Integer Bars: Learning Basic Algebra Using Integer Bars

Learning Basic Algebra Using Integer Bars

Introduction to Basic Algebra

Algebra uses equations in different forms to solve problems. We have already talked about equations in previous pages. When we combined 2 different bars to create a train we then showed how to write that in an equation. Example 1:

2 + 3 = 5

In algebra you would see an equation that includes a variable such as N to represent one of the numbers. Here is an example:

2 + N = 5

Usually, you have to find the value of   N  . We do this using the integer bars as shown below:

As you can see, the gray bar (the value of N) can be replaced with the light green bar which has a value of 3. To solve for   N  using algebra, we need to get   N  alone on one side of the equal side. We can do this by subtracting the value of 2 from both sides of the equation, as follows:

2 + N - 2 = 5 - 2

On the left side of the equal sign there is a value of 2 so when we subtract 2, they cancel each other which leaves N alone on this side of the equation.

N = 5 - 2
N = 3

Once again, we see the value of N is 3.

Example 2 uses multiplication which we have covered before. There are different ways to indicate multiplication in an equation. We can use the symbol x as we did in the multiplication lesson and as shown here:

2 x N= 6

The other ways to indicate multiplication are a dot "·", parentheses "()", or nothing as shown below:

2 · N = 6
2(N) = 6
2N = 6

Here is the multiplication example:

3N = 18

We can solve for N by using integers bars. We will use a random grey bar to represent the value of N. Since we have 3N, we will make a train of 3 grey bars on the left of the equal sign and a train of size 18 on the right side.

Since we have 3 bars on the left we need to find 3 bars of the same size that will match the size 18 bar.

The 3 bars that match the train of size 18 are the dark green or size 6 bars. Since we have 3 N's on the left and 3 size 6 bars on the right, the value of N turns out to be 6.

To solve this same equation algebraically, we need to have N by itself on one side of the equal sign. We can do this by dividing both sides of the original equation by 3.

3N = 18
(3N) / 3 = 18 / 3

On the left side of the equation, (3N) / 3, the 3 / 3 is 1 so we have 1N or just N. On the right side of the equation we just divide 18 / 3 (as we did in the division pages) and we get the final answer of N = 6. We get the same result by solving the equation using algebra as we did by using the integer bars.

Example 3 is more complicated because the variable, N, appears on both sides of the equal sign. First we will solve this using integer bars. Again, we will use a random grey bar to represent N.

3N - 5 = 2N + 2

We can start by subtracting 2N (2 grey bars) from both sides of the equation to get:

Then we add a 5 bar to both sides of the equation. On the left side of the equation we already have the subtraction of a 5 bar, so now adding a 5 bar cancels both 5 bars. On the right side we just add the 5 bar to the 2 bar.

Now we have a single grey bar alone on the left side of the equation. We add the 2 bar and the 5 bar on the right side to get the answer of a 7 bar.

So N = 7. Solving the same equation using algebra is done as follows:

3N - 5 = 2N + 2
3N - 5 - 2N = 2N + 2 - 2N
1N - 5 = 2
N - 5 + 5 = 2 + 5
N = 7

Exercises

7 + N = 12

2N + 5 = 15

10 = 4 + 3N

2N - 6 = N+ 2

4 + 4N = 6N - 2

   <- Click on this image to start the applet

Once you have solved all of the exercises, check your answers.

Learning About Charts Using Integer Bars
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Last Updated: Tuesday, 15-Jul-2003 23:45:51 GMT

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