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Solving a Linear System in Three Variables with a Solution  Problem 1
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Solving a system in three variables: Solving a system in three variables is pretty similar to solving a system in two, except we have this extra letter we’re dealing with. Actually how we need to go through it is to fist make a system in two variables which you know how to solve.
We can do that a number of ways; we can do it through substitution however that tends to get ugly and I prefer to do elimination only. So what we need to do is somehow take these 3 equations and 3 variables and make two equations in two. How we do that is we choose a variable to get rid off.
So by looking at these xs we have a 15 and a 2 and a 1, it’s going to be hard to get rid of these xs. The ys, a little bit easier because we have some common multiples however the zs, we have a 1, a 2 and a 2, those are going to be pretty easy to get rid off.
What we want to do is form two of these equations, pair them together to get rid of that z. Right away I can look at the middle and the bottom and see that if I add these two equations together I can get rid of z. So we pair these two up, x plus 4y plus 2z is equal to 16. 15x plus 6y minus 2z is equal to 12 and if we just add these together our zs cancel. I always throw a sign out in the front so I just keep track of what I’m actually doing. We add this together, x plus 15 x is 16x, 4y plus 6y is 10y, our zs cancel and we end up with 28.
So we have an equation in two variables; x and y. In order to solve this out, we need to make one more equation that doesn’t have a z as well. So going back to our original equations, we somehow need to pair two different equations to get rid of z. We can either pair the top and the bottom or the top two. I’m going to pair the top two, although you could pair the others just as well. Let’s pair these together.
In order to get rid of z, we need to then get this to be a two as well, so we need to multiply the top one by 2. Multiplying that by two we end up with 4x plus 2y minus 2z is equal to ten. And our middle equation stays the same; x plus 4y plus 2z is equal to 16. And again you want to add, so I’m going to throw my giant plus sign out in front, we add these up. 4x plus x is 5x, 2y plus 4y is 6y, our two zs cancel, 10 plus 16 is 26.
We now have a system that has two equations and two variables, our z is gone. What we want to do now is just like what we did when we have two equations and two unknowns. I prefer elimination so I'm going to go ahead and do that. This x looks a lot like a y, let’s clean this up a little bit.
Before I do that, what I notice is that all these numbers are even so we can actually make our life a little bit easier and divide by 2. So we end up with 8x plus 5y is equal to 14 and 5x plus 6y is equal to 26. So we want to get rid of 1 variable. You can either get rid of x or y, it doesn’t really make a difference, the numbers of y are a little bit smaller so I’m going to go ahead and get rid of those.
These come in multiples of 5 and 6, is 30, so I’m going to multiply the top by 6 and I’m going to multiply the bottom by 5. You don’t necessarily need the negative but I prefer to add I find that a lot less mistakes are made when you’re adding than when you’re subtracting and you forget to distribute that negative sign. So by multiplying through you end up with 48x plus 30y is equal to 84, 5 times 5 , 25x, minus 30y and 26 times 5 do that on the calculator just to make sure that we get it right, 130. Again we multiplied by negative so we could add, throw our giant negative, sorry, addition sign in front and solve this out. 48 minus 25 is 23x, our ys cancel and then we’re left with 84 plus 130 equals 46 telling us that x is 2.
So we found one point, what we still need to do is find the other two. So what we do is called back substitution. You plug this back into other equations so here I have x equals 2 and that can go into any equation that I have that has just x and another variable. So this one is right here, let’s plug that in there, so we can say, just coming over here 5 times what we had for our x, 2 plus 6y is equal to 26 and solve for y. 10 plus 6y is equal to 26, add the 10 over, 6y is equal to 36 giving us y is 6.
So we now have x and y, we still need to find z, so what we could do is take these two things and plug it into any of our beginning equations, way back over there. Let’s go back over here, and we can plug it into any equation that we want. Say the top one, has pretty small numbers, keep it as easy as possible. So we have 2, the x we found was 2 plus the y we found was 6 minus z is equal to 5. Simplifying everything, 4 plus 6 minus z is equal to 5, 2 minus z is equal to 5, negative z is equal to 3, z is equal to 3.
We went through everything, just putting all our answers together we end up with we had 2 for x, 6 for y and 3 for z. So long process but what we ended up doing was taking our system in 3 variables, combining a couple of equations to get rid off a single variable, did that twice so we had two equations in two variables, elimination like we did when we had two variables and then back substitution all the way through to end up with a single point.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
Sample Problems (4)
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Solving a Linear System in Three Variables with a Solution
Problem 1 8,660 viewsSolve the system:
2x + y − z = 5x + 4y + 2x = 1615x + 6y − 2x = 12 
Solving a Linear System in Three Variables with a Solution
Problem 2 4,735 viewsSolve the system:
4a + 2b − 3c = 6a − 4b + c = 4a + 2c = 2 
Solving a Linear System in Three Variables with a Solution
Problem 3 536 views 
Solving a Linear System in Three Variables with a Solution
Problem 4 538 views
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