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

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

Solve for x

$$x=2$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

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

Use the distributive property to multiply -x+1 by -3.

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

Subtract 3 from 1 to get -2.

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

Use the distributive property to multiply -3x by -x+1.

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

Use the distributive property to multiply -x+1 by 2.

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

Combine -3x and -2x to get -5x.

$$-2+3x=3x^{2}-5x+2$$

Subtract 3x^{2} from both sides.

$$-2+3x-3x^{2}=-5x+2$$

Add 5x to both sides.

$$-2+3x-3x^{2}+5x=2$$

Combine 3x and 5x to get 8x.

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

Subtract 2 from both sides.

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

Subtract 2 from -2 to get -4.

$$-4+8x-3x^{2}=0$$

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.

$$-3x^{2}+8x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 8 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-3\right)\left(-4\right)}}{2\left(-3\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-3\right)\left(-4\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-8±\sqrt{64+12\left(-4\right)}}{2\left(-3\right)}$$

Multiply 12 times -4.

$$x=\frac{-8±\sqrt{64-48}}{2\left(-3\right)}$$

Add 64 to -48.

$$x=\frac{-8±\sqrt{16}}{2\left(-3\right)}$$

Take the square root of 16.

$$x=\frac{-8±4}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-8±4}{-6}$$

Now solve the equation x=\frac{-8±4}{-6} when ± is plus. Add -8 to 4.

$$x=-\frac{4}{-6}$$

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

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

Now solve the equation x=\frac{-8±4}{-6} when ± is minus. Subtract 4 from -8.

$$x=-\frac{12}{-6}$$

Divide -12 by -6.

$$x=2$$

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

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

Use the distributive property to multiply -x+1 by -3.

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

Subtract 3 from 1 to get -2.

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

Use the distributive property to multiply -3x by -x+1.

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

Use the distributive property to multiply -x+1 by 2.

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

Combine -3x and -2x to get -5x.

$$-2+3x=3x^{2}-5x+2$$

Subtract 3x^{2} from both sides.

$$-2+3x-3x^{2}=-5x+2$$

Add 5x to both sides.

$$-2+3x-3x^{2}+5x=2$$

Combine 3x and 5x to get 8x.

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

Add 2 to both sides.

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

Add 2 and 2 to get 4.

$$8x-3x^{2}=4$$

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.

$$-3x^{2}+8x=4$$

Divide both sides by -3.

$$\frac{-3x^{2}+8x}{-3}=\frac{4}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{8}{-3}x=\frac{4}{-3}$$

Divide 8 by -3.

$$x^{2}-\frac{8}{3}x=\frac{4}{-3}$$

Divide 4 by -3.

$$x^{2}-\frac{8}{3}x=-\frac{4}{3}$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{3}x+\left(-\frac{4}{3}\right)^{2}=-\frac{4}{3}+\left(-\frac{4}{3}\right)^{2}$$

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

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=-\frac{4}{3}+\frac{16}{9}$$

Add -\frac{4}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{4}{9}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

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

Add \frac{4}{3} to both sides of the equation.

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