Solve 9x-4/x-2=(3+2/x-2)(3-2/x-2)

$$\frac { 9 x - 4 } { x - 2 } = ( 3 + \frac { 2 } { x - 2 } ) ( 3 - \frac { 2 } { x - 2 } )$$

Solve for x

$$x = \frac{12}{7} = 1\frac{5}{7} \approx 1.714285714$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x-2}{x-2}.

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

Since \frac{3\left(x-2\right)}{x-2} and \frac{2}{x-2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 3\left(x-2\right)+2.

$$9x-4=\frac{3x-6+2}{x-2}\left(3-\frac{2}{x-2}\right)\left(x-2\right)$$

Combine like terms in 3x-6+2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x-2}{x-2}.

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

Since \frac{3\left(x-2\right)}{x-2} and \frac{2}{x-2} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3\left(x-2\right)-2.

$$9x-4=\frac{3x-4}{x-2}\times \frac{3x-6-2}{x-2}\left(x-2\right)$$

Combine like terms in 3x-6-2.

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

Multiply \frac{3x-4}{x-2} times \frac{3x-8}{x-2} by multiplying numerator times numerator and denominator times denominator.

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

Express \frac{\left(3x-4\right)\left(3x-8\right)}{\left(x-2\right)\left(x-2\right)}\left(x-2\right) as a single fraction.

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

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

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

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

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

Subtract \frac{\left(3x-4\right)\left(3x-8\right)\left(x-2\right)}{x^{2}-4x+4} from both sides.

$$9x-4-\frac{\left(3x-4\right)\left(3x-8\right)\left(x-2\right)}{x^{2}-4x+4}=0$$

Use the distributive property to multiply 3x-4 by 3x-8 and combine like terms.

$$9x-4-\frac{\left(9x^{2}-36x+32\right)\left(x-2\right)}{x^{2}-4x+4}=0$$

Use the distributive property to multiply 9x^{2}-36x+32 by x-2 and combine like terms.

$$9x-4-\frac{9x^{3}-54x^{2}+104x-64}{x^{2}-4x+4}=0$$

Factor x^{2}-4x+4.

$$9x-4-\frac{9x^{3}-54x^{2}+104x-64}{\left(x-2\right)^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9x-4 times \frac{\left(x-2\right)^{2}}{\left(x-2\right)^{2}}.

$$\frac{\left(9x-4\right)\left(x-2\right)^{2}}{\left(x-2\right)^{2}}-\frac{9x^{3}-54x^{2}+104x-64}{\left(x-2\right)^{2}}=0$$

Since \frac{\left(9x-4\right)\left(x-2\right)^{2}}{\left(x-2\right)^{2}} and \frac{9x^{3}-54x^{2}+104x-64}{\left(x-2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(9x-4\right)\left(x-2\right)^{2}-\left(9x^{3}-54x^{2}+104x-64\right)}{\left(x-2\right)^{2}}=0$$

Do the multiplications in \left(9x-4\right)\left(x-2\right)^{2}-\left(9x^{3}-54x^{2}+104x-64\right).

$$\frac{9x^{3}-36x^{2}+36x-4x^{2}+16x-16-9x^{3}+54x^{2}-104x+64}{\left(x-2\right)^{2}}=0$$

Combine like terms in 9x^{3}-36x^{2}+36x-4x^{2}+16x-16-9x^{3}+54x^{2}-104x+64.

$$\frac{14x^{2}-52x+48}{\left(x-2\right)^{2}}=0$$

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)^{2}.

$$14x^{2}-52x+48=0$$

Divide both sides by 2.

$$7x^{2}-26x+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=-26 ab=7\times 24=168$$

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 168.

$$-1,-168 -2,-84 -3,-56 -4,-42 -6,-28 -7,-24 -8,-21 -12,-14$$

Calculate the sum for each pair.

$$-1-168=-169 -2-84=-86 -3-56=-59 -4-42=-46 -6-28=-34 -7-24=-31 -8-21=-29 -12-14=-26$$

The solution is the pair that gives sum -26.

$$a=-14 b=-12$$

Rewrite 7x^{2}-26x+24 as \left(7x^{2}-14x\right)+\left(-12x+24\right).

$$\left(7x^{2}-14x\right)+\left(-12x+24\right)$$

Factor out 7x in the first and -12 in the second group.

$$7x\left(x-2\right)-12\left(x-2\right)$$

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

$$\left(x-2\right)\left(7x-12\right)$$

To find equation solutions, solve x-2=0 and 7x-12=0.

$$x=2 x=\frac{12}{7}$$

Variable x cannot be equal to 2.

$$x=\frac{12}{7}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x-2}{x-2}.

