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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 First Degree (Linear) Equations
A firstdegree equation is one whose unknown variable is raised to a power of one. Another word for any firstdegree equation is a linear equation because when graphed using a table of values, a straight line is generated. Generally, to solve an equations means to find a value for the unknown variable that makes both sides of the equation true.
This skill is imperative to all types of math. Difficulty, however, increases as the variable increases in degree. For example, a firstdegree equation is a lot easier to solve than a seconddegree equation – known as a quadratic equation.
Here are some examples of a firstdegree equation
 5 = x + 1
 3(x + 4) = 2(x – 5)
Notice that in the examples, x is the only variable only used. For simplicity sake, we will limit ourselves to solving equations having just one variable. That said, every question has two sides:
The value of the variable that makes the sides of an equation equal to each other is called a solution. Finding the solution requires that you know how to manipulate expressions, which oftentimes involves combining like terms and expanding – two concepts extensively studied last unit. Depending on what the equation is modelling, the solution tells us something important that we would not otherwise know.
Let’s try solving an equation by trialanderror (guess and check), that way we don’t have to rely on any specific technique. Try finding a value of x for 5 = x + 1 that would make the right side and left side the same. It’s easy to see that the only solution here is 4; therefore, x = 4 satisfies the equation. We could have also found x = 4 algebraically by isolating the variable x to one side exclusively. Currently, the terms on the right side are x and +1. We can subtract –1 from both sides to eliminate the +1:
$5=x+1\phantom{\rule{0ex}{0ex}}5{\mathbf{\u2013}}{\mathbf{1}}=x\overline{)+1}\overline{)\mathbf{\u2013}\mathbf{1}}\phantom{\rule{0ex}{0ex}}4=x\phantom{\rule{0ex}{0ex}}$Another way to communicate what just happen is you moved the +1 over to the other side. Moving a positive term over the equal sign makes it negative, and the opposite is true if it were negative.
Equations Having Symbols of Grouping
Let’s try solving an equation containing parentheses:
5 = 2(x + 1)
You’ll notice now that our target variable is inside brackets. You could “liberate” the x in two main ways. The first way involves dividing both sides by 2, which will cancel out the 2 on the right side:
$\frac{{\displaystyle 5}}{{\displaystyle 2}}=\frac{{\displaystyle \overline{)2}\left(x+1\right)}}{{\displaystyle \overline{)2}}}\phantom{\rule{0ex}{0ex}}\frac{5}{2}=x+1\phantom{\rule{0ex}{0ex}}$You then bring the +1 over to left side, where is becomes:
$\frac{5}{2}\u20131=x\phantom{\rule{0ex}{0ex}}x=\frac{3}{2}\mathbf{or}\mathbf{}\mathbf{}\mathbf{}1.5\phantom{\rule{0ex}{0ex}}$The other way you could solve for x is by expanding right side first:
5 = 2x + 2
Then move the +2 over (it becomes negative):
5 – 2 = 2x
3 = 2x
x = 1.5
To check your answer, you can always substitute your answer back into the original equation to see if it works. Always use the original equation in case you manipulated the others incorrectly.
Sometimes you may have expand by the FOIL method. For example, if you’re given 2x² = 2(x + 1)(x + 2), the right side will be need to be fully expanded. The expanded version is shown:
$2{x}^{2}=\left(2x+2\right)\left(x+2\right)\phantom{\rule{0ex}{0ex}}2{x}^{2}=2{x}^{2}\mathbf{+}\mathbf{4}\mathit{x}\mathbf{+}\mathbf{2}\mathit{x}+4\phantom{\rule{0ex}{0ex}}2{x}^{2}=2{x}^{2}+6x+4\phantom{\rule{0ex}{0ex}}$The right side is now fully expanded, so it’s time to dedicate one side to terms containing variables only. This will enable you to collect like terms and eventually solve.
$\mathbf{2}{\mathit{x}}^{\mathbf{2}}\mathbf{\u2013}\mathbf{2}{\mathit{x}}^{\mathbf{2}}=6x+4\phantom{\rule{0ex}{0ex}}0=6x+4\phantom{\rule{0ex}{0ex}}\u20136x=4\phantom{\rule{0ex}{0ex}}x=\frac{4}{\u20136}\mathbf{reduces}\mathbf{}\mathbf{to}\mathbf{}\mathbf{}\mathbf{}\u2013\frac{\mathbf{2}}{\mathbf{3}}\phantom{\rule{0ex}{0ex}}$A more complicated and thorough explanation of an equation requiring polynomial expansion can be found in Question 1 below:
Equations Having the Unknown in Two or More Terms
Sometimes you’ll have to solve an equation where two or more terms contain the unknown variable:
2x + 3 = –3 – x
When this happens, you need to move all the terms containing the variable over to one side (choose the side that’s most convenient). By doing this, you can combine like terms then solve:
 Move –x to left side, becomes +x
 Move +3 to right side, becomes –3
2x + x = –3 – 3
3x = –6
 Divide both sides by 3
$x=\frac{\u20136}{3}\Rightarrow \u20132\phantom{\rule{0ex}{0ex}}$
Questions 1 and 2 in the video below relate to these type types of questions – you may skip 3 and 4.
Equations Having Fractions
Arguably, solving firstdegree equations containing fractions requires the most work. Luckily, there are several ways you could make the equation easier to work with. Let’s start with a simple example:
$\frac{x}{10}=5\phantom{\rule{0ex}{0ex}}$Of course you could solve this by guessing and checking. Although a more practical way would be to do this algebraiclly by multiplying both sides of the equation by the lowest common denominator (LCD). The only denominator here worth mentioning is the 10 underneath the x. By multiplying the left side by 10, the 10 cancels out in the denominator, and the 5 on the right side becomes 50:
$\overline{)10\xb7}\frac{\overline{)x}}{\overline{)10}}=5\xb710\phantom{\rule{0ex}{0ex}}x=50\phantom{\rule{0ex}{0ex}}$If you get something a little more complex, such as:
$\frac{2x}{5}=\frac{2}{7}\phantom{\rule{0ex}{0ex}}$This time the lowest common denominator is 35 (a tutorial on find the LCD can be found here). You now multiply every term by 35:
$35\xb7\left(\frac{2x}{5}=\frac{2}{7}\right)\phantom{\rule{0ex}{0ex}}35\xb7\frac{2x}{5}=35\xb7\frac{2}{7}\phantom{\rule{0ex}{0ex}}35\xf75=\mathbf{7}\mathbf{and}35\xf77=\mathbf{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\underset{\mathbf{7}}{\underset{\u23df}{\overline{)35}}}\xb7\frac{2x}{\overline{)5}}=\underset{\mathbf{5}}{\underset{\u23df}{\overline{)35}}}\xb7\frac{2}{\overline{)7}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}7\xb72x=5\xb72\phantom{\rule{0ex}{0ex}}14x=10\phantom{\rule{0ex}{0ex}}x=\frac{10}{14}\Rightarrow \frac{5}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$It looks more complicated than it actually is. Technically, everything that was done above can be summarized under one technique known as crossmultiplying. Cross multiplication involves multiplying each denominator to the opposite numerator; by doing this, the denominators disappear. The animation is shown below:
Let’s look at a few more examples where equations containing fractions are solved by this technique.