Using the balance model with negative terms in an equation
This article discusses how to use a balance to model simple linear equations in pre-algebra or algebra 1 when they also involve negative numbers. Please start at Part 1, as it explains how to use a balance with equations that only involve positive integers.
It is not as easy to portray negative numbers in an equation with the balance model, because in nature, we don't have "negative weight." But there is a way: you can think of the negatives as holes or empty spots that, when filled with positives, become "nothing" or zero.
x − 4 = 3 |
Add 4 circles to both sides to "fill in" the negative circles. |
x − 4 + 4 = 7 |
Now we have added 4 circles to both sides of the balance/equation. |
x = 7 |
The 4 circles "fill in" or cancel the negatives. So in the end there is nothing − or zero − left. |
Another example equation
Below, we will solve 2x + 3 = −5.
2x + 3 = −5 |
We want the left side to have blocks ONLY (the unknowns). Therefore we have to eliminate three circles from both sides. |
2x = −8 |
But since there are no circles to take away on the right side, we have to "add" negative circles. |
x = −4 |
As a last step, we take half of each side to arrive to the solution x = −4. |
Unknown is negative
The same thing can happen with the x's – they can be negative too! Study the following example:
2x − 3 | = | −x + 3 | To isolate x (to have it alone) on the left side, we need to eliminate −x on the right side. Since it is negative, we ADD x to both sides. |
2x − 3 + x | = | −x + 3 + x | Now −x and x cancel each other (on the right side). |
2x − 3 + x | = | 3 | Then we can add 2x and x on the left side to get 3x. |
3x − 3 | = | 3 | The next step is to eliminate −3 on the left side. For that end we add 3 to both sides. |
3x − 3 + 3 | = | 3 + 3 | Again, −3 and + 3 cancel each other. |
3x< /td> | = | 6 | As the last step, since there are three x's on the left side and we want to know how much one is worth, we divide both sides of the equation by three. |
x | = | 2 | This is the final solution. |
To check the solution, we substitute x = −4 to the equation 2x + 3 = −5:
2(−4) + 3 | = | −5 |
−8 + 3 | = | −5 |
−5 | = | −5 |
It checks, so the solution was correct.
More exercises
You can solve these equations with the help of the balance model or just with pencil and paper.
- x − 5 = 5
- 2x − 5 = 5
- 3x − 4 = 2x + 4
- 5x − 3 = −x + 3
- 5x + 3 = − 2x + 4
See also
Balance as a model of an equation
A lesson that precedes this one. It discusses how to use a balance to model simple equations in beginning algebra.
Scales problems - video lesson
In this video lesson for 4th or 5th grade, I solve 14 different balance problems, starting from the most simple and advancing to some that have double scales. Students learn the principles of dividing both sides of the equation by the same number and removing (subtracting) the same amount of both sides of the equation.