Solve x+7/x+2+7/x-2=2x+3/x-2

$$\frac { x + 7 } { x + 2 } + \frac { 7 } { x - 2 } = \frac { 2 x + 3 } { x - 2 }$$

Solve for x

$$x=3$$

Steps Using the Quadratic Formula

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

$$\left(x-2\right)\left(x+7\right)+\left(x+2\right)\times 7=\left(x+2\right)\left(2x+3\right)$$

Use the distributive property to multiply x-2 by x+7 and combine like terms.

$$x^{2}+5x-14+\left(x+2\right)\times 7=\left(x+2\right)\left(2x+3\right)$$

Use the distributive property to multiply x+2 by 7.

$$x^{2}+5x-14+7x+14=\left(x+2\right)\left(2x+3\right)$$

Combine 5x and 7x to get 12x.

$$x^{2}+12x-14+14=\left(x+2\right)\left(2x+3\right)$$

Add -14 and 14 to get 0.

$$x^{2}+12x=\left(x+2\right)\left(2x+3\right)$$

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

$$x^{2}+12x=2x^{2}+7x+6$$

Subtract 2x^{2} from both sides.

$$x^{2}+12x-2x^{2}=7x+6$$

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

$$-x^{2}+12x=7x+6$$

Subtract 7x from both sides.

$$-x^{2}+12x-7x=6$$

Combine 12x and -7x to get 5x.

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

Subtract 6 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\left(-6\right)}}{2\left(-1\right)}$$

Multiply 4 times -6.

$$x=\frac{-5±\sqrt{25-24}}{2\left(-1\right)}$$

Add 25 to -24.

$$x=\frac{-5±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$x=\frac{-5±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±1}{-2}$$

Now solve the equation x=\frac{-5±1}{-2} when ± is plus. Add -5 to 1.

$$x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x=2$$

Now solve the equation x=\frac{-5±1}{-2} when ± is minus. Subtract 1 from -5.

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

Divide -6 by -2.

$$x=3$$

The equation is now solved.

$$x=2 x=3$$

Variable x cannot be equal to 2.

$$x=3$$

Steps for Completing the Square

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

$$\left(x-2\right)\left(x+7\right)+\left(x+2\right)\times 7=\left(x+2\right)\left(2x+3\right)$$

Use the distributive property to multiply x-2 by x+7 and combine like terms.

$$x^{2}+5x-14+\left(x+2\right)\times 7=\left(x+2\right)\left(2x+3\right)$$

Use the distributive property to multiply x+2 by 7.

$$x^{2}+5x-14+7x+14=\left(x+2\right)\left(2x+3\right)$$

Combine 5x and 7x to get 12x.

$$x^{2}+12x-14+14=\left(x+2\right)\left(2x+3\right)$$

Add -14 and 14 to get 0.

$$x^{2}+12x=\left(x+2\right)\left(2x+3\right)$$

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

$$x^{2}+12x=2x^{2}+7x+6$$

Subtract 2x^{2} from both sides.

$$x^{2}+12x-2x^{2}=7x+6$$

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

$$-x^{2}+12x=7x+6$$

Subtract 7x from both sides.

$$-x^{2}+12x-7x=6$$

Combine 12x and -7x to get 5x.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 5 by -1.

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

Divide 6 by -1.

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

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.

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

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

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{1}{2} x-\frac{5}{2}=-\frac{1}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=3 x=2$$

Variable x cannot be equal to 2.

$$x=3$$