Solve n-4/n^2+2n-3-2n/n^2+2n-3=n/n-1

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

Solve for n

$$n=-2$$

Steps Using Factoring By Grouping

Variable n 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(n-1\right)\left(n+3\right), the least common multiple of n^{2}+2n-3,n-1.

$$n-4-2n=\left(n+3\right)n$$

Combine n and -2n to get -n.

$$-n-4=\left(n+3\right)n$$

Use the distributive property to multiply n+3 by n.

$$-n-4=n^{2}+3n$$

Subtract n^{2} from both sides.

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

Subtract 3n from both sides.

$$-n-4-n^{2}-3n=0$$

Combine -n and -3n to get -4n.

$$-4n-4-n^{2}=0$$

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

$$-n^{2}-4n-4=0$$

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

$$a+b=-4 ab=-\left(-4\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -4.

$$a=-2 b=-2$$

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

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

Factor out n in the first and 2 in the second group.

$$n\left(-n-2\right)+2\left(-n-2\right)$$

Factor out common term -n-2 by using distributive property.

$$\left(-n-2\right)\left(n+2\right)$$

To find equation solutions, solve -n-2=0 and n+2=0.

$$n=-2 n=-2$$

Steps Using the Quadratic Formula

Variable n 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(n-1\right)\left(n+3\right), the least common multiple of n^{2}+2n-3,n-1.

$$n-4-2n=\left(n+3\right)n$$

Combine n and -2n to get -n.

$$-n-4=\left(n+3\right)n$$

Use the distributive property to multiply n+3 by n.

$$-n-4=n^{2}+3n$$

Subtract n^{2} from both sides.

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

Subtract 3n from both sides.

$$-n-4-n^{2}-3n=0$$

Combine -n and -3n to get -4n.

$$-4n-4-n^{2}=0$$

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.

$$-n^{2}-4n-4=0$$

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

$$n=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Square -4.

$$n=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$n=\frac{-\left(-4\right)±\sqrt{16+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$n=\frac{-\left(-4\right)±\sqrt{16-16}}{2\left(-1\right)}$$

Add 16 to -16.

$$n=\frac{-\left(-4\right)±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$n=-\frac{-4}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$n=\frac{4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$n=\frac{4}{-2}$$

Divide 4 by -2.

$$n=-2$$

Steps for Completing the Square

Variable n 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(n-1\right)\left(n+3\right), the least common multiple of n^{2}+2n-3,n-1.

$$n-4-2n=\left(n+3\right)n$$

Combine n and -2n to get -n.

$$-n-4=\left(n+3\right)n$$

Use the distributive property to multiply n+3 by n.

$$-n-4=n^{2}+3n$$

Subtract n^{2} from both sides.

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

Subtract 3n from both sides.

$$-n-4-n^{2}-3n=0$$

Combine -n and -3n to get -4n.

$$-4n-4-n^{2}=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$-4n-n^{2}=4$$

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.

$$-n^{2}-4n=4$$

Divide both sides by -1.

$$\frac{-n^{2}-4n}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -4 by -1.

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

Divide 4 by -1.

$$n^{2}+4n=-4$$

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.

$$n^{2}+4n+2^{2}=-4+2^{2}$$

Square 2.

$$n^{2}+4n+4=-4+4$$

Add -4 to 4.

$$n^{2}+4n+4=0$$

Factor n^{2}+4n+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}.

$$\left(n+2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+2\right)^{2}}=\sqrt{0}$$

Simplify.

$$n+2=0 n+2=0$$

Subtract 2 from both sides of the equation.

$$n=-2 n=-2$$

The equation is now solved. Solutions are the same.

$$n=-2$$