2}+13x=13$$

Combine x and 13x to get 14x.

$$14x+1-2x^{2}=13$$

Subtract 13 from both sides.

$$14x+1-2x^{2}-13=0$$

Subtract 13 from 1 to get -12.

$$14x-12-2x^{2}=0$$

Divide both sides by 2.

$$7x-6-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+7x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=7 ab=-\left(-6\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 7.

$$a=6 b=1$$

Rewrite -x^{2}+7x-6 as \left(-x^{2}+6x\right)+\left(x-6\right).

$$\left(-x^{2}+6x\right)+\left(x-6\right)$$

Factor out -x in -x^{2}+6x.

$$-x\left(x-6\right)+x-6$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x+1\right)$$

To find equation solutions, solve x-6=0 and -x+1=0.

$$x=6 x=1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+5\right), the least common multiple of x^{2}+2x-15,x+5,x-3.

$$x+1=\left(x-3\right)\left(2x-6\right)-\left(x+5\right)$$

Use the distributive property to multiply x-3 by 2x-6 and combine like terms.

$$x+1=2x^{2}-12x+18-\left(x+5\right)$$

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

$$x+1=2x^{2}-12x+18-x-5$$

Combine -12x and -x to get -13x.

$$x+1=2x^{2}-13x+18-5$$

Subtract 5 from 18 to get 13.

$$x+1=2x^{2}-13x+13$$

Subtract 2x^{2} from both sides.

$$x+1-2x^{2}=-13x+13$$

Add 13x to both sides.

$$x+1-2x^{2}+13x=13$$

Combine x and 13x to get 14x.

$$14x+1-2x^{2}=13$$

Subtract 13 from both sides.

$$14x+1-2x^{2}-13=0$$

Subtract 13 from 1 to get -12.

$$14x-12-2x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-2x^{2}+14x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 14 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-14±\sqrt{196+8\left(-12\right)}}{2\left(-2\right)}$$

Multiply 8 times -12.

$$x=\frac{-14±\sqrt{196-96}}{2\left(-2\right)}$$

Add 196 to -96.

$$x=\frac{-14±\sqrt{100}}{2\left(-2\right)}$$

Take the square root of 100.

$$x=\frac{-14±10}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-14±10}{-4}$$

Now solve the equation x=\frac{-14±10}{-4} when ± is plus. Add -14 to 10.

$$x=-\frac{4}{-4}$$

Divide -4 by -4.

$$x=1$$

Now solve the equation x=\frac{-14±10}{-4} when ± is minus. Subtract 10 from -14.

$$x=-\frac{24}{-4}$$

Divide -24 by -4.

$$x=6$$

The equation is now solved.

$$x=1 x=6$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -5,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+5\right), the least common multiple of x^{2}+2x-15,x+5,x-3.

$$x+1=\left(x-3\right)\left(2x-6\right)-\left(x+5\right)$$

Use the distributive property to multiply x-3 by 2x-6 and combine like terms.

$$x+1=2x^{2}-12x+18-\left(x+5\right)$$

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

$$x+1=2x^{2}-12x+18-x-5$$

Combine -12x and -x to get -13x.

$$x+1=2x^{2}-13x+18-5$$

Subtract 5 from 18 to get 13.

$$x+1=2x^{2}-13x+13$$

Subtract 2x^{2} from both sides.

$$x+1-2x^{2}=-13x+13$$

Add 13x to both sides.

$$x+1-2x^{2}+13x=13$$

Combine x and 13x to get 14x.

$$14x+1-2x^{2}=13$$

Subtract 1 from both sides.

$$14x-2x^{2}=13-1$$

Subtract 1 from 13 to get 12.

$$14x-2x^{2}=12$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2x^{2}+14x=12$$

Divide both sides by -2.

$$\frac{-2x^{2}+14x}{-2}=\frac{12}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{14}{-2}x=\frac{12}{-2}$$

Divide 14 by -2.

$$x^{2}-7x=\frac{12}{-2}$$

Divide 12 by -2.

$$x^{2}-7x=-6$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-6+\frac{49}{4}$$

Add -6 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{25}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{5}{2} x-\frac{7}{2}=-\frac{5}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=6 x=1$$