 ###### 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 Linear System in Three Variables with no or Infinite Solutions - Problem 1

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 equations in three variables. So when we have a system of equations in three variables what we want to do is turn this into a system in two variables. So what we want to do is get rid of one variable. We do that by eliminating. You could substitute. In general, elimination is significantly easier and a lot less problems arise.

We want to choose a variable to get rid off. I look at my coefficients; coefficients of c are fairly big, coefficients of b are fairly big, coefficients of a are pretty small and pretty easy to deal with. What we want to do is combine a couple of these equations together to get rid of a. Looking at it, I see that I can pair the bottom two I have a coefficient of -1 and 1, if I just add these two equations together my a term disappears. So without rewriting this, I’m just going to add these two equations together. My –a and my a cancel out. 3 plus -2 is just going to leave us with a b, -2 plus 3c is just going to leave us with a c and -1 plus 1 is equal to zero. Now I have one equation b plus c is equal to zero.

Now we need to get rid of a from another pairing. We’ve already used the bottom two, which leaves us either with the top and the middle or the top and the bottom. It doesn’t really matter which one we do, in general I prefer to add, so I’m going to try want to get these to be opposite signs which means I’m going to want to multiply this middle one by 2 and then pair it with the top. If you do with the bottom one it would be just fine the same. What that means is the top one is going to stay the same.

We put 2a minus 4b plus 6c is equal to 5 and our middle term we’re multiplying by 2. -2a plus 6b minus 4c is equal to negative two. We want to get rid of our a term to match up the a term we already got rid of so we want to add. Our a’s cancel, -4b plus 6b is 2b, -4c minus 4c is 2c and 5 and -2 is equal to 3.

We now have two equations in two variables., we have this one we just found right here and the one we found over here. Again we want to use elimination. I see that I have 2b and 2c and b and c. So what I’m going to do is multiply this equation over here by 2 which leaves us with 2b plus 2c is equal to zero. In order to get rid of my b or my c term I need to then subtract. Our bs cancel, our cs cancel so what we’re actually left with on this side is zero and over here we’re left with 3. So we have zero is equal to 3. That is a false statement.

So what that tells us is that these points actually do not intersect at any set point. Two of them could intersect at a point, we don’t really have enough information to determine that but we know that all three planes do not come together at any single point or line or plane, so there really is no solution to this entire problem.