Solve 7/x-3+1=18/x^2-6x+9

$$\frac{ 7 }{ x-3 } +1 = \frac{ 18 }{ { x }^{ 2 } -6x+9 }$$

Solve for x

$$x=-6$$

$$x=5$$

Steps Using Factoring

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}, the least common multiple of x-3,x^{2}-6x+9.

$$\left(x-3\right)\times 7+\left(x-3\right)^{2}=18$$

Use the distributive property to multiply x-3 by 7.

$$7x-21+\left(x-3\right)^{2}=18$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$7x-21+x^{2}-6x+9=18$$

Combine 7x and -6x to get x.

$$x-21+x^{2}+9=18$$

Add -21 and 9 to get -12.

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

Subtract 18 from both sides.

$$x-12+x^{2}-18=0$$

Subtract 18 from -12 to get -30.

$$x-30+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+x-30=0$$

To solve the equation, factor x^{2}+x-30 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.

$$a+b=1 ab=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

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

$$\left(x-5\right)\left(x+6\right)$$

To find equation solutions, solve x-5=0 and x+6=0.

$$x=5 x=-6$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}, the least common multiple of x-3,x^{2}-6x+9.

$$\left(x-3\right)\times 7+\left(x-3\right)^{2}=18$$

Use the distributive property to multiply x-3 by 7.

$$7x-21+\left(x-3\right)^{2}=18$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$7x-21+x^{2}-6x+9=18$$

Combine 7x and -6x to get x.

$$x-21+x^{2}+9=18$$

Add -21 and 9 to get -12.

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

Subtract 18 from both sides.

$$x-12+x^{2}-18=0$$

Subtract 18 from -12 to get -30.

$$x-30+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-30\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

Rewrite x^{2}+x-30 as \left(x^{2}-5x\right)+\left(6x-30\right).

$$\left(x^{2}-5x\right)+\left(6x-30\right)$$

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

$$x\left(x-5\right)+6\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+6\right)$$

To find equation solutions, solve x-5=0 and x+6=0.

$$x=5 x=-6$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}, the least common multiple of x-3,x^{2}-6x+9.

$$\left(x-3\right)\times 7+\left(x-3\right)^{2}=18$$

Use the distributive property to multiply x-3 by 7.

$$7x-21+\left(x-3\right)^{2}=18$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$7x-21+x^{2}-6x+9=18$$

Combine 7x and -6x to get x.

$$x-21+x^{2}+9=18$$

Add -21 and 9 to get -12.

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

Subtract 18 from both sides.

$$x-12+x^{2}-18=0$$

Subtract 18 from -12 to get -30.

$$x-30+x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-30\right)}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-30\right)}}{2}$$

Multiply -4 times -30.

$$x=\frac{-1±\sqrt{1+120}}{2}$$

Add 1 to 120.

$$x=\frac{-1±\sqrt{121}}{2}$$

Take the square root of 121.

$$x=\frac{-1±11}{2}$$

Now solve the equation x=\frac{-1±11}{2} when ± is plus. Add -1 to 11.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{-1±11}{2} when ± is minus. Subtract 11 from -1.

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

Divide -12 by 2.

$$x=-6$$

The equation is now solved.

$$x=5 x=-6$$

Steps for Completing the Square

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)^{2}, the least common multiple of x-3,x^{2}-6x+9.

$$\left(x-3\right)\times 7+\left(x-3\right)^{2}=18$$

Use the distributive property to multiply x-3 by 7.

$$7x-21+\left(x-3\right)^{2}=18$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$7x-21+x^{2}-6x+9=18$$

Combine 7x and -6x to get x.

$$x-21+x^{2}+9=18$$

Add -21 and 9 to get -12.

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

Add 12 to both sides.

$$x+x^{2}=18+12$$

Add 18 and 12 to get 30.

$$x+x^{2}=30$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}+x=30$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=30+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=30+\frac{1}{4}$$

Add 30 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{121}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{11}{2} x+\frac{1}{2}=-\frac{11}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=5 x=-6$$