Monthly Archives: December 2018

Math 11 Week #15

Posted on by under Grade 11, Math 11, Pre-calc 11

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

Math 11 Week #14

Posted on by under Grade 11, Grade 11, Math 11, Pre-calc 11

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.

Math 11 Week #13

Posted on by under Grade 11, Math 11, Pre-calc 11

This week in math, we learned about reciprocal functions of parabolas. There are three forms of a reciprocal function graph for parabolas based on how many roots/where are the roots of a quadratic equation. They are similar to linear reciprocal functions, but a bit more complicated since there are can be up to two x-intercepts instead of just always having one with a linear function.

When we have a quadratic function with 2 x-intercepts like this: x^2+4x-5

The reciprocal function looks like this: 

All values of x when y=1 and y=-1 are turned into the invariant points. The vertical asymptotes would be the original x-intercepts of the parent function (-5 and 1) and horizontal would be y=0.

When we have a quadratic function with no roots because it stays either below or above the x axis like this: x^2+4x+5The reciprocal function would look like this: 

Since there are no roots, there would be no vertical asymptote and the horizontal asymptote would be y=0. The invariant point would be the reciprocal of the y value of the vertex.

Finally, if you had a quadratic function with one root like: y=x^2

The reciprocal function would look like this:

The invariant points would be where y=1 and and y=-1. The vertical asymptote would be the x value of the vertex and the horizontal asymptote would be y=0.

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