Solve 1/x-2-3/x+2=4/3

$$\frac { 1 } { x - 2 } - \frac { 3 } { x + 2 } = \frac { 4 } { 3 }$$

Solve for x

$$x=-4$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

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 3\left(x-2\right)\left(x+2\right), the least common multiple of x-2,x+2,3.

$$3x+6-\left(3x-6\right)\times 3=4\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply 3x-6 by 3.

$$3x+6-\left(9x-18\right)=4\left(x-2\right)\left(x+2\right)$$

To find the opposite of 9x-18, find the opposite of each term.

$$3x+6-9x+18=4\left(x-2\right)\left(x+2\right)$$

Combine 3x and -9x to get -6x.

$$-6x+6+18=4\left(x-2\right)\left(x+2\right)$$

Add 6 and 18 to get 24.

$$-6x+24=4\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply 4 by x-2.

$$-6x+24=\left(4x-8\right)\left(x+2\right)$$

Use the distributive property to multiply 4x-8 by x+2 and combine like terms.

$$-6x+24=4x^{2}-16$$

Subtract 4x^{2} from both sides.

$$-6x+24-4x^{2}=-16$$

Add 16 to both sides.

$$-6x+24-4x^{2}+16=0$$

Add 24 and 16 to get 40.

$$-6x+40-4x^{2}=0$$

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.

$$-4x^{2}-6x+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -6 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-4\right)\times 40}}{2\left(-4\right)}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-4\right)\times 40}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-6\right)±\sqrt{36+16\times 40}}{2\left(-4\right)}$$

Multiply 16 times 40.

$$x=\frac{-\left(-6\right)±\sqrt{36+640}}{2\left(-4\right)}$$

Add 36 to 640.

$$x=\frac{-\left(-6\right)±\sqrt{676}}{2\left(-4\right)}$$

Take the square root of 676.

$$x=\frac{-\left(-6\right)±26}{2\left(-4\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±26}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{6±26}{-8}$$

Now solve the equation x=\frac{6±26}{-8} when ± is plus. Add 6 to 26.

$$x=\frac{32}{-8}$$

Divide 32 by -8.

$$x=-4$$

Now solve the equation x=\frac{6±26}{-8} when ± is minus. Subtract 26 from 6.

$$x=-\frac{20}{-8}$$

Reduce the fraction \frac{-20}{-8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

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 3\left(x-2\right)\left(x+2\right), the least common multiple of x-2,x+2,3.

$$3x+6-\left(3x-6\right)\times 3=4\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply 3x-6 by 3.

$$3x+6-\left(9x-18\right)=4\left(x-2\right)\left(x+2\right)$$

To find the opposite of 9x-18, find the opposite of each term.

$$3x+6-9x+18=4\left(x-2\right)\left(x+2\right)$$

Combine 3x and -9x to get -6x.

$$-6x+6+18=4\left(x-2\right)\left(x+2\right)$$

Add 6 and 18 to get 24.

$$-6x+24=4\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply 4 by x-2.

$$-6x+24=\left(4x-8\right)\left(x+2\right)$$

Use the distributive property to multiply 4x-8 by x+2 and combine like terms.

$$-6x+24=4x^{2}-16$$

Subtract 4x^{2} from both sides.

$$-6x+24-4x^{2}=-16$$

Subtract 24 from both sides.

$$-6x-4x^{2}=-16-24$$

Subtract 24 from -16 to get -40.

$$-6x-4x^{2}=-40$$

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.

$$-4x^{2}-6x=-40$$

Divide both sides by -4.

$$\frac{-4x^{2}-6x}{-4}=-\frac{40}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{6}{-4}\right)x=-\frac{40}{-4}$$

Reduce the fraction \frac{-6}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{3}{2}x=-\frac{40}{-4}$$

Divide -40 by -4.

$$x^{2}+\frac{3}{2}x=10$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=10+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=10+\frac{9}{16}$$

Add 10 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{169}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{13}{4} x+\frac{3}{4}=-\frac{13}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

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