Math 11 Week #15

This week in pre-calculus, we learned about solving rational equations. Last week we learned how to solve expressions, but now I know how to find an actual solution for our variable. There are a couple ways to solve certain questions and in this post, I will be showing you my preferred ways: multiplying throughout by a common denominator and cross multiplying.

  • Multiplying throughout by a common denominator:
\frac{5}{x} - \frac{1}{3} =\frac{1}{x}

First, we need to find the lowest common denominator. To do that we look at all of our denominators and multiply them together, without repeating a certain number/variable. In this case, we wouldn’t want to use both x’s because we already are using one.

For this equation, we can see that our lowest common denominator is 3x ((3)(x)). Now that we know that, we multiply every term by 3x.

3x(\frac{5}{x}) - 3x(\frac{1}{3}) =3x(\frac{1}{x})

If one of the terms in a denominator of a fraction is the same as the LCD, we can cancel that term out, making it easier to solve.

3(\frac{5}{1}) - x(\frac{1}{1}) =3(\frac{1}{1})

 

3(5) - x(1) =3(1)

 

15 - x =3

Now we just solve it like a simple linear equation that it is.

12 = x

 

  • Cross-multiplication:

If we are given a question where we have 2 fractions that equal each other, we can use the cross multiplication method.

\frac{x}{2x-3} = \frac{3x}{x+11}

With this equation, we multiply the numerator of the first fraction and the denominator of the second fraction and vice versa.

x(x+11) = 2x-3(3x)

 

x^2 + 11x = 6x^2 -9x

 

0 = 6x^2 - x^2 - 11x -9x

 

0 = 5x^2 - 20x

 

5x(x-4)=0

 

x=0 and x=4

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