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

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

Solve for x

$$x = \frac{\sqrt{33} + 7}{4} \approx 3.186140662$$

$$x=\frac{7-\sqrt{33}}{4}\approx 0.313859338$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right), the least common multiple of x-2,1-x,x^{2}-3x+2.

$$\left(x-1\right)\left(x-1\right)-\left(2-x\right)x=3x-1$$

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

$$\left(x-1\right)^{2}-\left(2-x\right)x=3x-1$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-2x+1-\left(2-x\right)x=3x-1$$

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

$$x^{2}-2x+1-\left(2x-x^{2}\right)=3x-1$$

To find the opposite of 2x-x^{2}, find the opposite of each term.

$$x^{2}-2x+1-2x+x^{2}=3x-1$$

Combine -2x and -2x to get -4x.

$$x^{2}-4x+1+x^{2}=3x-1$$

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

$$2x^{2}-4x+1=3x-1$$

Subtract 3x from both sides.

$$2x^{2}-4x+1-3x=-1$$

Combine -4x and -3x to get -7x.

$$2x^{2}-7x+1=-1$$

Add 1 to both sides.

$$2x^{2}-7x+1+1=0$$

Add 1 and 1 to get 2.

$$2x^{2}-7x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 2}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\times 2}}{2\times 2}$$

Multiply -8 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-16}}{2\times 2}$$

Add 49 to -16.

$$x=\frac{-\left(-7\right)±\sqrt{33}}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{33}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±\sqrt{33}}{4}$$

Now solve the equation x=\frac{7±\sqrt{33}}{4} when ± is plus. Add 7 to \sqrt{33}.

$$x=\frac{\sqrt{33}+7}{4}$$

Now solve the equation x=\frac{7±\sqrt{33}}{4} when ± is minus. Subtract \sqrt{33} from 7.

$$x=\frac{7-\sqrt{33}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{33}+7}{4} x=\frac{7-\sqrt{33}}{4}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right), the least common multiple of x-2,1-x,x^{2}-3x+2.

$$\left(x-1\right)\left(x-1\right)-\left(2-x\right)x=3x-1$$

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

$$\left(x-1\right)^{2}-\left(2-x\right)x=3x-1$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-2x+1-\left(2-x\right)x=3x-1$$

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

$$x^{2}-2x+1-\left(2x-x^{2}\right)=3x-1$$

To find the opposite of 2x-x^{2}, find the opposite of each term.

$$x^{2}-2x+1-2x+x^{2}=3x-1$$

Combine -2x and -2x to get -4x.

$$x^{2}-4x+1+x^{2}=3x-1$$

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

$$2x^{2}-4x+1=3x-1$$

Subtract 3x from both sides.

$$2x^{2}-4x+1-3x=-1$$

Combine -4x and -3x to get -7x.

$$2x^{2}-7x+1=-1$$

Subtract 1 from both sides.

$$2x^{2}-7x=-1-1$$

Subtract 1 from -1 to get -2.

$$2x^{2}-7x=-2$$

Divide both sides by 2.

$$\frac{2x^{2}-7x}{2}=-\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{7}{2}x=-\frac{2}{2}$$

Divide -2 by 2.

$$x^{2}-\frac{7}{2}x=-1$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=-1+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-1+\frac{49}{16}$$

Add -1 to \frac{49}{16}.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{33}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{\sqrt{33}}{4} x-\frac{7}{4}=-\frac{\sqrt{33}}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{\sqrt{33}+7}{4} x=\frac{7-\sqrt{33}}{4}$$