Math 11 Week #14

This week in pre-calc we started our new unit with rational expressions. In 7.2, we learned all about multiplying these rational expressions. It is very similar to just multiplying regular fractions where you simplify as much as you can, and then solve the question. For example:

(\frac{16}{45})(\frac{25}{42})(\frac{21}{24})(\frac{12}{8})

Just to start, you can see that the 8 on the bottom can cancel out with the 16 on top, turning the 8 into a 1 and the 16 into a 2. You would continue doing this until you can’t anymore.

With this equation, to make it easier, you would try and simplify the most you can with the numerators and denominators. After simplifying, you would end up with a way easier equation to solve:

(\frac{1}{3})(\frac{5}{6})(\frac{1}{1})(\frac{1}{1}) =(\frac{5}{18})

Now with expressions with variables, it is very similar. With that though, there are restrictions. The denominator of an expression can never equal 0, so before solving, you must state the non-permissible variables.

For example with the expression:

(\frac{2x+10}{8x+16})(\frac{(x+2)^2}{x^2-25})

The first step is to state the non-permissible values and make sure the denominator ≠ 0. To do this you might need to factor out the denominator:

8x+16

8(x+2)

x≠-2

x^2-25

(x-5)(x+5)

x≠ 5, -5

Non-permissible values : x≠ 5, -5, 0

Next we need to factor out the entire expression:

(\frac{2(x+5)}{8(x+2)})(\frac{(x+2)(x+2)}{(x-5)(x+5)}

Next we can cancel out all of the equivalent values, such as (x+5) in the denominator and (x+5) in the numerator.

(\frac{2}{8})(\frac{(x+2)}{(x-5)}

Then simplify even more, since 2 can factor into 8.

(\frac{1}{4})(\frac{(x+2)}{(x-5)}

Now we solve and find our final answer because we can’t simplify the expression anymore:

\frac{x+2}{4(x-5)}

If we have a dividing question, all we do is flip the second expression and then treat is as a multiplication question – so we don’t really ever divide.

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