Solve 3x-1/x+1+x+1/2x-1=3

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

Solve for x

$$x=1$$

$$x=5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-1\right)\left(x+1\right), the least common multiple of x+1,2x-1.

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

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

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

Use the distributive property to multiply 2x-1 by 3x-1 and combine like terms.

$$6x^{2}-5x+1+\left(x+1\right)^{2}=3\left(2x-1\right)\left(x+1\right)$$

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

$$6x^{2}-5x+1+x^{2}+2x+1=3\left(2x-1\right)\left(x+1\right)$$

Combine 6x^{2} and x^{2} to get 7x^{2}.

$$7x^{2}-5x+1+2x+1=3\left(2x-1\right)\left(x+1\right)$$

Combine -5x and 2x to get -3x.

$$7x^{2}-3x+1+1=3\left(2x-1\right)\left(x+1\right)$$

Add 1 and 1 to get 2.

$$7x^{2}-3x+2=3\left(2x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 3 by 2x-1.

$$7x^{2}-3x+2=\left(6x-3\right)\left(x+1\right)$$

Use the distributive property to multiply 6x-3 by x+1 and combine like terms.

$$7x^{2}-3x+2=6x^{2}+3x-3$$

Subtract 6x^{2} from both sides.

$$7x^{2}-3x+2-6x^{2}=3x-3$$

Combine 7x^{2} and -6x^{2} to get x^{2}.

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

Subtract 3x from both sides.

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

Combine -3x and -3x to get -6x.

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

Add 3 to both sides.

$$x^{2}-6x+2+3=0$$

Add 2 and 3 to get 5.

$$x^{2}-6x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 5}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-\left(-6\right)±\sqrt{36-20}}{2}$$

Add 36 to -20.

$$x=\frac{-\left(-6\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-\left(-6\right)±4}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±4}{2}$$

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

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

Divide 10 by 2.

$$x=5$$

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

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

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=5 x=1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-1\right)\left(x+1\right), the least common multiple of x+1,2x-1.

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

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

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

Use the distributive property to multiply 2x-1 by 3x-1 and combine like terms.

$$6x^{2}-5x+1+\left(x+1\right)^{2}=3\left(2x-1\right)\left(x+1\right)$$

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

$$6x^{2}-5x+1+x^{2}+2x+1=3\left(2x-1\right)\left(x+1\right)$$

Combine 6x^{2} and x^{2} to get 7x^{2}.

$$7x^{2}-5x+1+2x+1=3\left(2x-1\right)\left(x+1\right)$$

Combine -5x and 2x to get -3x.

$$7x^{2}-3x+1+1=3\left(2x-1\right)\left(x+1\right)$$

Add 1 and 1 to get 2.

$$7x^{2}-3x+2=3\left(2x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 3 by 2x-1.

$$7x^{2}-3x+2=\left(6x-3\right)\left(x+1\right)$$

Use the distributive property to multiply 6x-3 by x+1 and combine like terms.

$$7x^{2}-3x+2=6x^{2}+3x-3$$

Subtract 6x^{2} from both sides.

$$7x^{2}-3x+2-6x^{2}=3x-3$$

Combine 7x^{2} and -6x^{2} to get x^{2}.

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

Subtract 3x from both sides.

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

Combine -3x and -3x to get -6x.

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

Subtract 2 from both sides.

$$x^{2}-6x=-3-2$$

Subtract 2 from -3 to get -5.

$$x^{2}-6x=-5$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-5+9$$

Add -5 to 9.

$$x^{2}-6x+9=4$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-3=2 x-3=-2$$

Add 3 to both sides of the equation.

$$x=5 x=1$$