 ###### Carl Horowitz

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!

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Solving a System of Linear Equations in Two Variables - Problem 3

# Solving a System of Linear Equations in Two Variables - Problem 2

Carl Horowitz ###### Carl Horowitz

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!

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Solving a system of linear equations in two variables. When we’re solving a system of linear equations we have two tools at our disposal; we have substitution and we have elimination.

Substitution is we’re just taking a variable from one equation and plugging it into the other one. Looking at the problem behind me, see if you can figure out if substitution is a good option. What I see if every single variable has a coefficient. So every x and every y has a coefficient and there are no common factors amongst the other terms with those coefficients. What I mean by that is basically if I wanted to solve for this x, well I would have to divide by two but I’m going to introduce a fraction in there. If I want to solve for any variable, we’re going to get fractions.

Most people tend not to like fractions so I would recommend trying to avoid substitution in this case. If you like fractions feel free to go right ahead, I tend not to be such a fan so I’m going to try the other method which is elimination.

The principal behind elimination is basically to get the same coefficient on any of our variables so that way when we add or subtract our equations together those coefficients disappear and we’re left with a single variable. So in doing that what we can do is multiply one or both equations by a number in order to get the coefficients the same.

Looking at this we have the option of either trying to get rid of the xs or the ys. We don’t really have too many common factors, we have 2 and -5, the smallest thing they have in common is 10 or 4 and 10, the smallest thing they have in common is 20. So it doesn’t really matter which one we want to get rid of I’m just going to say let’s go for the xs. We could go for the ys as well.

So multiplying the xs, we need both coefficients to be 10. Let’s multiply the top equation by 5, multiply the bottom equation by 2. In general I try to get my coefficients equal and opposite, I find that students make a lot less mistakes when you’re adding as opposed to subtracting. Subtracting you have to distribute that negative through and things can go wrong. I tend to find it a little bit easier to get your coefficients equal and opposite but you could subtract as well and as long as you do your signs alright you’ll be fine.

Distributing this 5 through, we have 10x minus 20y is equal to 15. Distributing the 2 through, -10x plus 20y is equal to 14. So now I always include the sign that I’m doing, makes sure I keep everything straight. So I want to add so my 10 and -10 cancel, but what happens is the 10s cancel, the 20s cancel so what I’m really left with is zero on this side and then 29 on the other. Now what does that mean?

We have zero is equal to 29. That’s not a true statement so what that tells me is that there is no point on these two lines that actually intersect. So this is false which tells me that there are no solutions. In other words these two lines are parallel. They're parallel, two parallel lines are never going to cross. Whenever you get a false solution, whenever you get zero is equal to 29 or some sort of statement that doesn’t make sense, typically that means you have no solution, the lines aren’t going to intersect.

So taking a system of linear equations, trying to figure out which method we want to use, substitution or elimination, to then solving and interpreting our results. In this case we have a false statement which means we have no solution.