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

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

Solve for x

$$x=5$$

$$x=0$$

Steps Using the Quadratic Formula

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

$$\left(2x+2\right)\left(x-2\right)=\left(x-1\right)\left(x+4\right)$$

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

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

Use the distributive property to multiply x-1 by x+4 and combine like terms.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 3x from both sides.

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

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

$$x^{2}-5x-4=-4$$

Add 4 to both sides.

$$x^{2}-5x-4+4=0$$

Add -4 and 4 to get 0.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-5\right)^{2}.

$$x=\frac{-\left(-5\right)±5}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±5}{2}$$

Now solve the equation x=\frac{5±5}{2} when ± is plus. Add 5 to 5.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{5±5}{2} when ± is minus. Subtract 5 from 5.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=5 x=0$$

Steps for Completing the Square

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

$$\left(2x+2\right)\left(x-2\right)=\left(x-1\right)\left(x+4\right)$$

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

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

Use the distributive property to multiply x-1 by x+4 and combine like terms.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 3x from both sides.

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

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

$$x^{2}-5x-4=-4$$

Add 4 to both sides.

$$x^{2}-5x=-4+4$$

Add -4 and 4 to get 0.

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

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{25}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

Add \frac{5}{2} to both sides of the equation.

$$x=5 x=0$$