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

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

Solve for x

$$x=-1$$

Steps Using the Quadratic Formula

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

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

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

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

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

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

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

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

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

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

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

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

Subtract 3x from both sides.

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

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

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

Subtract 6 from both sides.

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

Subtract 6 from -3 to get -9.

$$3x^{2}-6x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 3\left(-9\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-6\right)±\sqrt{36-12\left(-9\right)}}{2\times 3}$$

Multiply -12 times -9.

$$x=\frac{-\left(-6\right)±\sqrt{36+108}}{2\times 3}$$

Add 36 to 108.

$$x=\frac{-\left(-6\right)±\sqrt{144}}{2\times 3}$$

Take the square root of 144.

$$x=\frac{-\left(-6\right)±12}{2\times 3}$$

The opposite of -6 is 6.

$$x=\frac{6±12}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{6±12}{6}$$

Now solve the equation x=\frac{6±12}{6} when ± is plus. Add 6 to 12.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{6±12}{6} when ± is minus. Subtract 12 from 6.

$$x=-\frac{6}{6}$$

Divide -6 by 6.

$$x=-1$$

The equation is now solved.

$$x=3 x=-1$$

Variable x cannot be equal to 3.

$$x=-1$$

Steps for Completing the Square

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

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

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

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

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

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

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

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

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

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

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

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

Subtract 3x from both sides.

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

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

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

Add 3 to both sides.

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

Add 6 and 3 to get 9.

$$3x^{2}-6x=9$$

Divide both sides by 3.

$$\frac{3x^{2}-6x}{3}=\frac{9}{3}$$

Dividing by 3 undoes the multiplication by 3.

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

Divide -6 by 3.

$$x^{2}-2x=\frac{9}{3}$$

Divide 9 by 3.

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

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.

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

Add 3 to 1.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=2 x-1=-2$$

Add 1 to both sides of the equation.

$$x=3 x=-1$$

Variable x cannot be equal to 3.

$$x=-1$$