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

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

Solve for x

$$x=\sqrt{5}+3\approx 5.236067977$$

$$x=3-\sqrt{5}\approx 0.763932023$$

Steps Using the Quadratic Formula

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

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

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

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

Consider \left(x-2\right)\left(x+2\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 2.

$$3x^{2}-5x-2=x^{2}-4+\left(x+2\right)\left(x-1\right)$$

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

$$3x^{2}-5x-2=x^{2}-4+x^{2}+x-2$$

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

$$3x^{2}-5x-2=2x^{2}-4+x-2$$

Subtract 2 from -4 to get -6.

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

Subtract 2x^{2} from both sides.

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

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

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

Subtract -6 from both sides.

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

The opposite of -6 is 6.

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

Subtract x from both sides.

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

Add -2 and 6 to get 4.

$$x^{2}-5x+4-x=0$$

Combine -5x and -x to get -6x.

$$x^{2}-6x+4=0$$

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

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

Square -6.

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

Multiply -4 times 4.

$$x=\frac{-\left(-6\right)±\sqrt{36-16}}{2}$$

Add 36 to -16.

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

Take the square root of 20.

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

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{6±2\sqrt{5}}{2} when ± is plus. Add 6 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}+6}{2}$$

Divide 6+2\sqrt{5} by 2.

$$x=\sqrt{5}+3$$

Now solve the equation x=\frac{6±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from 6.

$$x=\frac{6-2\sqrt{5}}{2}$$

Divide 6-2\sqrt{5} by 2.

$$x=3-\sqrt{5}$$

The equation is now solved.

$$x=\sqrt{5}+3 x=3-\sqrt{5}$$

Steps for Completing the Square

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

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

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

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

Consider \left(x-2\right)\left(x+2\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 2.

$$3x^{2}-5x-2=x^{2}-4+\left(x+2\right)\left(x-1\right)$$

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

$$3x^{2}-5x-2=x^{2}-4+x^{2}+x-2$$

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

$$3x^{2}-5x-2=2x^{2}-4+x-2$$

Subtract 2 from -4 to get -6.

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

Subtract 2x^{2} from both sides.

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

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

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

Subtract x from both sides.

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

Combine -5x and -x to get -6x.

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

Add 2 to both sides.

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

Add -6 and 2 to get -4.

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

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.

$$x^{2}-6x+\left(-3\right)^{2}=-4+\left(-3\right)^{2}$$

Square -3.

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

Add -4 to 9.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-3=\sqrt{5} x-3=-\sqrt{5}$$

Add 3 to both sides of the equation.

$$x=\sqrt{5}+3 x=3-\sqrt{5}$$