Solve 9/2+x+7/2-x=10

$$\frac { 9 } { 2 + x } + \frac { 7 } { 2 - x } = 10$$

Solve for x

$$x=-\frac{4}{5}=-0.8$$

$$x=1$$

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 2+x,2-x.

$$\left(x-2\right)\times 9-\left(2+x\right)\times 7=10\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply x-2 by 9.

$$9x-18-\left(2+x\right)\times 7=10\left(x-2\right)\left(x+2\right)$$

Multiply -1 and 7 to get -7.

$$9x-18-7\left(2+x\right)=10\left(x-2\right)\left(x+2\right)$$

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

$$9x-18-14-7x=10\left(x-2\right)\left(x+2\right)$$

Subtract 14 from -18 to get -32.

$$9x-32-7x=10\left(x-2\right)\left(x+2\right)$$

Combine 9x and -7x to get 2x.

$$2x-32=10\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply 10 by x-2.

$$2x-32=\left(10x-20\right)\left(x+2\right)$$

Use the distributive property to multiply 10x-20 by x+2 and combine like terms.

$$2x-32=10x^{2}-40$$

Subtract 10x^{2} from both sides.

$$2x-32-10x^{2}=-40$$

Add 40 to both sides.

$$2x-32-10x^{2}+40=0$$

Add -32 and 40 to get 8.

$$2x+8-10x^{2}=0$$

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.

$$-10x^{2}+2x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 2 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-10\right)\times 8}}{2\left(-10\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-10\right)\times 8}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-2±\sqrt{4+40\times 8}}{2\left(-10\right)}$$

Multiply 40 times 8.

$$x=\frac{-2±\sqrt{4+320}}{2\left(-10\right)}$$

Add 4 to 320.

$$x=\frac{-2±\sqrt{324}}{2\left(-10\right)}$$

Take the square root of 324.

$$x=\frac{-2±18}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-2±18}{-20}$$

Now solve the equation x=\frac{-2±18}{-20} when ± is plus. Add -2 to 18.

$$x=\frac{16}{-20}$$

Reduce the fraction \frac{16}{-20} to lowest terms by extracting and canceling out 4.

$$x=-\frac{4}{5}$$

Now solve the equation x=\frac{-2±18}{-20} when ± is minus. Subtract 18 from -2.

$$x=-\frac{20}{-20}$$

Divide -20 by -20.

$$x=1$$

The equation is now solved.

$$x=-\frac{4}{5} x=1$$

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 2+x,2-x.

$$\left(x-2\right)\times 9-\left(2+x\right)\times 7=10\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply x-2 by 9.

$$9x-18-\left(2+x\right)\times 7=10\left(x-2\right)\left(x+2\right)$$

Multiply -1 and 7 to get -7.

$$9x-18-7\left(2+x\right)=10\left(x-2\right)\left(x+2\right)$$

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

$$9x-18-14-7x=10\left(x-2\right)\left(x+2\right)$$

Subtract 14 from -18 to get -32.

$$9x-32-7x=10\left(x-2\right)\left(x+2\right)$$

Combine 9x and -7x to get 2x.

$$2x-32=10\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply 10 by x-2.

$$2x-32=\left(10x-20\right)\left(x+2\right)$$

Use the distributive property to multiply 10x-20 by x+2 and combine like terms.

$$2x-32=10x^{2}-40$$

Subtract 10x^{2} from both sides.

$$2x-32-10x^{2}=-40$$

Add 32 to both sides.

$$2x-10x^{2}=-40+32$$

Add -40 and 32 to get -8.

$$2x-10x^{2}=-8$$

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.

$$-10x^{2}+2x=-8$$

Divide both sides by -10.

$$\frac{-10x^{2}+2x}{-10}=-\frac{8}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{2}{-10}x=-\frac{8}{-10}$$

Reduce the fraction \frac{2}{-10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{5}x=-\frac{8}{-10}$$

Reduce the fraction \frac{-8}{-10} to lowest terms by extracting and canceling out 2.

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

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{4}{5}+\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{4}{5}+\frac{1}{100}$$

Add \frac{4}{5} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{81}{100}$$

Factor x^{2}-\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{81}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Simplify.

$$x-\frac{1}{10}=\frac{9}{10} x-\frac{1}{10}=-\frac{9}{10}$$

Add \frac{1}{10} to both sides of the equation.

$$x=1 x=-\frac{4}{5}$$