Week 15 – Math 10

This week in Math 10 I learned how to solve word problems using time, speed, and distance applications.

Imagine this is the word problem:

A student drove 1245km from Edmonton to Vancouver in 16\frac{1}{2} hours. This included a one hour stop in Golden Ears and a 30 minute stop in Kamloops. She averaged 100km/h on the divided highways and 75km/h on the non-divided mountainous roads. How much time did she spend on the divided highways?

To solve this, there are two key factors we must know; how to create a grid, and common sense.

Because distance, speed, and time are used in this word problem, we are using it to find how much time was spent on the divided highways.

We are first going to fill in everything we know point-blank about the world problem. It says they averaged 100km/h on the highway and 75km/h on the mountains. We also know the total distance driven, which is 1245km. It also says in the word problem that it took 16\frac{1}{2} hours to drive from Edmonton to Vancouver, so why are there 15 hours on the grid and not 16\frac{1}{2}? This is where common sense comes in. There were 16\frac{1}{2} hours total throughout the whole trip, however, because of the one hour and half hour stops, she spent 15 hours purely driving.

Now we fill in what we don’t know, which is represented by x and y. We don’t know the time that was spent driving on the highway or the mountains.

Now we add to create an equation. Because x and y represent the time on the highway and mountain, we will be replacing the hours in the speed with X. The result is 100x and 75y.

Now we simply add up the columns, giving us the equations:

x + y = 15

100x + 75y = 1245

At this point, answer the equations using substitution, elimination, or graphing (don’t).

Week 14 – Math 10

This week in Math 10 I learned about substitution.

Substitution is inserting an equation into another equation. An example of this would be:

x + 4y = -3 and 3x – 7y = 29.

First, pick what equation to put in the other. In this case, I selected x + 4y = -3 to go into 3x – 7y = 29.

Now we must isolate X by moving 4y over, giving us x = -4y – 3.

We now can insert.

3(-4y -3) -7y = 29

Use distributive property on the 3, which gives us:

-12y – 9 – 7y = 29.

Move the -9 over to isolate the y’s.

-12y – 7y = 29 + 9

-19y = 38

Divide by -19 on both sides.

\frac{-19y}{-19} = \frac{38}{-19}

y = -2

Week 13 – Math 10

This week in Math 10 I learned how to get the X and Y intercepts from a General Form equation.

General Form is pretty useless. It doesn’t tell you much about anything; however, it is good for finding X and Y intercepts.

Example: 3x – 5y + 60 = 0

To find the X intercept, first we must cancel out the Y, giving us 3x + 60 = 0.

Then we need to cancel out the 60; we do this by subtracting 60, which we need to do to both sides of the equation.

3x +60 = 0

-60  -60

3x = -60

Then we need to isolate X. To do this, we divide 3x and -60 by 3.

\frac{3x}{3} = \frac{-60}{3}

x = -20


To get the Y-intercept, simply do the same thing, but instead of canceling out the Y, cancel out the X. Cancel out the integers and do so to both sides, the same thing; Then, divide both sides by the Y value instead of X.

It will look like this:

3x – 5y + 60 = 0


-5y + 60 = 0

-60    -60

-5y = -60

\frac{-5y}{-5} = \frac{-60}{-5}

Y = 12

Week 12 – Math 10

This week in Math 10 I learned about slopes.

Slopes are the equation of a line on a graph. Slope = \frac{rise}{run}, where rise = y and run = x.

For example, if a slope is \frac{1}{2}, that means if you go up 1 unit then across 2 units (lines are always seen from left to right). It will look like a staircase. To create the line, just draw a line following the staircase as shown below:

If a slope is negative, for example \frac{-1}{-2}, instead of the staircase/line going right and up, it goes left and down.

Week 11 – Math 10

This week in Math 10 I learned about Mapping Notation.

Mapping Notation is a way to elaborate and show a function, that I find easy to understand.



C = 12n +75

In Mapping Notation, this equation would look like this:

n —> 12 + 75

The arrow means “maps onto”. In other words, the equation is saying that the input (n) maps onto the output (12 + 75). n maps specifically on to 12, but not 75 as seen from the example, which results back to the example at the beginning: C = 12n + 75.

Week 10 – Math 10

This week in Math 10 I learned how to solve functions with total value.

Total value is shown by using two | | on both sides of an equation, encasing it like a bracket. |example|.

Total value equations always equal to a positive number. This can help in verifying answers because if the answer you got is negative, you know you did something wrong.

An example of a total value equation in function notation would be:

f(x) = 2 |4x – 3| + 7