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

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

Solve for x

$$x=1$$

$$x=-1$$

Steps by Finding Square Root

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

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

Use the distributive property to multiply 1-3x by 1-2x and combine like terms.

$$1-5x+6x^{2}=\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply x-3 by x-2 and combine like terms.

$$1-5x+6x^{2}=x^{2}-5x+6$$

Subtract x^{2} from both sides.

$$1-5x+6x^{2}-x^{2}=-5x+6$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$1-5x+5x^{2}=-5x+6$$

Add 5x to both sides.

$$1-5x+5x^{2}+5x=6$$

Combine -5x and 5x to get 0.

$$1+5x^{2}=6$$

Subtract 1 from both sides.

$$5x^{2}=6-1$$

Subtract 1 from 6 to get 5.

$$5x^{2}=5$$

Divide both sides by 5.

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

Divide 5 by 5 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply 1-3x by 1-2x and combine like terms.

$$1-5x+6x^{2}=\left(x-3\right)\left(x-2\right)$$

Use the distributive property to multiply x-3 by x-2 and combine like terms.

$$1-5x+6x^{2}=x^{2}-5x+6$$

Subtract x^{2} from both sides.

$$1-5x+6x^{2}-x^{2}=-5x+6$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$1-5x+5x^{2}=-5x+6$$

Add 5x to both sides.

$$1-5x+5x^{2}+5x=6$$

Combine -5x and 5x to get 0.

$$1+5x^{2}=6$$

Subtract 6 from both sides.

$$1+5x^{2}-6=0$$

Subtract 6 from 1 to get -5.

$$-5+5x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$5x^{2}-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 0 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 5\left(-5\right)}}{2\times 5}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 5\left(-5\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{0±\sqrt{-20\left(-5\right)}}{2\times 5}$$

Multiply -20 times -5.

$$x=\frac{0±\sqrt{100}}{2\times 5}$$

Take the square root of 100.

$$x=\frac{0±10}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{0±10}{10}$$

Now solve the equation x=\frac{0±10}{10} when ± is plus. Divide 10 by 10.

$$x=1$$

Now solve the equation x=\frac{0±10}{10} when ± is minus. Divide -10 by 10.

$$x=-1$$

The equation is now solved.

$$x=1 x=-1$$