Solve L=15/L-2

$$L = \frac { 15 } { L - 2 }$$

Solve for L

$$L=-3$$

$$L=5$$

Steps Using Factoring

Subtract \frac{15}{L-2} from both sides.

$$L-\frac{15}{L-2}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply L times \frac{L-2}{L-2}.

$$\frac{L\left(L-2\right)}{L-2}-\frac{15}{L-2}=0$$

Since \frac{L\left(L-2\right)}{L-2} and \frac{15}{L-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{L\left(L-2\right)-15}{L-2}=0$$

Do the multiplications in L\left(L-2\right)-15.

$$\frac{L^{2}-2L-15}{L-2}=0$$

Variable L cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by L-2.

$$L^{2}-2L-15=0$$

To solve the equation, factor L^{2}-2L-15 using formula L^{2}+\left(a+b\right)L+ab=\left(L+a\right)\left(L+b\right). To find a and b, set up a system to be solved.

$$a+b=-2 ab=-15$$

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 -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite factored expression \left(L+a\right)\left(L+b\right) using the obtained values.

$$\left(L-5\right)\left(L+3\right)$$

To find equation solutions, solve L-5=0 and L+3=0.

$$L=5 L=-3$$

Steps Using Factoring By Grouping

Subtract \frac{15}{L-2} from both sides.

$$L-\frac{15}{L-2}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply L times \frac{L-2}{L-2}.

$$\frac{L\left(L-2\right)}{L-2}-\frac{15}{L-2}=0$$

Since \frac{L\left(L-2\right)}{L-2} and \frac{15}{L-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{L\left(L-2\right)-15}{L-2}=0$$

Do the multiplications in L\left(L-2\right)-15.

$$\frac{L^{2}-2L-15}{L-2}=0$$

Variable L cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by L-2.

$$L^{2}-2L-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as L^{2}+aL+bL-15. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-15\right)=-15$$

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 -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite L^{2}-2L-15 as \left(L^{2}-5L\right)+\left(3L-15\right).

$$\left(L^{2}-5L\right)+\left(3L-15\right)$$

Factor out L in the first and 3 in the second group.

$$L\left(L-5\right)+3\left(L-5\right)$$

Factor out common term L-5 by using distributive property.

$$\left(L-5\right)\left(L+3\right)$$

To find equation solutions, solve L-5=0 and L+3=0.

$$L=5 L=-3$$

Steps Using the Quadratic Formula

Subtract \frac{15}{L-2} from both sides.

$$L-\frac{15}{L-2}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply L times \frac{L-2}{L-2}.

$$\frac{L\left(L-2\right)}{L-2}-\frac{15}{L-2}=0$$

Since \frac{L\left(L-2\right)}{L-2} and \frac{15}{L-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{L\left(L-2\right)-15}{L-2}=0$$

Do the multiplications in L\left(L-2\right)-15.

$$\frac{L^{2}-2L-15}{L-2}=0$$

Variable L cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by L-2.

$$L^{2}-2L-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$L=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-15\right)}}{2}$$

Square -2.

$$L=\frac{-\left(-2\right)±\sqrt{4-4\left(-15\right)}}{2}$$

Multiply -4 times -15.

$$L=\frac{-\left(-2\right)±\sqrt{4+60}}{2}$$

Add 4 to 60.

$$L=\frac{-\left(-2\right)±\sqrt{64}}{2}$$

Take the square root of 64.

$$L=\frac{-\left(-2\right)±8}{2}$$

The opposite of -2 is 2.

$$L=\frac{2±8}{2}$$

Now solve the equation L=\frac{2±8}{2} when ± is plus. Add 2 to 8.

$$L=\frac{10}{2}$$

Divide 10 by 2.

$$L=5$$

Now solve the equation L=\frac{2±8}{2} when ± is minus. Subtract 8 from 2.

$$L=-\frac{6}{2}$$

Divide -6 by 2.

$$L=-3$$

The equation is now solved.

$$L=5 L=-3$$

Steps for Completing the Square

Subtract \frac{15}{L-2} from both sides.

$$L-\frac{15}{L-2}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply L times \frac{L-2}{L-2}.

$$\frac{L\left(L-2\right)}{L-2}-\frac{15}{L-2}=0$$

Since \frac{L\left(L-2\right)}{L-2} and \frac{15}{L-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{L\left(L-2\right)-15}{L-2}=0$$

Do the multiplications in L\left(L-2\right)-15.

$$\frac{L^{2}-2L-15}{L-2}=0$$

Variable L cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by L-2.

$$L^{2}-2L-15=0$$

Add 15 to both sides. Anything plus zero gives itself.

$$L^{2}-2L=15$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$L^{2}-2L+1=15+1$$

Add 15 to 1.

$$L^{2}-2L+1=16$$

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

$$\left(L-1\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(L-1\right)^{2}}=\sqrt{16}$$

Simplify.

$$L-1=4 L-1=-4$$

Add 1 to both sides of the equation.

$$L=5 L=-3$$