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Hi, this is Julie Harland and I'm your math gal.
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Please visit my website at yourmathgal.com where you could search for any of my videos organized by topic.
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We're going to solve this system of linear equations with three variables using the substitution method.
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We already solved it in another video using the addition or elimination method,
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but I wanted to do the same problem showing you another way to do it.
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I label the three equations A, B, and C, so it's easier to refer to which equation I'm looking at.
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We look at these and they say, is it easy to solve for any of the variables here?
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In the second equation, it's easy to solve for X because the coefficient of X is 1.
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I'm going to take equation B here, and I'm going to solve for X.
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If we take that equation and solve for X, notice I would have to add to Y and subtract to Z from the other side,
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from both sides to get X equals to Y minus to Z plus 4.
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Hopefully that makes sense to you.
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I always like to put a star. This is something I'm going to go refer back to later.
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Once I know what Y is and I know what Z is, I'll be able to plug it in and figure out what X is.
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The idea of the substitution method is once you solve for one variable in one of the equations,
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you're going to plug it back into the other equations,
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and you'll be able to eliminate X from the other equations.
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We'll take equation A, and we're going to plug that in for X.
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Instead of writing X, I'm going to write 2y minus 2Z plus 4.
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That's 3x plus 2y plus C equals 1.
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We could simplify this side a little bit more.
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Let's do the distributive property. We'll add light terms, and then we'll get the variables at one side constants on the other.
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We've got 6y minus 6z plus 12 plus 2y plus Z equals 1.
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We're going to continue. We have 8y minus 5z plus 12 equals 1.
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Then we just subtract 12 from both sides to get 8y.
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I'm sorry. That was supposed to be 5z right here.
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8y minus 5z equals negative 11.
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I now have a new equation, and I'm going to go ahead and just call that D for now.
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I have a new equation once I plugged in this for X.
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I plugged it into equation A.
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I started without the equation B, and I solved for X.
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I'm plugging that into equation A.
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I also need to plug this 2y minus 2Z plus 4 in for X in equation C.
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That's my next step. We're going to take equation C, and we're going to do the same thing.
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We have 2X, so I need to write in place of X, 2y minus 2Z plus 4.
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Plus 4y plus 3Z equals 9.
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If you watched the way I did this using the addition method, I think it's a little bit easier to be honest.
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You have to keep track of a lot of things here. You have to have a lot of space going across each line.
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This is how you would do the substitution method. It kind of gets kind of lengthy each equation.
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But it's still doable.
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Now we distribute the 2, 4y minus 4Z plus 8 plus 4y plus 3Z equals 9.
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Then we're going to combine like terms here. We have 8y.
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The negative 4z plus 3z is going to be a minusc plus 8 equals 9.
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Then if I subtract 8 from both sides, we have 8y minusc equals negative 1.
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I'm sorry, equals positive 1.
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Subtracting 8 from both sides.
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Okay, so I have a new equation.
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C now looks a little bit different once I've replaced X with 2y minus 2Z plus 4.
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Now I have a new equation. Now, equation D and equation E each have only 2 variables, y and c.
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There's no X term.
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So now I can go back to, how do I solve those 2 equations for y and c?
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And this you've done before, you could again use the elimination or the substitution method.
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So let's go on from here. So here are our two new equations that I call D and E.
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And now we need to solve this system of equations to find out what y and c are using either substitution or the addition method.
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You could use substitution, but when I see this 8y and another 8y, it looks pretty easy just to multiply one of these equations by negative 1 and use the addition method.
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So that's the one I'm going to use. And it doesn't matter which equation you multiply by negative 1, I'm going to multiply the top one by negative 1.
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Now I'll tell you why I decided to do that. It's because then I would have a plus by Z and I know I'm going to have a positive number in front of the Z.
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But you could do it the other way or you could solve this system using substitution.
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Alright, so this gives me when I multiply the top equation by negative 1, each term gets multiplied by negative 1. Negative 8y plus 5z equals 11.
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And 8y minus c equals 1 because I didn't alter the second equation.
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And then when I add the y's cancel, so I have 5z minus c which is 4z equals 12 and dividing both sides by 4, Z is 3.