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

Since \frac{3\left(x-2\right)}{x-2} and \frac{2}{x-2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 3\left(x-2\right)+2.

$$9x-4=\frac{3x-6+2}{x-2}\left(3-\frac{2}{x-2}\right)\left(x-2\right)$$

Combine like terms in 3x-6+2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x-2}{x-2}.

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

Since \frac{3\left(x-2\right)}{x-2} and \frac{2}{x-2} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3\left(x-2\right)-2.

$$9x-4=\frac{3x-4}{x-2}\times \frac{3x-6-2}{x-2}\left(x-2\right)$$

Combine like terms in 3x-6-2.

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

Multiply \frac{3x-4}{x-2} times \frac{3x-8}{x-2} by multiplying numerator times numerator and denominator times denominator.

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

Express \frac{\left(3x-4\right)\left(3x-8\right)}{\left(x-2\right)\left(x-2\right)}\left(x-2\right) as a single fraction.

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

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

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

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

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

Subtract \frac{\left(3x-4\right)\left(3x-8\right)\left(x-2\right)}{x^{2}-4x+4} from both sides.

$$9x-4-\frac{\left(3x-4\right)\left(3x-8\right)\left(x-2\right)}{x^{2}-4x+4}=0$$

Use the distributive property to multiply 3x-4 by 3x-8 and combine like terms.

$$9x-4-\frac{\left(9x^{2}-36x+32\right)\left(x-2\right)}{x^{2}-4x+4}=0$$

Use the distributive property to multiply 9x^{2}-36x+32 by x-2 and combine like terms.

$$9x-4-\frac{9x^{3}-54x^{2}+104x-64}{x^{2}-4x+4}=0$$

Factor x^{2}-4x+4.

$$9x-4-\frac{9x^{3}-54x^{2}+104x-64}{\left(x-2\right)^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9x-4 times \frac{\left(x-2\right)^{2}}{\left(x-2\right)^{2}}.

$$\frac{\left(9x-4\right)\left(x-2\right)^{2}}{\left(x-2\right)^{2}}-\frac{9x^{3}-54x^{2}+104x-64}{\left(x-2\right)^{2}}=0$$

Since \frac{\left(9x-4\right)\left(x-2\right)^{2}}{\left(x-2\right)^{2}} and \frac{9x^{3}-54x^{2}+104x-64}{\left(x-2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(9x-4\right)\left(x-2\right)^{2}-\left(9x^{3}-54x^{2}+104x-64\right)}{\left(x-2\right)^{2}}=0$$

Do the multiplications in \left(9x-4\right)\left(x-2\right)^{2}-\left(9x^{3}-54x^{2}+104x-64\right).

$$\frac{9x^{3}-36x^{2}+36x-4x^{2}+16x-16-9x^{3}+54x^{2}-104x+64}{\left(x-2\right)^{2}}=0$$

Combine like terms in 9x^{3}-36x^{2}+36x-4x^{2}+16x-16-9x^{3}+54x^{2}-104x+64.

$$\frac{14x^{2}-52x+48}{\left(x-2\right)^{2}}=0$$

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)^{2}.

$$14x^{2}-52x+48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 14 for a, -52 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-52\right)±\sqrt{\left(-52\right)^{2}-4\times 14\times 48}}{2\times 14}$$

Square -52.

$$x=\frac{-\left(-52\right)±\sqrt{2704-4\times 14\times 48}}{2\times 14}$$

Multiply -4 times 14.

$$x=\frac{-\left(-52\right)±\sqrt{2704-56\times 48}}{2\times 14}$$

Multiply -56 times 48.

$$x=\frac{-\left(-52\right)±\sqrt{2704-2688}}{2\times 14}$$

Add 2704 to -2688.

$$x=\frac{-\left(-52\right)±\sqrt{16}}{2\times 14}$$

Take the square root of 16.

$$x=\frac{-\left(-52\right)±4}{2\times 14}$$

The opposite of -52 is 52.

$$x=\frac{52±4}{2\times 14}$$

Multiply 2 times 14.

$$x=\frac{52±4}{28}$$

Now solve the equation x=\frac{52±4}{28} when ± is plus. Add 52 to 4.

$$x=\frac{56}{28}$$

Divide 56 by 28.

$$x=2$$

Now solve the equation x=\frac{52±4}{28} when ± is minus. Subtract 4 from 52.

$$x=\frac{48}{28}$$

Reduce the fraction \frac{48}{28} to lowest terms by extracting and canceling out 4.

$$x=\frac{12}{7}$$

The equation is now solved.

$$x=2 x=\frac{12}{7}$$

Variable x cannot be equal to 2.

$$x=\frac{12}{7}$$

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by x-2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x-2}{x-2}.

