Solve x^2-5x/x-2=6/2-x

$$\frac { x ^ { 2 } - 5 x } { x - 2 } = \frac { 6 } { 2 - x }$$

Solve for x

$$x=3$$

Steps Using Factoring

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2, the least common multiple of x-2,2-x.

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

Add 6 to both sides.

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

To solve the equation, factor x^{2}-5x+6 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-5 ab=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

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

To find equation solutions, solve x-3=0 and x-2=0.

$$x=3 x=2$$

Variable x cannot be equal to 2.

$$x=3$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2, the least common multiple of x-2,2-x.

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

Add 6 to both sides.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\times 6=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite x^{2}-5x+6 as \left(x^{2}-3x\right)+\left(-2x+6\right).

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

Factor out x in the first and -2 in the second group.

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

Factor out common term x-3 by using distributive property.

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

To find equation solutions, solve x-3=0 and x-2=0.

$$x=3 x=2$$

Variable x cannot be equal to 2.

$$x=3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2, the least common multiple of x-2,2-x.

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

Add 6 to both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 6}}{2}$$

Multiply -4 times 6.

$$x=\frac{-\left(-5\right)±\sqrt{25-24}}{2}$$

Add 25 to -24.

$$x=\frac{-\left(-5\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-\left(-5\right)±1}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±1}{2}$$

Now solve the equation x=\frac{5±1}{2} when ± is plus. Add 5 to 1.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{5±1}{2} when ± is minus. Subtract 1 from 5.

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

Divide 4 by 2.

$$x=2$$

The equation is now solved.

$$x=3 x=2$$

Variable x cannot be equal to 2.

$$x=3$$

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2, the least common multiple of x-2,2-x.

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

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.

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

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

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Add -6 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{1}{2} x-\frac{5}{2}=-\frac{1}{2}$$

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

$$x=3 x=2$$

Variable x cannot be equal to 2.

$$x=3$$