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

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

Solve for x

$$x=3$$

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 2,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x-2\right), the least common multiple of x-4,x-2.

$$\left(x-2\right)\left(x-2\right)-\left(x-4\right)=-2\left(x-4\right)\left(x-2\right)$$

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

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

To find the opposite of x-4, find the opposite of each term.

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

Combine -4x and -x to get -5x.

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

Add 4 and 4 to get 8.

$$x^{2}-5x+8=-2\left(x-4\right)\left(x-2\right)$$

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

$$x^{2}-5x+8=\left(-2x+8\right)\left(x-2\right)$$

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

$$x^{2}-5x+8=-2x^{2}+12x-16$$

Add 2x^{2} to both sides.

$$x^{2}-5x+8+2x^{2}=12x-16$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-5x+8=12x-16$$

Subtract 12x from both sides.

$$3x^{2}-5x+8-12x=-16$$

Combine -5x and -12x to get -17x.

$$3x^{2}-17x+8=-16$$

Add 16 to both sides.

$$3x^{2}-17x+8+16=0$$

Add 8 and 16 to get 24.

$$3x^{2}-17x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -17 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 3\times 24}}{2\times 3}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 3\times 24}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-17\right)±\sqrt{289-12\times 24}}{2\times 3}$$

Multiply -12 times 24.

$$x=\frac{-\left(-17\right)±\sqrt{289-288}}{2\times 3}$$

Add 289 to -288.

$$x=\frac{-\left(-17\right)±\sqrt{1}}{2\times 3}$$

Take the square root of 1.

$$x=\frac{-\left(-17\right)±1}{2\times 3}$$

The opposite of -17 is 17.

$$x=\frac{17±1}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{17±1}{6}$$

Now solve the equation x=\frac{17±1}{6} when ± is plus. Add 17 to 1.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{17±1}{6} when ± is minus. Subtract 1 from 17.

$$x=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{8}{3}$$

The equation is now solved.

$$x=3 x=\frac{8}{3}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 2,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x-2\right), the least common multiple of x-4,x-2.

$$\left(x-2\right)\left(x-2\right)-\left(x-4\right)=-2\left(x-4\right)\left(x-2\right)$$

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

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

To find the opposite of x-4, find the opposite of each term.

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

Combine -4x and -x to get -5x.

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

Add 4 and 4 to get 8.

$$x^{2}-5x+8=-2\left(x-4\right)\left(x-2\right)$$

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

$$x^{2}-5x+8=\left(-2x+8\right)\left(x-2\right)$$

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

$$x^{2}-5x+8=-2x^{2}+12x-16$$

Add 2x^{2} to both sides.

$$x^{2}-5x+8+2x^{2}=12x-16$$

Combine x^{2} and 2x^{2} to get 3x^{2}.

$$3x^{2}-5x+8=12x-16$$

Subtract 12x from both sides.

$$3x^{2}-5x+8-12x=-16$$

Combine -5x and -12x to get -17x.

$$3x^{2}-17x+8=-16$$

Subtract 8 from both sides.

$$3x^{2}-17x=-16-8$$

Subtract 8 from -16 to get -24.

$$3x^{2}-17x=-24$$

Divide both sides by 3.

$$\frac{3x^{2}-17x}{3}=-\frac{24}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{17}{3}x=-\frac{24}{3}$$

Divide -24 by 3.

$$x^{2}-\frac{17}{3}x=-8$$

Divide -\frac{17}{3}, the coefficient of the x term, by 2 to get -\frac{17}{6}. Then add the square of -\frac{17}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{3}x+\left(-\frac{17}{6}\right)^{2}=-8+\left(-\frac{17}{6}\right)^{2}$$

Square -\frac{17}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{3}x+\frac{289}{36}=-8+\frac{289}{36}$$

Add -8 to \frac{289}{36}.

$$x^{2}-\frac{17}{3}x+\frac{289}{36}=\frac{1}{36}$$

Factor x^{2}-\frac{17}{3}x+\frac{289}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{17}{6}=\frac{1}{6} x-\frac{17}{6}=-\frac{1}{6}$$

Add \frac{17}{6} to both sides of the equation.

$$x=3 x=\frac{8}{3}$$