# Algebra with one variable We have recently added a couple of Algebra quizzes to gniruT in the Grades 3 - 7 section. Because this could be a bit of a stretch we though we should work through a few examples and explain our thinking.

This is a gentle introduction to Algebra with one unknown. Algebra is an important foundation for higher level math and this simple introduction is a great way to get started. If you know basic arithmetic you are ready to go.

Let’s look at some fundamentals first.

Both sides of an equation have exactly the same value, they are equal.
For example 5 = 5 and 2 + 3 = 5 are both equations and taking that one step further we can also write 2 + 3 = 1 * 5

Now let’s expand on that further and introduce an unknown.
For example 2 + __ = 1 * 5 and when asked to fill in the blank we can easily work out the answer is 3.
Now let’s substitute the blank for a variable and use the letter y to represent that variable.
We now have 2 + y = 1 * 5 and that is what a lot of algebra equations look like. Now this is great but we already know that the answer is 3. Put that aside for the moment and let’s use this simple example to work through.

Solve 2 + y = 1 * 5 for y is the same as asking what does y equal.
To figure out what y equals we need to isolate it. As we work through the problems we want to be thinking can we simplify and what is the next step to get y by itself.
First of all let’s simplify: 2 + y = 5
Now we want to isolate y and because both sides of the equation are equal whatever we do to one side we have to do to the other side. We can see that on the left side a 2 is added to y that we must get rid of to isolate y.
Let’s subtract 2 from both sides: 2 + y - 2 = 5 - 2
Let’s simplify, the 2 and -2 cancel out because 2 - 2 = 0 and the 5 -2 = 3
So we are left with y = 3 and that is it, we just solved it.

Ok so that was pretty easy but before we move on to a slightly harder problem a couple of important notes

• The logic we applied above to remove the 2 and isolate the y also applies for multiplication and division, whatever you do to one side you have to do to the other. For example if it was 2 multiplied by y then to isolate the y we could divide both sides by 2.

• You may see some questions with 5y or 5 * y these both mean 5 multiply y.

• It is normally easier to first isolate the terms by subtracting or adding before dividing or multiplying. But as long as you do the same thing on both sides of the equal sign you can do it in any order you like.

Let’s try a more challenging one

Solve a * 5 - 23 = 2 for a.
The first thing we should ask is can we simplify and in this case the answer is no because the 5 is multiplied by a ( a * 5) we cannot do 5 -23, we still have to follow the BODMAS rule.
So let’s start to isolate y and we can add 23 to both sides to cancel out the -23 on the left hand side.
This becomes: a * 5 - 23 + 23 = 2 + 23
Simplify: a * 5 = 25
Next we can divide by 5 on both sides: a * 5 / 5 = 25 / 5
Simplify: 5/5 = 1 and 25 / 5 = 5
This becomes: a * 1 = 5 and because anything multiplied by 1 remains unchanged we get a = 5
That’s it we just crushed another one.

Now let’s work a word problem
If a laptop costs twice as much as an iPad and the total for both is \$1350 what is the price of the iPad?

We want to calculate what the price of the iPad is but all we know is that the total is \$1350 and the laptop is 2 times the price of the iPad.

As a first step let’s introduce a variable and say the price of the iPad is equal to P

We know that the laptop price is 2 times the iPad price so we can say that the laptop costs 2 * P

This is great so now we have a variable for both of them and we also know that the price of the iPad and the price of the laptop equals \$1350
So we can put an equation together and we get: P + 2P = 1350
We can simplify to: 3P = 1350
Now all we have to do is isolate P by dividing both sides by 3
Which is P * 3 / 3 = 1350 / 3
The * 3 and / 3 cancel out and we are left with P = 450

So the price of the iPad is \$450 and the price of the laptop is \$900

In this last example we will work a problem with an unknown on both sides.
Solve 3Q + 6 = 6Q + 3 for Q.
With this type of problem the first thing we want to do is remove one of the variables from one of the sides.
a. We can either subtract 3Q or 6Q from both sides, let’s do 3Q so we don’t have to deal with negatives. 3Q - 3Q + 6 = 6Q - 3Q + 3
b. Simplify the left side: The 3Q - 3Q cancels out so we are left with 6 = 6Q - 3Q + 3
c. Simplify the right side: 6Q - 3Q equals 3Q so we are left with 6 = 3Q + 3
d. Now we subtract 3 from both sides: 6 - 3 = 3Q + 3 -3
e. Simplifies to: 3 = 3Q
f. Because 3Q is 3 multiply Q we now need to divide both sides by 3 to isolate the Q so this gives 3 / 3 = 3 /3 * Q
g. This simplifies to 1 = Q or Q =1

That’s it, have fun with the quizzes.