 42 lessons
 0 quizzes
 10 week duration

Measurements
An introduction to numerical computation. Emphasis is placed on scientific and engineering notation, the rule of significant figures, and converting between SI and Imperial units.

Fractions, Percentage, Ratios and Proportion
Emphasis here is placed on understanding fractions, percent, and using ratios to compare quantities and set up proportions to solve problems.

Introduction to Algebra

Factoring

Solving Equations

Functions and Graphs

Geometry
This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.
 Identify, measure, and calculate different types of straight lines and angles
 Calculate the interior angles of polygons
 Solve problems involving a variety of different types of triangles
 Calculate the area of a variety of different types of quadrilaterals
 Solve problems involving circles
 Calculate the areas and volumes of different solids
Solve Equations Containing Fractions
Depending on the number of terms in the equation, these questions can go from being simple to extremely difficult. Easiest of these types is when you have a single term on the left side and a single term on the right. This of course was discussed earlier in this unit when we looked at equations having fractions, and the technique we use is called crossmultiplication.
However, this method has its limitations when when you have 3 terms, let say:
$\frac{a}{3}\u20138=6a\phantom{\rule{0ex}{0ex}}$To solve for a, you can either combine the first two terms together, then crossmultiply (method 1), or find a common denominator – it’s 3 – and multiply each term by 3 to reduce it into an equation without fractions (method 2).
Let’s focus on method 1. Combing the first two terms we get:
$\frac{a}{3}\u20138\phantom{\rule{0ex}{0ex}}=\frac{a\u201324}{3}\phantom{\rule{0ex}{0ex}}$Now our equation is suitable for crossmultiplication:
$\frac{a\u201324}{3}=\frac{6a}{1}\phantom{\rule{0ex}{0ex}}3\left(6a\right)=a\u201324\phantom{\rule{0ex}{0ex}}18a=a\u201324\phantom{\rule{0ex}{0ex}}18a\u2013a=\u201324\phantom{\rule{0ex}{0ex}}17a=\u201324\phantom{\rule{0ex}{0ex}}a=\frac{\u201324}{17}\phantom{\rule{0ex}{0ex}}$Had you gone with method 2, the solution would be the same:
Let’s try another example where method 1 or 2 need to be applied first.
$\frac{x\u20132}{4}=\frac{2x\u20134}{2}\u20136\phantom{\rule{0ex}{0ex}}$Starting with method 1, notice that the right side has two terms that can be combined.
$\cdots =\frac{2x\u20134}{2}\u20136\phantom{\rule{0ex}{0ex}}=\frac{2x\u20134\u20132\left(6\right)}{2}\phantom{\rule{0ex}{0ex}}=\frac{2x\u20134\u201312}{2}\phantom{\rule{0ex}{0ex}}=\frac{2x\u201316}{2}\phantom{\rule{0ex}{0ex}}$Now the equation becomes:
$\frac{x\u20132}{4}=\frac{2x\u201316}{2}\phantom{\rule{0ex}{0ex}}\overline{)crossmultiply}\phantom{\rule{0ex}{0ex}}2\left(x\u20132\right)=4\left(2x\u201316\right)\phantom{\rule{0ex}{0ex}}\overline{)expandbothsides}\phantom{\rule{0ex}{0ex}}2x\u20134=8x\u201364\phantom{\rule{0ex}{0ex}}\overline{)collectliketerms}\phantom{\rule{0ex}{0ex}}2x\u20138x=\u201364+4\phantom{\rule{0ex}{0ex}}\u20136x=\u201360\phantom{\rule{0ex}{0ex}}\overline{)solve}\phantom{\rule{0ex}{0ex}}x=\frac{\u201360}{\u20136}=10\phantom{\rule{0ex}{0ex}}$To see it done using method 2, watch the video below:
Try solving this last example on your own before continuing to another related concept involving algebraic fractions.
$\frac{3x\u20135}{2}=\frac{2\left(x\u20131\right)}{3}+4\phantom{\rule{0ex}{0ex}}$
Solving fractions when the variable’s in the denominator
Unlike previous examples, the ones you’ll see here have variables in the denominator position. Again, both methods work, but the video shows method 2 only which involves finding the lowest common denominator (LCD). Once the LCD is determined, you multiple the whole equation by the LCD – this will eliminate the every denominator.
Just as you saw when you first simplified expressions through factoring, you may need to do the same when dealing with equations. This is why you learned it with simple expressions first before incorporating the concept to solving.