Solve 30/x^2-9=x/x-3+2/x-3

$$\frac { 30 } { x ^ { 2 } - 9 } = \frac { x } { x - 3 } + \frac { 2 } { x - 3 }$$

Solve for x

$$x=-8$$

Steps Using Factoring

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

$$30=\left(x+3\right)x+\left(x+3\right)\times 2$$

Use the distributive property to multiply x+3 by x.

$$30=x^{2}+3x+\left(x+3\right)\times 2$$

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

$$30=x^{2}+3x+2x+6$$

Combine 3x and 2x to get 5x.

$$30=x^{2}+5x+6$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+5x+6=30$$

Subtract 30 from both sides.

$$x^{2}+5x+6-30=0$$

Subtract 30 from 6 to get -24.

$$x^{2}+5x-24=0$$

To solve the equation, factor x^{2}+5x-24 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=5 ab=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 5.

$$a=-3 b=8$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-3\right)\left(x+8\right)$$

To find equation solutions, solve x-3=0 and x+8=0.

$$x=3 x=-8$$

Variable x cannot be equal to 3.

$$x=-8$$

Steps Using Factoring By Grouping

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

$$30=\left(x+3\right)x+\left(x+3\right)\times 2$$

Use the distributive property to multiply x+3 by x.

$$30=x^{2}+3x+\left(x+3\right)\times 2$$

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

$$30=x^{2}+3x+2x+6$$

Combine 3x and 2x to get 5x.

$$30=x^{2}+5x+6$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+5x+6=30$$

Subtract 30 from both sides.

$$x^{2}+5x+6-30=0$$

Subtract 30 from 6 to get -24.

$$x^{2}+5x-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-24\right)=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 5.

$$a=-3 b=8$$

Rewrite x^{2}+5x-24 as \left(x^{2}-3x\right)+\left(8x-24\right).

$$\left(x^{2}-3x\right)+\left(8x-24\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(x-3\right)+8\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+8\right)$$

To find equation solutions, solve x-3=0 and x+8=0.

$$x=3 x=-8$$

Variable x cannot be equal to 3.

$$x=-8$$

Steps Using the Quadratic Formula

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

$$30=\left(x+3\right)x+\left(x+3\right)\times 2$$

Use the distributive property to multiply x+3 by x.

$$30=x^{2}+3x+\left(x+3\right)\times 2$$

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

$$30=x^{2}+3x+2x+6$$

Combine 3x and 2x to get 5x.

$$30=x^{2}+5x+6$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+5x+6=30$$

Subtract 30 from both sides.

$$x^{2}+5x+6-30=0$$

Subtract 30 from 6 to get -24.

$$x^{2}+5x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-24\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-24\right)}}{2}$$

Multiply -4 times -24.

$$x=\frac{-5±\sqrt{25+96}}{2}$$

Add 25 to 96.

$$x=\frac{-5±\sqrt{121}}{2}$$

Take the square root of 121.

$$x=\frac{-5±11}{2}$$

Now solve the equation x=\frac{-5±11}{2} when ± is plus. Add -5 to 11.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-5±11}{2} when ± is minus. Subtract 11 from -5.

$$x=-\frac{16}{2}$$

Divide -16 by 2.

$$x=-8$$

The equation is now solved.

$$x=3 x=-8$$

Variable x cannot be equal to 3.

$$x=-8$$

Steps for Completing the Square

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

$$30=\left(x+3\right)x+\left(x+3\right)\times 2$$

Use the distributive property to multiply x+3 by x.

$$30=x^{2}+3x+\left(x+3\right)\times 2$$

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

$$30=x^{2}+3x+2x+6$$

Combine 3x and 2x to get 5x.

$$30=x^{2}+5x+6$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+5x+6=30$$

Subtract 6 from both sides.

$$x^{2}+5x=30-6$$

Subtract 6 from 30 to get 24.

$$x^{2}+5x=24$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=24+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=24+\frac{25}{4}$$

Add 24 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{121}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{11}{2} x+\frac{5}{2}=-\frac{11}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=3 x=-8$$

Variable x cannot be equal to 3.

$$x=-8$$