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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\}