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

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

Solve for x

$$x=-\frac{3}{4}=-0.75$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by x\left(x-6\right), the least common multiple of x-6,x^{2}-6x.

$$x\left(x-6\right)+x\times 7x=9$$

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

$$x^{2}-6x+x\times 7x=9$$

Multiply x and x to get x^{2}.

$$x^{2}-6x+x^{2}\times 7=9$$

Combine x^{2} and x^{2}\times 7 to get 8x^{2}.

$$8x^{2}-6x=9$$

Subtract 9 from both sides.

$$8x^{2}-6x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=-6 ab=8\left(-9\right)=-72$$

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 -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -6.

$$a=-12 b=6$$

Rewrite 8x^{2}-6x-9 as \left(8x^{2}-12x\right)+\left(6x-9\right).

$$\left(8x^{2}-12x\right)+\left(6x-9\right)$$

Factor out 4x in the first and 3 in the second group.

$$4x\left(2x-3\right)+3\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(4x+3\right)$$

To find equation solutions, solve 2x-3=0 and 4x+3=0.

$$x=\frac{3}{2} x=-\frac{3}{4}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by x\left(x-6\right), the least common multiple of x-6,x^{2}-6x.

$$x\left(x-6\right)+x\times 7x=9$$

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

$$x^{2}-6x+x\times 7x=9$$

Multiply x and x to get x^{2}.

$$x^{2}-6x+x^{2}\times 7=9$$

Combine x^{2} and x^{2}\times 7 to get 8x^{2}.

$$8x^{2}-6x=9$$

Subtract 9 from both sides.

$$8x^{2}-6x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8\left(-9\right)}}{2\times 8}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 8\left(-9\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-6\right)±\sqrt{36-32\left(-9\right)}}{2\times 8}$$

Multiply -32 times -9.

$$x=\frac{-\left(-6\right)±\sqrt{36+288}}{2\times 8}$$

Add 36 to 288.

$$x=\frac{-\left(-6\right)±\sqrt{324}}{2\times 8}$$

Take the square root of 324.

$$x=\frac{-\left(-6\right)±18}{2\times 8}$$

The opposite of -6 is 6.

$$x=\frac{6±18}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{6±18}{16}$$

Now solve the equation x=\frac{6±18}{16} when ± is plus. Add 6 to 18.

$$x=\frac{24}{16}$$

Reduce the fraction \frac{24}{16} to lowest terms by extracting and canceling out 8.

$$x=\frac{3}{2}$$

Now solve the equation x=\frac{6±18}{16} when ± is minus. Subtract 18 from 6.

$$x=-\frac{12}{16}$$

Reduce the fraction \frac{-12}{16} to lowest terms by extracting and canceling out 4.

$$x=-\frac{3}{4}$$

The equation is now solved.

$$x=\frac{3}{2} x=-\frac{3}{4}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,6 since division by zero is not defined. Multiply both sides of the equation by x\left(x-6\right), the least common multiple of x-6,x^{2}-6x.

$$x\left(x-6\right)+x\times 7x=9$$

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

$$x^{2}-6x+x\times 7x=9$$

Multiply x and x to get x^{2}.

$$x^{2}-6x+x^{2}\times 7=9$$

Combine x^{2} and x^{2}\times 7 to get 8x^{2}.

$$8x^{2}-6x=9$$

Divide both sides by 8.

$$\frac{8x^{2}-6x}{8}=\frac{9}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{6}{8}\right)x=\frac{9}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{3}{4}x=\frac{9}{8}$$

Divide -\frac{3}{4}, the coefficient of the x term, by 2 to get -\frac{3}{8}. Then add the square of -\frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{4}x+\left(-\frac{3}{8}\right)^{2}=\frac{9}{8}+\left(-\frac{3}{8}\right)^{2}$$

Square -\frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{9}{8}+\frac{9}{64}$$

Add \frac{9}{8} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{81}{64}$$

Factor x^{2}-\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{8}\right)^{2}=\frac{81}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{8}\right)^{2}}=\sqrt{\frac{81}{64}}$$

Simplify.

$$x-\frac{3}{8}=\frac{9}{8} x-\frac{3}{8}=-\frac{9}{8}$$

Add \frac{3}{8} to both sides of the equation.

$$x=\frac{3}{2} x=-\frac{3}{4}$$