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

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

Solve for x

$$x = \frac{\sqrt{11} + 3}{2} \approx 3.158312395$$

$$x=\frac{3-\sqrt{11}}{2}\approx -0.158312395$$

Steps Using the Quadratic Formula

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

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

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

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

Use the distributive property to multiply 3x-1 by x-3 and combine like terms.

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

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

$$3x^{2}-10x+3=x^{2}-4x+4$$

Subtract x^{2} from both sides.

$$3x^{2}-10x+3-x^{2}=-4x+4$$

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

$$2x^{2}-10x+3=-4x+4$$

Add 4x to both sides.

$$2x^{2}-10x+3+4x=4$$

Combine -10x and 4x to get -6x.

$$2x^{2}-6x+3=4$$

Subtract 4 from both sides.

$$2x^{2}-6x+3-4=0$$

Subtract 4 from 3 to get -1.

$$2x^{2}-6x-1=0$$

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

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

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 2\left(-1\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-6\right)±\sqrt{36-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$x=\frac{-\left(-6\right)±\sqrt{36+8}}{2\times 2}$$

Add 36 to 8.

$$x=\frac{-\left(-6\right)±\sqrt{44}}{2\times 2}$$

Take the square root of 44.

$$x=\frac{-\left(-6\right)±2\sqrt{11}}{2\times 2}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{11}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{6±2\sqrt{11}}{4}$$

Now solve the equation x=\frac{6±2\sqrt{11}}{4} when ± is plus. Add 6 to 2\sqrt{11}.

$$x=\frac{2\sqrt{11}+6}{4}$$

Divide 6+2\sqrt{11} by 4.

$$x=\frac{\sqrt{11}+3}{2}$$

Now solve the equation x=\frac{6±2\sqrt{11}}{4} when ± is minus. Subtract 2\sqrt{11} from 6.

$$x=\frac{6-2\sqrt{11}}{4}$$

Divide 6-2\sqrt{11} by 4.

$$x=\frac{3-\sqrt{11}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{11}+3}{2} x=\frac{3-\sqrt{11}}{2}$$

Steps for Completing the Square

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

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

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

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

Use the distributive property to multiply 3x-1 by x-3 and combine like terms.

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

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

$$3x^{2}-10x+3=x^{2}-4x+4$$

Subtract x^{2} from both sides.

$$3x^{2}-10x+3-x^{2}=-4x+4$$

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

$$2x^{2}-10x+3=-4x+4$$

Add 4x to both sides.

$$2x^{2}-10x+3+4x=4$$

Combine -10x and 4x to get -6x.

$$2x^{2}-6x+3=4$$

Subtract 3 from both sides.

$$2x^{2}-6x=4-3$$

Subtract 3 from 4 to get 1.

$$2x^{2}-6x=1$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -6 by 2.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{2}+\frac{9}{4}$$

Add \frac{1}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=\frac{11}{4}$$

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

$$\left(x-\frac{3}{2}\right)^{2}=\frac{11}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{11}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{11}}{2} x-\frac{3}{2}=-\frac{\sqrt{11}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{11}+3}{2} x=\frac{3-\sqrt{11}}{2}$$