Solve x+2/x+6=x^2+18x/x^2+3x-18-x-6/x-3

$$\frac { x + 2 } { x + 6 } = \frac { x ^ { 2 } + 18 x } { x ^ { 2 } + 3 x - 18 } - \frac { x - 6 } { x - 3 }$$

Solve for x

$$x=-2$$

$$x=21$$

Steps Using Factoring

Variable x cannot be equal to any of the values -6,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+6\right), the least common multiple of x+6,x^{2}+3x-18,x-3.

$$\left(x-3\right)\left(x+2\right)=x^{2}+18x-\left(x+6\right)\left(x-6\right)$$

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

$$x^{2}-x-6=x^{2}+18x-\left(x+6\right)\left(x-6\right)$$

Consider \left(x+6\right)\left(x-6\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 6.

$$x^{2}-x-6=x^{2}+18x-\left(x^{2}-36\right)$$

To find the opposite of x^{2}-36, find the opposite of each term.

$$x^{2}-x-6=x^{2}+18x-x^{2}+36$$

Combine x^{2} and -x^{2} to get 0.

$$x^{2}-x-6=18x+36$$

Subtract 18x from both sides.

$$x^{2}-x-6-18x=36$$

Combine -x and -18x to get -19x.

$$x^{2}-19x-6=36$$

Subtract 36 from both sides.

$$x^{2}-19x-6-36=0$$

Subtract 36 from -6 to get -42.

$$x^{2}-19x-42=0$$

To solve the equation, factor x^{2}-19x-42 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.

$$a+b=-19 ab=-42$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -42.

$$1,-42 2,-21 3,-14 6,-7$$

Calculate the sum for each pair.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

The solution is the pair that gives sum -19.

$$a=-21 b=2$$

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

$$\left(x-21\right)\left(x+2\right)$$

To find equation solutions, solve x-21=0 and x+2=0.

$$x=21 x=-2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -6,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+6\right), the least common multiple of x+6,x^{2}+3x-18,x-3.

$$\left(x-3\right)\left(x+2\right)=x^{2}+18x-\left(x+6\right)\left(x-6\right)$$

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

$$x^{2}-x-6=x^{2}+18x-\left(x+6\right)\left(x-6\right)$$

Consider \left(x+6\right)\left(x-6\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 6.

$$x^{2}-x-6=x^{2}+18x-\left(x^{2}-36\right)$$

To find the opposite of x^{2}-36, find the opposite of each term.

$$x^{2}-x-6=x^{2}+18x-x^{2}+36$$

Combine x^{2} and -x^{2} to get 0.

$$x^{2}-x-6=18x+36$$

Subtract 18x from both sides.

$$x^{2}-x-6-18x=36$$

Combine -x and -18x to get -19x.

$$x^{2}-19x-6=36$$

Subtract 36 from both sides.

$$x^{2}-19x-6-36=0$$

Subtract 36 from -6 to get -42.

$$x^{2}-19x-42=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-42. To find a and b, set up a system to be solved.

$$a+b=-19 ab=1\left(-42\right)=-42$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -42.

$$1,-42 2,-21 3,-14 6,-7$$

Calculate the sum for each pair.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

The solution is the pair that gives sum -19.

$$a=-21 b=2$$

Rewrite x^{2}-19x-42 as \left(x^{2}-21x\right)+\left(2x-42\right).

$$\left(x^{2}-21x\right)+\left(2x-42\right)$$

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

$$x\left(x-21\right)+2\left(x-21\right)$$

Factor out common term x-21 by using distributive property.

$$\left(x-21\right)\left(x+2\right)$$

To find equation solutions, solve x-21=0 and x+2=0.

$$x=21 x=-2$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -6,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+6\right), the least common multiple of x+6,x^{2}+3x-18,x-3.

$$\left(x-3\right)\left(x+2\right)=x^{2}+18x-\left(x+6\right)\left(x-6\right)$$

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

$$x^{2}-x-6=x^{2}+18x-\left(x+6\right)\left(x-6\right)$$

Consider \left(x+6\right)\left(x-6\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 6.

$$x^{2}-x-6=x^{2}+18x-\left(x^{2}-36\right)$$

To find the opposite of x^{2}-36, find the opposite of each term.

$$x^{2}-x-6=x^{2}+18x-x^{2}+36$$

Combine x^{2} and -x^{2} to get 0.

$$x^{2}-x-6=18x+36$$

Subtract 18x from both sides.

$$x^{2}-x-6-18x=36$$

Combine -x and -18x to get -19x.

$$x^{2}-19x-6=36$$

Subtract 36 from both sides.

$$x^{2}-19x-6-36=0$$

Subtract 36 from -6 to get -42.

$$x^{2}-19x-42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -19 for b, and -42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\left(-42\right)}}{2}$$

Square -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\left(-42\right)}}{2}$$

Multiply -4 times -42.

$$x=\frac{-\left(-19\right)±\sqrt{361+168}}{2}$$

Add 361 to 168.

$$x=\frac{-\left(-19\right)±\sqrt{529}}{2}$$

Take the square root of 529.

$$x=\frac{-\left(-19\right)±23}{2}$$

The opposite of -19 is 19.

$$x=\frac{19±23}{2}$$

Now solve the equation x=\frac{19±23}{2} when ± is plus. Add 19 to 23.

$$x=\frac{42}{2}$$

Divide 42 by 2.

$$x=21$$

Now solve the equation x=\frac{19±23}{2} when ± is minus. Subtract 23 from 19.

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

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=21 x=-2$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -6,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+6\right), the least common multiple of x+6,x^{2}+3x-18,x-3.

$$\left(x-3\right)\left(x+2\right)=x^{2}+18x-\left(x+6\right)\left(x-6\right)$$

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

$$x^{2}-x-6=x^{2}+18x-\left(x+6\right)\left(x-6\right)$$

Consider \left(x+6\right)\left(x-6\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 6.

$$x^{2}-x-6=x^{2}+18x-\left(x^{2}-36\right)$$

To find the opposite of x^{2}-36, find the opposite of each term.

$$x^{2}-x-6=x^{2}+18x-x^{2}+36$$

Combine x^{2} and -x^{2} to get 0.

$$x^{2}-x-6=18x+36$$

Subtract 18x from both sides.

$$x^{2}-x-6-18x=36$$

Combine -x and -18x to get -19x.

$$x^{2}-19x-6=36$$

Add 6 to both sides.

$$x^{2}-19x=36+6$$

Add 36 and 6 to get 42.

$$x^{2}-19x=42$$

Divide -19, the coefficient of the x term, by 2 to get -\frac{19}{2}. Then add the square of -\frac{19}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-19x+\left(-\frac{19}{2}\right)^{2}=42+\left(-\frac{19}{2}\right)^{2}$$

Square -\frac{19}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-19x+\frac{361}{4}=42+\frac{361}{4}$$

Add 42 to \frac{361}{4}.

$$x^{2}-19x+\frac{361}{4}=\frac{529}{4}$$

Factor x^{2}-19x+\frac{361}{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}.

$$\left(x-\frac{19}{2}\right)^{2}=\frac{529}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{19}{2}\right)^{2}}=\sqrt{\frac{529}{4}}$$

Simplify.

$$x-\frac{19}{2}=\frac{23}{2} x-\frac{19}{2}=-\frac{23}{2}$$

Add \frac{19}{2} to both sides of the equation.

$$x=21 x=-2$$