Solving Linear Systems by Elimination
There’s a trick when solving a system of equations that can be used in certain situations as a bit of a shortcut. This trick is available because of how the rules of algebra naturally work, but it can be a little difficult to understand at first. Consider the following system of equations:
2x + y = 4
-2x + 2y = 2
Let’s take the first equation, 2x + y = 4. We can add something to the left side of the equation as long as we add something of equal size to the right side of the equation. So what if we added -2x + 2y to the left side of the equation, and added 2 to the right side of the equation?
Because -2x + 2y = 2, this is acceptable by the rules of algebra since we are adding the same value to both sides (they’re the same because we’re told they’re equal). Now notice what happens to our equations:
2x + y = 4
2x + y + (-2x + 2y) = 4 + 2
3y = 6
y = 2
Because there was 2x in the first equation and -2x in the second equation, they quickly eliminate each other when we use this approach, and we’re able to quickly solve for y, where we get y = 2. Substitution will then tell us that 2x + 2 = 4, and x = 1. Instead of just using this with addition, we can also use elimination with subtraction like with the following system of equations:
3x + 6y = 12
3x – y = 9
If we start with 3x + 6y = 12, we can subtract 3x – y on the left side, and subtract 9 on the right side. Remember that we can do this because 3x – y = 9, as given by the second equation. When we subtract 3x – y from 3x + 6y, then it’s obvious that the 3x portion of each equation is going to be eliminated, as follows:
3x + 6y = 12
3x + 6y – (3x – y) = 12 – 9
3x + 6y – 3x + y = 3
7y = 3
y = 3/7
And we can substitute back in to solve for the corresponding value for x.