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

Since \frac{3\left(x-2\right)}{x-2} and \frac{2}{x-2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 3\left(x-2\right)+2.

$$9x-4=\frac{3x-6+2}{x-2}\left(3-\frac{2}{x-2}\right)\left(x-2\right)$$

Combine like terms in 3x-6+2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x-2}{x-2}.

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

Since \frac{3\left(x-2\right)}{x-2} and \frac{2}{x-2} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3\left(x-2\right)-2.

$$9x-4=\frac{3x-4}{x-2}\times \frac{3x-6-2}{x-2}\left(x-2\right)$$

Combine like terms in 3x-6-2.

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

Multiply \frac{3x-4}{x-2} times \frac{3x-8}{x-2} by multiplying numerator times numerator and denominator times denominator.

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

Express \frac{\left(3x-4\right)\left(3x-8\right)}{\left(x-2\right)\left(x-2\right)}\left(x-2\right) as a single fraction.

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

Multiply x-2 and x-2 to get \left(x-2\right)^{2}.

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

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

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

Subtract \frac{\left(3x-4\right)\left(3x-8\right)\left(x-2\right)}{x^{2}-4x+4} from both sides.

$$9x-4-\frac{\left(3x-4\right)\left(3x-8\right)\left(x-2\right)}{x^{2}-4x+4}=0$$

Use the distributive property to multiply 3x-4 by 3x-8 and combine like terms.

$$9x-4-\frac{\left(9x^{2}-36x+32\right)\left(x-2\right)}{x^{2}-4x+4}=0$$

Use the distributive property to multiply 9x^{2}-36x+32 by x-2 and combine like terms.

$$9x-4-\frac{9x^{3}-54x^{2}+104x-64}{x^{2}-4x+4}=0$$

Factor x^{2}-4x+4.

$$9x-4-\frac{9x^{3}-54x^{2}+104x-64}{\left(x-2\right)^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 9x-4 times \frac{\left(x-2\right)^{2}}{\left(x-2\right)^{2}}.

$$\frac{\left(9x-4\right)\left(x-2\right)^{2}}{\left(x-2\right)^{2}}-\frac{9x^{3}-54x^{2}+104x-64}{\left(x-2\right)^{2}}=0$$

Since \frac{\left(9x-4\right)\left(x-2\right)^{2}}{\left(x-2\right)^{2}} and \frac{9x^{3}-54x^{2}+104x-64}{\left(x-2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(9x-4\right)\left(x-2\right)^{2}-\left(9x^{3}-54x^{2}+104x-64\right)}{\left(x-2\right)^{2}}=0$$

Do the multiplications in \left(9x-4\right)\left(x-2\right)^{2}-\left(9x^{3}-54x^{2}+104x-64\right).

$$\frac{9x^{3}-36x^{2}+36x-4x^{2}+16x-16-9x^{3}+54x^{2}-104x+64}{\left(x-2\right)^{2}}=0$$

Combine like terms in 9x^{3}-36x^{2}+36x-4x^{2}+16x-16-9x^{3}+54x^{2}-104x+64.

$$\frac{14x^{2}-52x+48}{\left(x-2\right)^{2}}=0$$

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)^{2}.

$$14x^{2}-52x+48=0$$

Subtract 48 from both sides. Anything subtracted from zero gives its negation.

$$14x^{2}-52x=-48$$

Divide both sides by 14.

$$\frac{14x^{2}-52x}{14}=-\frac{48}{14}$$

Dividing by 14 undoes the multiplication by 14.

$$x^{2}+\left(-\frac{52}{14}\right)x=-\frac{48}{14}$$

Reduce the fraction \frac{-52}{14} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{26}{7}x=-\frac{48}{14}$$

Reduce the fraction \frac{-48}{14} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{26}{7}x=-\frac{24}{7}$$

Divide -\frac{26}{7}, the coefficient of the x term, by 2 to get -\frac{13}{7}. Then add the square of -\frac{13}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{26}{7}x+\left(-\frac{13}{7}\right)^{2}=-\frac{24}{7}+\left(-\frac{13}{7}\right)^{2}$$

Square -\frac{13}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{26}{7}x+\frac{169}{49}=-\frac{24}{7}+\frac{169}{49}$$

Add -\frac{24}{7} to \frac{169}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{26}{7}x+\frac{169}{49}=\frac{1}{49}$$

Factor x^{2}-\frac{26}{7}x+\frac{169}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{13}{7}\right)^{2}=\frac{1}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{7}\right)^{2}}=\sqrt{\frac{1}{49}}$$

Simplify.

$$x-\frac{13}{7}=\frac{1}{7} x-\frac{13}{7}=-\frac{1}{7}$$

Add \frac{13}{7} to both sides of the equation.

$$x=2 x=\frac{12}{7}$$

Variable x cannot be equal to 2.

$$x=\frac{12}{7}$$