Solve x/x-2+2/x+3=10/x^2+x-6

$$\frac { x } { x - 2 } + \frac { 2 } { x + 3 } = \frac { 10 } { x ^ { 2 } + x - 6 }$$

Solve for x

$$x=-7$$

Steps Using Factoring

Variable x cannot be equal to any of the values -3,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+3\right), the least common multiple of x-2,x+3,x^{2}+x-6.

$$\left(x+3\right)x+\left(x-2\right)\times 2=10$$

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

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

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

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

Combine 3x and 2x to get 5x.

$$x^{2}+5x-4=10$$

Subtract 10 from both sides.

$$x^{2}+5x-4-10=0$$

Subtract 10 from -4 to get -14.

$$x^{2}+5x-14=0$$

To solve the equation, factor x^{2}+5x-14 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.

$$a+b=5 ab=-14$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -14.

$$-1,14 -2,7$$

Calculate the sum for each pair.

$$-1+14=13 -2+7=5$$

The solution is the pair that gives sum 5.

$$a=-2 b=7$$

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

$$\left(x-2\right)\left(x+7\right)$$

To find equation solutions, solve x-2=0 and x+7=0.

$$x=2 x=-7$$

Variable x cannot be equal to 2.

$$x=-7$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -3,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+3\right), the least common multiple of x-2,x+3,x^{2}+x-6.

$$\left(x+3\right)x+\left(x-2\right)\times 2=10$$

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

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

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

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

Combine 3x and 2x to get 5x.

$$x^{2}+5x-4=10$$

Subtract 10 from both sides.

$$x^{2}+5x-4-10=0$$

Subtract 10 from -4 to get -14.

$$x^{2}+5x-14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-14. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-14\right)=-14$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -14.

$$-1,14 -2,7$$

Calculate the sum for each pair.

$$-1+14=13 -2+7=5$$

The solution is the pair that gives sum 5.

$$a=-2 b=7$$

Rewrite x^{2}+5x-14 as \left(x^{2}-2x\right)+\left(7x-14\right).

$$\left(x^{2}-2x\right)+\left(7x-14\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x-2\right)+7\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x+7\right)$$

To find equation solutions, solve x-2=0 and x+7=0.

$$x=2 x=-7$$

Variable x cannot be equal to 2.

$$x=-7$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+3\right), the least common multiple of x-2,x+3,x^{2}+x-6.

$$\left(x+3\right)x+\left(x-2\right)\times 2=10$$

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

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

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

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

Combine 3x and 2x to get 5x.

$$x^{2}+5x-4=10$$

Subtract 10 from both sides.

$$x^{2}+5x-4-10=0$$

Subtract 10 from -4 to get -14.

$$x^{2}+5x-14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-14\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-14\right)}}{2}$$

Multiply -4 times -14.

$$x=\frac{-5±\sqrt{25+56}}{2}$$

Add 25 to 56.

$$x=\frac{-5±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-5±9}{2}$$

Now solve the equation x=\frac{-5±9}{2} when ± is plus. Add -5 to 9.

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

Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{-5±9}{2} when ± is minus. Subtract 9 from -5.

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

The equation is now solved.

$$x=2 x=-7$$

Variable x cannot be equal to 2.

$$x=-7$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+3\right), the least common multiple of x-2,x+3,x^{2}+x-6.

$$\left(x+3\right)x+\left(x-2\right)\times 2=10$$

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

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

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

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

Combine 3x and 2x to get 5x.

$$x^{2}+5x-4=10$$

Add 4 to both sides.

$$x^{2}+5x=10+4$$

Add 10 and 4 to get 14.

$$x^{2}+5x=14$$

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.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=14+\left(\frac{5}{2}\right)^{2}$$

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

$$x^{2}+5x+\frac{25}{4}=14+\frac{25}{4}$$

Add 14 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{81}{4}$$

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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{9}{2} x+\frac{5}{2}=-\frac{9}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=2 x=-7$$

Variable x cannot be equal to 2.

$$x=-7$$