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

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

Solve for x

$$x=-1$$

$$x=5$$

Steps Using the Quadratic Formula

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

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

Multiply x+3 and x+3 to get \left(x+3\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=\left(x-1\right)\left(3x+1\right)$$

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

$$x^{2}+6x+9=3x^{2}-2x-1$$

Subtract 3x^{2} from both sides.

$$x^{2}+6x+9-3x^{2}=-2x-1$$

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

$$-2x^{2}+6x+9=-2x-1$$

Add 2x to both sides.

$$-2x^{2}+6x+9+2x=-1$$

Combine 6x and 2x to get 8x.

$$-2x^{2}+8x+9=-1$$

Add 1 to both sides.

$$-2x^{2}+8x+9+1=0$$

Add 9 and 1 to get 10.

$$-2x^{2}+8x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 8 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-2\right)\times 10}}{2\left(-2\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-2\right)\times 10}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-8±\sqrt{64+8\times 10}}{2\left(-2\right)}$$

Multiply 8 times 10.

$$x=\frac{-8±\sqrt{64+80}}{2\left(-2\right)}$$

Add 64 to 80.

$$x=\frac{-8±\sqrt{144}}{2\left(-2\right)}$$

Take the square root of 144.

$$x=\frac{-8±12}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-8±12}{-4}$$

Now solve the equation x=\frac{-8±12}{-4} when ± is plus. Add -8 to 12.

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

Divide 4 by -4.

$$x=-1$$

Now solve the equation x=\frac{-8±12}{-4} when ± is minus. Subtract 12 from -8.

$$x=-\frac{20}{-4}$$

Divide -20 by -4.

$$x=5$$

The equation is now solved.

$$x=-1 x=5$$

Steps for Completing the Square

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

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

Multiply x+3 and x+3 to get \left(x+3\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=\left(x-1\right)\left(3x+1\right)$$

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

$$x^{2}+6x+9=3x^{2}-2x-1$$

Subtract 3x^{2} from both sides.

$$x^{2}+6x+9-3x^{2}=-2x-1$$

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

$$-2x^{2}+6x+9=-2x-1$$

Add 2x to both sides.

$$-2x^{2}+6x+9+2x=-1$$

Combine 6x and 2x to get 8x.

$$-2x^{2}+8x+9=-1$$

Subtract 9 from both sides.

$$-2x^{2}+8x=-1-9$$

Subtract 9 from -1 to get -10.

$$-2x^{2}+8x=-10$$

Divide both sides by -2.

$$\frac{-2x^{2}+8x}{-2}=-\frac{10}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{8}{-2}x=-\frac{10}{-2}$$

Divide 8 by -2.

$$x^{2}-4x=-\frac{10}{-2}$$

Divide -10 by -2.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Add 5 to 4.

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

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{9}$$

Simplify.

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

Add 2 to both sides of the equation.

$$x=5 x=-1$$