Solve x/(x+3)-1/(x+6)=3/x^2+9x+18

$$\frac { x } { ( x + 3 ) } - \frac { 1 } { ( x + 6 ) } = \frac { 3 } { x ^ { 2 } + 9 x + 18 }$$

Solve for x

$$x=1$$

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+3,x+6,x^{2}+9x+18.

$$\left(x+6\right)x-\left(x+3\right)=3$$

Use the distributive property to multiply x+6 by x.

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

To find the opposite of x+3, find the opposite of each term.

$$x^{2}+6x-x-3=3$$

Combine 6x and -x to get 5x.

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

Subtract 3 from both sides.

$$x^{2}+5x-3-3=0$$

Subtract 3 from -3 to get -6.

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

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 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 -6.

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

Calculate the sum for each pair.

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

The solution is the pair that gives sum 5.

$$a=-1 b=6$$

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

$$\left(x-1\right)\left(x+6\right)$$

To find equation solutions, solve x-1=0 and x+6=0.

$$x=1 x=-6$$

Variable x cannot be equal to -6.

$$x=1$$

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+3,x+6,x^{2}+9x+18.

$$\left(x+6\right)x-\left(x+3\right)=3$$

Use the distributive property to multiply x+6 by x.

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

To find the opposite of x+3, find the opposite of each term.

$$x^{2}+6x-x-3=3$$

Combine 6x and -x to get 5x.

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

Subtract 3 from both sides.

$$x^{2}+5x-3-3=0$$

Subtract 3 from -3 to get -6.

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

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.

$$a+b=5 ab=1\left(-6\right)=-6$$

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 -6.

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

Calculate the sum for each pair.

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

The solution is the pair that gives sum 5.

$$a=-1 b=6$$

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

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

Factor out x in the first and 6 in the second group.

$$x\left(x-1\right)+6\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+6\right)$$

To find equation solutions, solve x-1=0 and x+6=0.

$$x=1 x=-6$$

Variable x cannot be equal to -6.

$$x=1$$

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+3,x+6,x^{2}+9x+18.

$$\left(x+6\right)x-\left(x+3\right)=3$$

Use the distributive property to multiply x+6 by x.

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

To find the opposite of x+3, find the opposite of each term.

$$x^{2}+6x-x-3=3$$

Combine 6x and -x to get 5x.

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

Subtract 3 from both sides.

$$x^{2}+5x-3-3=0$$

Subtract 3 from -3 to get -6.

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

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}.

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

Square 5.

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

Multiply -4 times -6.

$$x=\frac{-5±\sqrt{25+24}}{2}$$

Add 25 to 24.

$$x=\frac{-5±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-5±7}{2}$$

Now solve the equation x=\frac{-5±7}{2} when ± is plus. Add -5 to 7.

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

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{-5±7}{2} when ± is minus. Subtract 7 from -5.

$$x=-\frac{12}{2}$$

Divide -12 by 2.

$$x=-6$$

The equation is now solved.

$$x=1 x=-6$$

Variable x cannot be equal to -6.

$$x=1$$

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+3,x+6,x^{2}+9x+18.

$$\left(x+6\right)x-\left(x+3\right)=3$$

Use the distributive property to multiply x+6 by x.

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

To find the opposite of x+3, find the opposite of each term.

$$x^{2}+6x-x-3=3$$

Combine 6x and -x to get 5x.

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

Add 3 to both sides.

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

Add 3 and 3 to get 6.

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

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}.

$$x^{2}+5x+\frac{25}{4}=\frac{49}{4}$$

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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{7}{2} x+\frac{5}{2}=-\frac{7}{2}$$

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

$$x=1 x=-6$$

Variable x cannot be equal to -6.

$$x=1$$