Linear independence is not taught in any GMAT curriculum or book; However, it is very useful to simplify Data Sufficiency problems into counting problems:
- How many variables do we have?
- How many equations do we have?
Many GMAT Data Sufficiency questions are as simple as “How many variables do we have? Do we have that many equations?” Sometimes just from experience, when you see two particular equations consisting of two variables, you can tell there will be one unique solution. This likely means the information is sufficient since we can conclude only one answer. So let’s look at an example:
x + 2y = 12
6x – y = 10
How do we typically solve this equation? First, we would like to eliminate one variable. The typical method would be to double the second equation then add them up to eliminate y. And that should give us only one unique solution. Now let’s change the question up a little bit.:
x + 2y = 122
6x – y = 199
Do we still have only one unique solution? Indeed, we repeat the same steps and they are still effective, giving us only one solution. This thought process is crucial in solving Data Sufficiency problems, as questions encourage the test taker to not touch the equations at all, but instead to use logical procedures to figure out how many possible solutions would exist. The key is to check for linear independency.
If we have two equations and two variables, it’s possible to have one unique solution. The same applies to 3 equations 3 variables, X equations X variables. As the number X goes higher it’s increasingly encouraged to not solve for the equations. Whether we can find a unique solution in these cases will depend on if they are linearly independent. In the case of X equations and X variables, linearly independent equations will always have one unique solution.
Linear independence has two definitions to check.:
Linear means we cannot have any variables multiplying or dividing each other. If we have two variables x and y, having quadratic terms such as x^2, y^2, x*y would mean it is not linear. Expressions such as x/y = 3 would be linear, however, since we can turn this into x = 3y. This is very quick to check after becoming familiar with the definition of linear.
(x + 2)*(x – 2) = 1
This is not linear. If we multiply this out, we will find an x^2 term. This will have two solutions.
x/y = x + 2
This is not linear. After bringing y on the right side we will find an x*y term.
x + 10y + 20z = -10x + 5y – 3z + 10
This is linear. There are no quadratic or higher powers of terms.
Note, however, that even if we don’t have linear equations it is still possible to have one unique solution with X variables X equations.
Next, we need to check if the equations are independent. This is also not too difficult to check. Independent equations means that all the equations “serve a purpose,” and we do not have any extraneous information in the equations. If we have only two equations, we only need to check that they aren’t the same equation after simplification.
3x – 3y = 30
x = y + 10
At first sight, one might think these are two different equations, but if we simplify the first equation we can see it is equivalent to x – y = 10, or x = y + 10. Therefore, these two equations are exactly the same. Then what’s the big deal? Well our two equations are effectively only one equation now. The information concluded from these two equations, is only x = y + 10, which of course has infinitely many solutions.
Here’s another typical example of not independent with three equations.:
x + 2y + 3z = 14
y – z = -1
x + 3y + 2z = 13
Observing the equations above, can you find anything special about them? The hint is to observe the numbers on the right, 14 – 1 = 13. That should be suspicious for us, so try adding the first and second equation. We find that the third equation happens to be the first one plus the second one. In this case we say the third equation is not independent, because it is based on the first and second equation. Therefore, these are effectively only two equations with three variables, which gives up infinitely many solutions once more.
Finally, let’s combine everything we learned and see if you can check for linear independence now.:
x + 2y = 10
y – 2x = -10
It is linear, and equations don’t cancel so it is independent.
10W:L = 5:2
5W: (2L) = 5:8
This is linear, but not independent. The first equation gives W:L = something, the second equation also gives W:L = something.
| x – y | = 3
x^2 – xy = 3
The second equation is not linear. The first equation is still linear but it means (x-y) is either 3 or -3. They are independent, however, and we should solve to see if we can find any unique solution. The second equation turns into x * (x – y) = 3, and we know (x – y) is either 3 or -3. That gives up two solutions (x = 1, y = -2), or (x = -1, y = 2). Depending on the question, this could still be sufficient (e.g. what is |x|?).
What was the number of parcels picked up by a certain mail truck on Tuesday?
(1) On Tuesday, the mail truck picked up 15 fewer than three times as many envelopes as parcels.
(2) On Tuesday, the mail truck picked up 5 more parcels than envelopes.
Solution: (1) gives a linear equation. (2) gives a different linear equation. Combining would be sufficient since there’s one unique solution. We choose C.
There is a lot more to linear independence; however, just understanding the definition would help you save time on the easy algebra questions, while simplifying more complicated algebraic questions. With this technique, hopefully you will start to prefer seeing equations in the problems!