Solve 1-5/x+2=7/x^2-7x-18

$$1 - \frac { 5 } { x + 2 } = \frac { 7 } { x ^ { 2 } - 7 x - 18 }$$

Solve for x

$$x=2$$

$$x=10$$

Steps Using Factoring

Variable x cannot be equal to any of the values -2,9 since division by zero is not defined. Multiply both sides of the equation by \left(x-9\right)\left(x+2\right), the least common multiple of x+2,x^{2}-7x-18.

$$\left(x-9\right)\left(x+2\right)-\left(x-9\right)\times 5=7$$

Use the distributive property to multiply x-9 by x+2 and combine like terms.

$$x^{2}-7x-18-\left(x-9\right)\times 5=7$$

Use the distributive property to multiply x-9 by 5.

$$x^{2}-7x-18-\left(5x-45\right)=7$$

To find the opposite of 5x-45, find the opposite of each term.

$$x^{2}-7x-18-5x+45=7$$

Combine -7x and -5x to get -12x.

$$x^{2}-12x-18+45=7$$

Add -18 and 45 to get 27.

$$x^{2}-12x+27=7$$

Subtract 7 from both sides.

$$x^{2}-12x+27-7=0$$

Subtract 7 from 27 to get 20.

$$x^{2}-12x+20=0$$

To solve the equation, factor x^{2}-12x+20 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.

$$a+b=-12 ab=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -12.

$$a=-10 b=-2$$

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

$$\left(x-10\right)\left(x-2\right)$$

To find equation solutions, solve x-10=0 and x-2=0.

$$x=10 x=2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -2,9 since division by zero is not defined. Multiply both sides of the equation by \left(x-9\right)\left(x+2\right), the least common multiple of x+2,x^{2}-7x-18.

$$\left(x-9\right)\left(x+2\right)-\left(x-9\right)\times 5=7$$

Use the distributive property to multiply x-9 by x+2 and combine like terms.

$$x^{2}-7x-18-\left(x-9\right)\times 5=7$$

Use the distributive property to multiply x-9 by 5.

$$x^{2}-7x-18-\left(5x-45\right)=7$$

To find the opposite of 5x-45, find the opposite of each term.

$$x^{2}-7x-18-5x+45=7$$

Combine -7x and -5x to get -12x.

$$x^{2}-12x-18+45=7$$

Add -18 and 45 to get 27.

$$x^{2}-12x+27=7$$

Subtract 7 from both sides.

$$x^{2}-12x+27-7=0$$

Subtract 7 from 27 to get 20.

$$x^{2}-12x+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\times 20=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -12.

$$a=-10 b=-2$$

Rewrite x^{2}-12x+20 as \left(x^{2}-10x\right)+\left(-2x+20\right).

$$\left(x^{2}-10x\right)+\left(-2x+20\right)$$

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

$$x\left(x-10\right)-2\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x-2\right)$$

To find equation solutions, solve x-10=0 and x-2=0.

$$x=10 x=2$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,9 since division by zero is not defined. Multiply both sides of the equation by \left(x-9\right)\left(x+2\right), the least common multiple of x+2,x^{2}-7x-18.

$$\left(x-9\right)\left(x+2\right)-\left(x-9\right)\times 5=7$$

Use the distributive property to multiply x-9 by x+2 and combine like terms.

$$x^{2}-7x-18-\left(x-9\right)\times 5=7$$

Use the distributive property to multiply x-9 by 5.

$$x^{2}-7x-18-\left(5x-45\right)=7$$

To find the opposite of 5x-45, find the opposite of each term.

$$x^{2}-7x-18-5x+45=7$$

Combine -7x and -5x to get -12x.

$$x^{2}-12x-18+45=7$$

Add -18 and 45 to get 27.

$$x^{2}-12x+27=7$$

Subtract 7 from both sides.

$$x^{2}-12x+27-7=0$$

Subtract 7 from 27 to get 20.

$$x^{2}-12x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -12 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 20}}{2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 20}}{2}$$

Multiply -4 times 20.

$$x=\frac{-\left(-12\right)±\sqrt{144-80}}{2}$$

Add 144 to -80.

$$x=\frac{-\left(-12\right)±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-\left(-12\right)±8}{2}$$

The opposite of -12 is 12.

$$x=\frac{12±8}{2}$$

Now solve the equation x=\frac{12±8}{2} when ± is plus. Add 12 to 8.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{12±8}{2} when ± is minus. Subtract 8 from 12.

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

Divide 4 by 2.

$$x=2$$

The equation is now solved.

$$x=10 x=2$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,9 since division by zero is not defined. Multiply both sides of the equation by \left(x-9\right)\left(x+2\right), the least common multiple of x+2,x^{2}-7x-18.

$$\left(x-9\right)\left(x+2\right)-\left(x-9\right)\times 5=7$$

Use the distributive property to multiply x-9 by x+2 and combine like terms.

$$x^{2}-7x-18-\left(x-9\right)\times 5=7$$

Use the distributive property to multiply x-9 by 5.

$$x^{2}-7x-18-\left(5x-45\right)=7$$

To find the opposite of 5x-45, find the opposite of each term.

$$x^{2}-7x-18-5x+45=7$$

Combine -7x and -5x to get -12x.

$$x^{2}-12x-18+45=7$$

Add -18 and 45 to get 27.

$$x^{2}-12x+27=7$$

Subtract 27 from both sides.

$$x^{2}-12x=7-27$$

Subtract 27 from 7 to get -20.

$$x^{2}-12x=-20$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=-20+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=-20+36$$

Add -20 to 36.

$$x^{2}-12x+36=16$$

Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-6\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-6=4 x-6=-4$$

Add 6 to both sides of the equation.

$$x=10 x=2$$