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

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

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

Steps Using Factoring By Grouping

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

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

Use the distributive property to multiply x+1 by -2.

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

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

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

Combine x and -2x to get -x.

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

Consider \left(x-1\right)\left(x+1\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 1.

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

Subtract x^{2} from both sides.

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

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

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

Add 1 to both sides.

$$-x+3-3x^{2}+1=0$$

Add 3 and 1 to get 4.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-3\times 4=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=3 b=-4$$

Rewrite -3x^{2}-x+4 as \left(-3x^{2}+3x\right)+\left(-4x+4\right).

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

Factor out 3x in the first and 4 in the second group.

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

Factor out common term -x+1 by using distributive property.

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

To find equation solutions, solve -x+1=0 and 3x+4=0.

$$x=1 x=-\frac{4}{3}$$

Variable x cannot be equal to 1.

$$x=-\frac{4}{3}$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply x+1 by -2.

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

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

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

Combine x and -2x to get -x.

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

Consider \left(x-1\right)\left(x+1\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 1.

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

Subtract x^{2} from both sides.

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

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

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

Add 1 to both sides.

$$-x+3-3x^{2}+1=0$$

Add 3 and 1 to get 4.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -1 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-3\right)\times 4}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+12\times 4}}{2\left(-3\right)}$$

Multiply 12 times 4.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2\left(-3\right)}$$

Add 1 to 48.

$$x=\frac{-\left(-1\right)±\sqrt{49}}{2\left(-3\right)}$$

Take the square root of 49.

$$x=\frac{-\left(-1\right)±7}{2\left(-3\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±7}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{1±7}{-6}$$

Now solve the equation x=\frac{1±7}{-6} when ± is plus. Add 1 to 7.

$$x=\frac{8}{-6}$$

Reduce the fraction \frac{8}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{4}{3}$$

Now solve the equation x=\frac{1±7}{-6} when ± is minus. Subtract 7 from 1.

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

Divide -6 by -6.

$$x=1$$

The equation is now solved.

$$x=-\frac{4}{3} x=1$$

Variable x cannot be equal to 1.

$$x=-\frac{4}{3}$$

Steps for Completing the Square

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

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

Use the distributive property to multiply x+1 by -2.

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

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

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

Combine x and -2x to get -x.

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

Consider \left(x-1\right)\left(x+1\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 1.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 3 from both sides.

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

Subtract 3 from -1 to get -4.

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

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.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide -1 by -3.

$$x^{2}+\frac{1}{3}x=-\frac{4}{-3}$$

Divide -4 by -3.

$$x^{2}+\frac{1}{3}x=\frac{4}{3}$$

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=\frac{4}{3}+\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{4}{3}+\frac{1}{36}$$

Add \frac{4}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{49}{36}$$

Factor x^{2}+\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{7}{6} x+\frac{1}{6}=-\frac{7}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$x=1 x=-\frac{4}{3}$$

Variable x cannot be equal to 1.

$$x=-\frac{4}{3}$$