Solve 3(x-1)/x+1-2(x+1)/x-1=5

$$\frac{ 3(x-1) }{ x+1 } - \frac{ 2(x+1) }{ x-1 } =5$$

Solve for x

$$x=-3$$

$$x=\frac{1}{2}=0.5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right), the least common multiple of x+1,x-1.

$$\left(x-1\right)\times 3\left(x-1\right)-\left(x+1\right)\times 2\left(x+1\right)=5\left(x-1\right)\left(x+1\right)$$

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

$$\left(x-1\right)^{2}\times 3-\left(x+1\right)\times 2\left(x+1\right)=5\left(x-1\right)\left(x+1\right)$$

Multiply x+1 and x+1 to get \left(x+1\right)^{2}.

$$\left(x-1\right)^{2}\times 3-\left(x+1\right)^{2}\times 2=5\left(x-1\right)\left(x+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$\left(x-1\right)^{2}\times 3-\left(x^{2}+2x+1\right)\times 2=5\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply x^{2}+2x+1 by 2.

$$\left(x-1\right)^{2}\times 3-\left(2x^{2}+4x+2\right)=5\left(x-1\right)\left(x+1\right)$$

To find the opposite of 2x^{2}+4x+2, find the opposite of each term.

$$\left(x-1\right)^{2}\times 3-2x^{2}-4x-2=5\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 5 by x-1.

$$\left(x-1\right)^{2}\times 3-2x^{2}-4x-2=\left(5x-5\right)\left(x+1\right)$$

Use the distributive property to multiply 5x-5 by x+1 and combine like terms.

$$\left(x-1\right)^{2}\times 3-2x^{2}-4x-2=5x^{2}-5$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$\left(x^{2}-2x+1\right)\times 3-2x^{2}-4x-2=5x^{2}-5$$

Use the distributive property to multiply x^{2}-2x+1 by 3.

$$3x^{2}-6x+3-2x^{2}-4x-2=5x^{2}-5$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}-6x+3-4x-2=5x^{2}-5$$

Combine -6x and -4x to get -10x.

$$x^{2}-10x+3-2=5x^{2}-5$$

Subtract 2 from 3 to get 1.

$$x^{2}-10x+1=5x^{2}-5$$

Subtract 5x^{2} from both sides.

$$x^{2}-10x+1-5x^{2}=-5$$

Combine x^{2} and -5x^{2} to get -4x^{2}.

$$-4x^{2}-10x+1=-5$$

Add 5 to both sides.

$$-4x^{2}-10x+1+5=0$$

Add 1 and 5 to get 6.

$$-4x^{2}-10x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -10 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-4\right)\times 6}}{2\left(-4\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-4\right)\times 6}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-10\right)±\sqrt{100+16\times 6}}{2\left(-4\right)}$$

Multiply 16 times 6.

$$x=\frac{-\left(-10\right)±\sqrt{100+96}}{2\left(-4\right)}$$

Add 100 to 96.

$$x=\frac{-\left(-10\right)±\sqrt{196}}{2\left(-4\right)}$$

Take the square root of 196.

$$x=\frac{-\left(-10\right)±14}{2\left(-4\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±14}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{10±14}{-8}$$

Now solve the equation x=\frac{10±14}{-8} when ± is plus. Add 10 to 14.

$$x=\frac{24}{-8}$$

Divide 24 by -8.

$$x=-3$$

Now solve the equation x=\frac{10±14}{-8} when ± is minus. Subtract 14 from 10.

$$x=-\frac{4}{-8}$$

Reduce the fraction \frac{-4}{-8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right), the least common multiple of x+1,x-1.

$$\left(x-1\right)\times 3\left(x-1\right)-\left(x+1\right)\times 2\left(x+1\right)=5\left(x-1\right)\left(x+1\right)$$

Multiply x-1 and x-1 to get \left(x-1\right)^{2}.

$$\left(x-1\right)^{2}\times 3-\left(x+1\right)\times 2\left(x+1\right)=5\left(x-1\right)\left(x+1\right)$$

Multiply x+1 and x+1 to get \left(x+1\right)^{2}.

$$\left(x-1\right)^{2}\times 3-\left(x+1\right)^{2}\times 2=5\left(x-1\right)\left(x+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$\left(x-1\right)^{2}\times 3-\left(x^{2}+2x+1\right)\times 2=5\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply x^{2}+2x+1 by 2.

$$\left(x-1\right)^{2}\times 3-\left(2x^{2}+4x+2\right)=5\left(x-1\right)\left(x+1\right)$$

To find the opposite of 2x^{2}+4x+2, find the opposite of each term.

$$\left(x-1\right)^{2}\times 3-2x^{2}-4x-2=5\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 5 by x-1.

$$\left(x-1\right)^{2}\times 3-2x^{2}-4x-2=\left(5x-5\right)\left(x+1\right)$$

Use the distributive property to multiply 5x-5 by x+1 and combine like terms.

$$\left(x-1\right)^{2}\times 3-2x^{2}-4x-2=5x^{2}-5$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$\left(x^{2}-2x+1\right)\times 3-2x^{2}-4x-2=5x^{2}-5$$

Use the distributive property to multiply x^{2}-2x+1 by 3.

$$3x^{2}-6x+3-2x^{2}-4x-2=5x^{2}-5$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}-6x+3-4x-2=5x^{2}-5$$

Combine -6x and -4x to get -10x.

$$x^{2}-10x+3-2=5x^{2}-5$$

Subtract 2 from 3 to get 1.

$$x^{2}-10x+1=5x^{2}-5$$

Subtract 5x^{2} from both sides.

$$x^{2}-10x+1-5x^{2}=-5$$

Combine x^{2} and -5x^{2} to get -4x^{2}.

$$-4x^{2}-10x+1=-5$$

Subtract 1 from both sides.

$$-4x^{2}-10x=-5-1$$

Subtract 1 from -5 to get -6.

$$-4x^{2}-10x=-6$$

Divide both sides by -4.

$$\frac{-4x^{2}-10x}{-4}=-\frac{6}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{10}{-4}\right)x=-\frac{6}{-4}$$

Reduce the fraction \frac{-10}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{5}{2}x=-\frac{6}{-4}$$

Reduce the fraction \frac{-6}{-4} to lowest terms by extracting and canceling out 2.

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

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=\frac{3}{2}+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{3}{2}+\frac{25}{16}$$

Add \frac{3}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{49}{16}$$

Factor x^{2}+\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{7}{4} x+\frac{5}{4}=-\frac{7}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

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