 ###### 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 a Solution - 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 in three variables. So solving a system in 3 variables is a lot like solving a system in two except obviously we have another variable we’re dealing with. For this particular example, we have three equations, however in one of them we are missing a variable. So in this bottom equation as you can see we don’t have a b-term. What we want to do is we actually want to make two equations in two variables. The nice thing about that is we already have one of them. We already have this equation down here has no b-term, what we need to do is combine the other two equations to get an equation without b as well.

We want to somehow combine these two to get rid of b. Using elimination we can see we have a 2 and a -4 so if we multiply the top equation by 2, wind up both with 4 and we can get rid off our b term. So let’s multiply this top by 2. That top equation then turns into 8a plus 4b, minus 6c is equal to 12, middle equation then stays the same, a minus 4b plus c is equal to -4. I always throw a giant plus or minus sign in front just to make sure I’m keeping track of what I’m doing. Here I have 4 and -4 so if I add these they cancel off, so we throw a giant plus sign in front.

Use elimination. 8a plus a, 9a, our 4bs cancel out, -6c plus c is -5c, 12 minus 4 is 8. So we now two equations; we have the one that we started with and this one we just made. Let’s rewrite the one we started with right below here.

So our first equation was –a plus 2c is equal to 2. We now have two equations in two variables. Using elimination or substitution we need to get rid of one of them. I’m partial to elimination so what I see is I can easily multiply this bottom equation by 9 and I can get rid of my a term. So let’s go through that.

Multiply this by 9, our top equation still is the same, so we’re still left with 9a minus 5c is equal to 8 and our bottom equation now turns into -9a plus 18c is equal to 18. Again make sure our sign is out in front, we want to cancel this off so we’re going to be adding. Our a cancel, -5 plus 18 is 13c, 18 plus 8 is 26, so we end up with c is equal to 2.

So we found one number, we still have two to go, we still have a and b. So what we can do is take this c and plug it back into any equation. I see that this equation right here before we multiplied by 9 is pretty straight forward. So what we have then is –a plus 2 times 2 is equal to 2. –a plus 4 is equal to 2, subtract 4, -a is equal to -2, a is equal to 2. So we now have c and a, both equal to 2.

Last thing we need to do is substitute it back into one of our original equations to find b. We copy one of our original equations right over here; it’s a pretty straight forward equation so let’s plug everything in. We have a minus 4b plus c is equal to -4. Just plug in 2 and 2 for a and c. 2 plus 4b plus 2 is equal to -4. 2 plus 2, 4 minus 4b is equal to -4, subtract the 4 over, -4b is equal to -8, giving us our third variable as 2 as well.

So what that tells us is our answer to our equation is the point (2, 2, 2). So going through our steps of getting our two equations in two variables. For this particular problem we already had one equation in two variables so we only had to put the other two together to eliminate that same variable creating a system, solving it out, getting our answer.