University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
So there are two ways of solving a system of linear equations in two variables. We can either use substitution or we can use elimination and most of the time it's up to us to choose which way we want. Every once in while teachers will ask us to say okay use substitution or use elimination but most of the time we just get a system and we have to solve it somehow and that’s up to us.
What we’re going to do is look at this particular problem. We have two lines and we’re trying to figure out where they intersect. I could choose either one and for this particular example either one is actually a pretty good option. The reason I say that is using elimination basically what we do is we want to somehow add or subtract these equations to get rid of a term.
Our ys are already equal and opposite which means if I just add these two equations together my ys disappear. That’s easy enough. For substitution what we want to do is we want to get one of our variables by itself and what I see in this equation is the x on the bottom equation is already by itself so it’s pretty easy to solve for it if I just subtract the 2y over I’ve solved for x. So for this particular example, either way is perfectly fine.
I’m going to do substitution. I don’t really know why but it’s just the way I’m going to do it for this particular problem which means I want to solve for x in the bottom. Subtract the 2y over, x is equal to 3 minus 2y and then from there all we do is take the x that we solved for here and plug it into the other equation. It’s really important to plug it into the other equation, if you plug it back into the equation you started with, everything is going to disappear; we’re going to end up with zero equals zero. So you always solve in for a variable in one equation and plug it back into the other.
So now all we do is solve this equation up. We have 4 times x, but this x is the same as this one over here, 3 minus 2y, minus 2y is equal to 7. Distribute out, 12 minus 8y minus 2y is equal to 7. Combine like terms, 12 minus 10y is equal to 7. Subtract the 12 over, -10y is equal to -5, divide by -10 leaving us with y is equal to ½.
So we found our y value but our solution is actually going to be a point on the coordinate plane so what we need to do is find the x coordinate. We need to take the y and plug back into any of the equations we have and typically the easiest thing to do is plug it into the equation that we’ve already solve for our variable. So if I take this y, plug it back into the equation I already solved for I end up with x is equal to 3 minus 2 times ½. 2 times ½ is 1, 3 minus 1 is 2. So I end with x is equal to 2 giving me the solution (2,½).
So solving a system of linear equations by substitution, I used substitution because I already have a variable by itself so it’s really easy to solve. This problem I could have done elimination just as well but taking the substitution all the way through and finding my answer.
Unit
Systems of Linear Equations