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

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

Solve for x

$$x=-5$$

$$x=-1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by x+3.

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

Use the distributive property to multiply x+3 by \left(-x+1\right)^{-1}.

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

Use the distributive property to multiply x\left(-x+1\right)^{-1}+3\left(-x+1\right)^{-1} by x+2 and combine like terms.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 1 from both sides.

$$\left(-x+1\right)^{-1}x^{2}+5x\left(-x+1\right)^{-1}+6\left(-x+1\right)^{-1}-1=0$$

Reorder the terms.

$$\frac{1}{-x+1}x^{2}+5\times \frac{1}{-x+1}x-1+6\times \frac{1}{-x+1}=0$$

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

$$1x^{2}+5\times 1x+\left(-x+1\right)\left(-1\right)+6\times 1=0$$

Do the multiplications.

$$1x^{2}+5x+\left(-x+1\right)\left(-1\right)+6=0$$

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

$$1x^{2}+5x+x-1+6=0$$

Combine 5x and x to get 6x.

$$1x^{2}+6x-1+6=0$$

Add -1 and 6 to get 5.

$$1x^{2}+6x+5=0$$

Reorder the terms.

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

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}.

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

Square 6.

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

Multiply -4 times 5.

$$x=\frac{-6±\sqrt{36-20}}{2}$$

Add 36 to -20.

$$x=\frac{-6±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-6±4}{2}$$

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

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

Divide -2 by 2.

$$x=-1$$

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

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

Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=-1 x=-5$$

Steps for Completing the Square

Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by x+3.

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

Use the distributive property to multiply x+3 by \left(-x+1\right)^{-1}.

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

Use the distributive property to multiply x\left(-x+1\right)^{-1}+3\left(-x+1\right)^{-1} by x+2 and combine like terms.

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

Swap sides so that all variable terms are on the left hand side.

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

Reorder the terms.

$$\frac{1}{-x+1}x^{2}+5\times \frac{1}{-x+1}x+6\times \frac{1}{-x+1}=1$$

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -x+1.

$$1x^{2}+5\times 1x+6\times 1=-x+1$$

Do the multiplications.

$$1x^{2}+5x+6=-x+1$$

Add x to both sides.

$$1x^{2}+5x+6+x=1$$

Combine 5x and x to get 6x.

$$1x^{2}+6x+6=1$$

Subtract 6 from both sides.

$$1x^{2}+6x=1-6$$

Subtract 6 from 1 to get -5.

$$1x^{2}+6x=-5$$

Reorder the terms.

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

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.

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

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.

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

Subtract 3 from both sides of the equation.

$$x=-1 x=-5$$