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

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

Solve for x

$$x=-2$$

$$x=0$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{3}{2},-1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(2x+3\right), the least common multiple of x+1,2x+3.

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

Use the distributive property to multiply 2x+3 by 2.

$$4x+6-\left(x+1\right)\times 3=\left(x+1\right)\left(2x+3\right)$$

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

$$4x+6-\left(3x+3\right)=\left(x+1\right)\left(2x+3\right)$$

To find the opposite of 3x+3, find the opposite of each term.

$$4x+6-3x-3=\left(x+1\right)\left(2x+3\right)$$

Combine 4x and -3x to get x.

$$x+6-3=\left(x+1\right)\left(2x+3\right)$$

Subtract 3 from 6 to get 3.

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

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

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

Subtract 2x^{2} from both sides.

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

Subtract 5x from both sides.

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

Combine x and -5x to get -4x.

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

Subtract 3 from both sides.

$$-4x+3-2x^{2}-3=0$$

Subtract 3 from 3 to get 0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Take the square root of \left(-4\right)^{2}.

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

The opposite of -4 is 4.

$$x=\frac{4±4}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{4±4}{-4}$$

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

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

Divide 8 by -4.

$$x=-2$$

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

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

The equation is now solved.

$$x=-2 x=0$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{3}{2},-1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(2x+3\right), the least common multiple of x+1,2x+3.

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

Use the distributive property to multiply 2x+3 by 2.

$$4x+6-\left(x+1\right)\times 3=\left(x+1\right)\left(2x+3\right)$$

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

$$4x+6-\left(3x+3\right)=\left(x+1\right)\left(2x+3\right)$$

To find the opposite of 3x+3, find the opposite of each term.

$$4x+6-3x-3=\left(x+1\right)\left(2x+3\right)$$

Combine 4x and -3x to get x.

$$x+6-3=\left(x+1\right)\left(2x+3\right)$$

Subtract 3 from 6 to get 3.

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

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

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

Subtract 2x^{2} from both sides.

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

Subtract 5x from both sides.

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

Combine x and -5x to get -4x.

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

Subtract 3 from both sides.

$$-4x-2x^{2}=3-3$$

Subtract 3 from 3 to get 0.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -2.

$$\frac{-2x^{2}-4x}{-2}=\frac{0}{-2}$$

Dividing by -2 undoes the multiplication by -2.

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

Divide -4 by -2.

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

Divide 0 by -2.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square 1.

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

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=1 x+1=-1$$

Subtract 1 from both sides of the equation.

$$x=0 x=-2$$