# Solving Rational Equations

## Solving Rational Equations by Getting LCD

In the first video, we are going to solve rational equations by getting the LCD (Least Common Denominator).

For Example: $\frac{1}{x+1}=\frac{1}{x^2+x}+\frac{5}{x}$

Find the LCD, which is $x^2+x$. Multiple each expression to get its denominator to $x^2+x$
$\frac{(x)1}{(x)x+1}=\frac{1}{x^2+x}+\frac{(x+1)5}{(x+1)x}$

Multiply
$\frac{x}{x^2+x}=\frac{1}{x^2+x}+\frac{5x+5}{x^2+x}$

Now that the denominators are the same, we can add across
$x=1+5x+5$

Solve the equation for the value of x. Combine like terms
$x=5x+6$

Move x to one side of the equation
$-4x=+6$

Isolate x and we have
$x=-\frac{3}{2}$

## Solving Rational Equations by Multiplying by LCD

In the second video, we are going to solve rational equations by multiplying by the LCD (Least Common Denominator).

For example: $\frac{1}{2x-12}+\frac{1}{2x^2-4x-48}=\frac{1}{x-6}$

Determine the LCD of the three denominators, which is 2(x-6)(x+4). Multiply the numerators by the LCD
$\frac{2(x-6)(x+4)1}{2x-12}+\frac{2(x-6)(x+4)1}{2x^2-4x-48}=\frac{2(x-6)(x+4)1}{x-6}$

Reduce each expression
$\frac{(x+4)1}{1}+\frac{1}{1}=\frac{2(x+4)1}{1}$

Multiply
$x+4+1=2(x+4)$

Use the distributive property
$x+4+1=2x+8$

Combine like terms
$x+5=2x+8$

Move x to one side of the equation by subtracting x
$5=x+8$

Subtract 8 from both sides to isolate x, and we have
$-3=x$

## Solving Rational Equations by using the Quadratic Formula

In the third video, we are going to solve rational equations by using the quadratic formula.

For example: $\frac{5b^2-10b}{b^2-16}+\frac{b-3}{b-4}=\frac{3}{b^2-16}$

Determine the LCD, which is $(b-4)(b+9)$, and multiply it with each numerator
$\frac{(b-4)(b+9)5b^2-10b}{b^2-16}+\frac{(b-4)(b+9)b-3}{b-4}=\frac{(b-4)(b+9)3}{b^2-16}$

Reduce each expressions
$5b^2-10b+(b-3)(b-4)=3$

Use the distributive property
$5b^2-10b+b^2+4b-3b-12=3$

Combine like terms
$6b^2-9b-12=3$

Subtract 3 from both sides of the equation so that it is equal to 0
$6b^2-9b-15=0$

Divide by 3
$2b^2-3b-5=0$

Use the quadratic formula to find x. First, identify the values of ab, and c
$a=2, b=-3, c=-5$

Substitute the values into the quadratic formula and solve
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\frac{3\pm\sqrt{(-3)^2-4(2)(-5)}}{2(2)}$
$\frac{3\pm\sqrt{9-4(2)(-5)}}{2(2)}$
$\frac{3\pm\sqrt{9+40}}{4}$
$\frac{3\pm\sqrt{49}}{4}$
$\frac{3\pm7}{4}$

Solve $\frac{3\pm7}{4}$
$\frac{3+7}{4}$
$\frac{10}{4}$
$\frac{5}{2}$

$\frac{3-7}{4}$
$\frac{4}{4}$
$\frac{1}{1}$
$1$

x is $\frac{5}{2}$ and $-1$, so the solution set is
$\{\frac{5}{2}, -1\}$

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