Solve x-2/x+3-30/x^2-9=3/8

$$\frac { x - 2 } { x + 3 } - \frac { 30 } { x ^ { 2 } - 9 } = \frac { 3 } { 8 }$$

Solve for x

$$x=11$$

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 8\left(x-3\right)\left(x+3\right), the least common multiple of x+3,x^{2}-9,8.

$$\left(8x-24\right)\left(x-2\right)-8\times 30=3\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply 8x-24 by x-2 and combine like terms.

$$8x^{2}-40x+48-8\times 30=3\left(x-3\right)\left(x+3\right)$$

Multiply -8 and 30 to get -240.

$$8x^{2}-40x+48-240=3\left(x-3\right)\left(x+3\right)$$

Subtract 240 from 48 to get -192.

$$8x^{2}-40x-192=3\left(x-3\right)\left(x+3\right)$$

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

$$8x^{2}-40x-192=\left(3x-9\right)\left(x+3\right)$$

Use the distributive property to multiply 3x-9 by x+3 and combine like terms.

$$8x^{2}-40x-192=3x^{2}-27$$

Subtract 3x^{2} from both sides.

$$8x^{2}-40x-192-3x^{2}=-27$$

Combine 8x^{2} and -3x^{2} to get 5x^{2}.

$$5x^{2}-40x-192=-27$$

Add 27 to both sides.

$$5x^{2}-40x-192+27=0$$

Add -192 and 27 to get -165.

$$5x^{2}-40x-165=0$$

Divide both sides by 5.

$$x^{2}-8x-33=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-33. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\left(-33\right)=-33$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -33.

$$1,-33 3,-11$$

Calculate the sum for each pair.

$$1-33=-32 3-11=-8$$

The solution is the pair that gives sum -8.

$$a=-11 b=3$$

Rewrite x^{2}-8x-33 as \left(x^{2}-11x\right)+\left(3x-33\right).

$$\left(x^{2}-11x\right)+\left(3x-33\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-11\right)+3\left(x-11\right)$$

Factor out common term x-11 by using distributive property.

$$\left(x-11\right)\left(x+3\right)$$

To find equation solutions, solve x-11=0 and x+3=0.

$$x=11 x=-3$$

Variable x cannot be equal to -3.

$$x=11$$

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 8\left(x-3\right)\left(x+3\right), the least common multiple of x+3,x^{2}-9,8.

$$\left(8x-24\right)\left(x-2\right)-8\times 30=3\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply 8x-24 by x-2 and combine like terms.

$$8x^{2}-40x+48-8\times 30=3\left(x-3\right)\left(x+3\right)$$

Multiply -8 and 30 to get -240.

$$8x^{2}-40x+48-240=3\left(x-3\right)\left(x+3\right)$$

Subtract 240 from 48 to get -192.

$$8x^{2}-40x-192=3\left(x-3\right)\left(x+3\right)$$

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

$$8x^{2}-40x-192=\left(3x-9\right)\left(x+3\right)$$

Use the distributive property to multiply 3x-9 by x+3 and combine like terms.

$$8x^{2}-40x-192=3x^{2}-27$$

Subtract 3x^{2} from both sides.

$$8x^{2}-40x-192-3x^{2}=-27$$

Combine 8x^{2} and -3x^{2} to get 5x^{2}.

$$5x^{2}-40x-192=-27$$

Add 27 to both sides.

$$5x^{2}-40x-192+27=0$$

Add -192 and 27 to get -165.

$$5x^{2}-40x-165=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -40 for b, and -165 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 5\left(-165\right)}}{2\times 5}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 5\left(-165\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-40\right)±\sqrt{1600-20\left(-165\right)}}{2\times 5}$$

Multiply -20 times -165.

$$x=\frac{-\left(-40\right)±\sqrt{1600+3300}}{2\times 5}$$

Add 1600 to 3300.

$$x=\frac{-\left(-40\right)±\sqrt{4900}}{2\times 5}$$

Take the square root of 4900.

$$x=\frac{-\left(-40\right)±70}{2\times 5}$$

The opposite of -40 is 40.

$$x=\frac{40±70}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{40±70}{10}$$

Now solve the equation x=\frac{40±70}{10} when ± is plus. Add 40 to 70.

$$x=\frac{110}{10}$$

Divide 110 by 10.

$$x=11$$

Now solve the equation x=\frac{40±70}{10} when ± is minus. Subtract 70 from 40.

$$x=-\frac{30}{10}$$

Divide -30 by 10.

$$x=-3$$

The equation is now solved.

$$x=11 x=-3$$

Variable x cannot be equal to -3.

$$x=11$$

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 8\left(x-3\right)\left(x+3\right), the least common multiple of x+3,x^{2}-9,8.

$$\left(8x-24\right)\left(x-2\right)-8\times 30=3\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply 8x-24 by x-2 and combine like terms.

$$8x^{2}-40x+48-8\times 30=3\left(x-3\right)\left(x+3\right)$$

Multiply -8 and 30 to get -240.

$$8x^{2}-40x+48-240=3\left(x-3\right)\left(x+3\right)$$

Subtract 240 from 48 to get -192.

$$8x^{2}-40x-192=3\left(x-3\right)\left(x+3\right)$$

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

$$8x^{2}-40x-192=\left(3x-9\right)\left(x+3\right)$$

Use the distributive property to multiply 3x-9 by x+3 and combine like terms.

$$8x^{2}-40x-192=3x^{2}-27$$

Subtract 3x^{2} from both sides.

$$8x^{2}-40x-192-3x^{2}=-27$$

Combine 8x^{2} and -3x^{2} to get 5x^{2}.

$$5x^{2}-40x-192=-27$$

Add 192 to both sides.

$$5x^{2}-40x=-27+192$$

Add -27 and 192 to get 165.

$$5x^{2}-40x=165$$

Divide both sides by 5.

$$\frac{5x^{2}-40x}{5}=\frac{165}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{40}{5}\right)x=\frac{165}{5}$$

Divide -40 by 5.

$$x^{2}-8x=\frac{165}{5}$$

Divide 165 by 5.

$$x^{2}-8x=33$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=33+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=33+16$$

Add 33 to 16.

$$x^{2}-8x+16=49$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{49}$$

Simplify.

$$x-4=7 x-4=-7$$

Add 4 to both sides of the equation.

$$x=11 x=-3$$

Variable x cannot be equal to -3.

$$x=11$$