Solve (2/x+1-2x-3/x-1)div1/x+1+2x=3

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

Solve for x

$$x=2$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+1 and x-1 is \left(x-1\right)\left(x+1\right). Multiply \frac{2}{x+1} times \frac{x-1}{x-1}. Multiply \frac{2x-3}{x-1} times \frac{x+1}{x+1}.

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

Since \frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)} and \frac{\left(2x-3\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2\left(x-1\right)-\left(2x-3\right)\left(x+1\right).

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

Combine like terms in 2x-2-2x^{2}-2x+3x+3.

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

Variable x cannot be equal to -1 since division by zero is not defined. Divide \frac{3x+1-2x^{2}}{\left(x-1\right)\left(x+1\right)} by \frac{1}{x+1} by multiplying \frac{3x+1-2x^{2}}{\left(x-1\right)\left(x+1\right)} by the reciprocal of \frac{1}{x+1}.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x times \frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}.

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

Since \frac{\left(3x+1-2x^{2}\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)} and \frac{2x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(3x+1-2x^{2}\right)\left(x+1\right)+2x\left(x-1\right)\left(x+1\right).

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

Combine like terms in 3x^{2}+3x+x+1-2x^{3}-2x^{2}+2x^{3}+2x^{2}-2x^{2}-2x.

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

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Subtract 3 from both sides.

$$\frac{x^{2}+2x+1}{x^{2}-1}-3=0$$

Factor x^{2}-1.

$$\frac{x^{2}+2x+1}{\left(x-1\right)\left(x+1\right)}-3=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}.

$$\frac{x^{2}+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{3\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=0$$

Since \frac{x^{2}+2x+1}{\left(x-1\right)\left(x+1\right)} and \frac{3\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}+2x+1-3\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=0$$

Do the multiplications in x^{2}+2x+1-3\left(x-1\right)\left(x+1\right).

$$\frac{x^{2}+2x+1-3x^{2}-3x+3x+3}{\left(x-1\right)\left(x+1\right)}=0$$

Combine like terms in x^{2}+2x+1-3x^{2}-3x+3x+3.

$$\frac{-2x^{2}+2x+4}{\left(x-1\right)\left(x+1\right)}=0$$

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).

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

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

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

Square 2.

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

Multiply -4 times -2.

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

Multiply 8 times 4.

$$x=\frac{-2±\sqrt{4+32}}{2\left(-2\right)}$$

Add 4 to 32.

$$x=\frac{-2±\sqrt{36}}{2\left(-2\right)}$$

Take the square root of 36.

$$x=\frac{-2±6}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-2±6}{-4}$$

Now solve the equation x=\frac{-2±6}{-4} when ± is plus. Add -2 to 6.

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

Divide 4 by -4.

$$x=-1$$

Now solve the equation x=\frac{-2±6}{-4} when ± is minus. Subtract 6 from -2.

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

Divide -8 by -4.

$$x=2$$

The equation is now solved.

$$x=-1 x=2$$

Variable x cannot be equal to -1.

$$x=2$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+1 and x-1 is \left(x-1\right)\left(x+1\right). Multiply \frac{2}{x+1} times \frac{x-1}{x-1}. Multiply \frac{2x-3}{x-1} times \frac{x+1}{x+1}.

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

Since \frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)} and \frac{\left(2x-3\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2\left(x-1\right)-\left(2x-3\right)\left(x+1\right).

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

Combine like terms in 2x-2-2x^{2}-2x+3x+3.

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

Variable x cannot be equal to -1 since division by zero is not defined. Divide \frac{3x+1-2x^{2}}{\left(x-1\right)\left(x+1\right)} by \frac{1}{x+1} by multiplying \frac{3x+1-2x^{2}}{\left(x-1\right)\left(x+1\right)} by the reciprocal of \frac{1}{x+1}.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x times \frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}.

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

Since \frac{\left(3x+1-2x^{2}\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)} and \frac{2x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(3x+1-2x^{2}\right)\left(x+1\right)+2x\left(x-1\right)\left(x+1\right).

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

Combine like terms in 3x^{2}+3x+x+1-2x^{3}-2x^{2}+2x^{3}+2x^{2}-2x^{2}-2x.

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

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

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).

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

Use the distributive property to multiply 3 by x-1.

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

Use the distributive property to multiply 3x-3 by x+1 and combine like terms.

$$x^{2}+2x+1=3x^{2}-3$$

Subtract 3x^{2} from both sides.

$$x^{2}+2x+1-3x^{2}=-3$$

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

$$-2x^{2}+2x+1=-3$$

Subtract 1 from both sides.

$$-2x^{2}+2x=-3-1$$

Subtract 1 from -3 to get -4.

$$-2x^{2}+2x=-4$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 2 by -2.

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

Divide -4 by -2.

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

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.

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

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

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

Add 2 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{9}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

$$x=2 x=-1$$

Variable x cannot be equal to -1.

$$x=2$$