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So that is the solution for Z. So we can go back to either of these two equations, and plug in 3 for Z and go ahead and solve for y now.
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So I'm going to choose this second equation. I'm going to plug in 3 for Z. So I have 8y minus 3, right? I'm putting in 3 for Z equals 1.
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So adding 3 to both sides, we have 8y equals 4 and then dividing both sides by 8.
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We've got y equals 1.5. So now I know what y is. So I know y and Z. Now I've got to go back to one of the original three equations with three variables to figure out what X is.
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Or remember I did this three substitution. So I'm going to go back to where I said what X was in terms of y and Z.
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So remember so far I don't know what X is but I know that y is 1.5 and Z is 3. So let's look back up and see what we had here.
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And I started off. Here it is. We're going to look right there. So that's the equation I'm going to use, plugging in my values of y and Z.
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Alright so I write this equation. X equals 2y minus 2z plus 4. And I'm going to plug in 1.5 for y and 3 for Z. And then I'll find out what X is. So I have 2 times 1.5 minus 2 times 3 plus 4.
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So X equals 1 minus 6 plus 4. So that gives you a negative 1. Now I know what the X value is as well. So X is negative 1.
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So it looks like this here is the correct solution. But of course you don't know until you plug those values for X, Y, and Z into all three of the original equations.
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So let's look at how the check works. So now we're looking at three, the three original equations. So we're going to have to check to see that negative 1.5.3 works in all three equations.
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So I like to do it with a T bar. And so you're going to plug in negative 1 for X, 1.5 for Y, 3 for Z in the first equation. Simplify it and make sure you get the same number on both sides. And then you're going to do that for B. And then you're going to do that for C.
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So go ahead and put the video on pause and try that. And then I'll show you what it looks like. Okay, so we're going to plug it in. We're going to have 3 times negative 1 plus 2 times 1.5 plus 3, which is negative 3 plus 1 plus 3, which is 1.
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1 on both sides. So when I plugged in those values for X, Y, and Z in the first equation, it's true. So now we're going to go to the second equation. Negative 1 minus 2 times 1.5 plus 2 times 3 is negative 1 minus 1 plus 6, which is 4. We have 4 on the other side. So that checks.
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And the last one, we're going to plug in 2 times negative 1 plus 4 times 1.5 plus 3 times 3. Negative 2 plus 2 plus 9. That's a 9 as well. So it's true that when I plugged in those values of negative 1 for X, 1.5 for Y in 3 for Z, that's the correct solution.
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And so that order pair is the solution to this system. And I actually used a combination of the substitution and the addition method. I started off with the substitution method until I had got 2 variables. And then it seemed easier to use the addition method.
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But let's just go back up to that step where I had these two equations. 8, Y minus 5, Z equals negative 11 and 8, Y minus C equals 1. You could have done this using substitution. So if you want to see how that works, go ahead and keep watching the video. We're going to get the same answer.
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So we started off with these three equations, A, B, and C. And I solve for X here. From B, I was able to solve for X and write this equation. And then remember, we substituted that back into both A and C. And we got these two new equations.
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So this is 8, Y, minus 5, Z equals negative 11 and 8, Y, minus C equals 1. And now I can take equation E and solve for Z if I wanted. So if I solve for Z, what would I do?
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I would subtract one from both sides, giving me 8, Y, minus 1 and add Z to both sides. Or if you like to see it written as Z equals 8 minus Z equals 8, Y minus 1. I've now altered equation E so that I could use substitution.
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So see the first substitution, X equals 2, Y minus 2, Z plus 4, second substitution. Z equals 8, Y, minus 1. I plug that in for Z into, sorry, not that equation, the one above it, into the one I didn't use substitution on.
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Alright, so now I take equation D and I plug that in. So I have 8, Y, minus 5, times Z, which is now 8, Y, minus 1.
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So that equals negative 11 and I solve that equation. 8, Y, minus 40, Y, plus 5 equals negative 11. So that gives me negative 32, Y equals negative 16.
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See, divide by negative 32. So we have Y equals 1 half. So I finally solved for Y. Now we're going to just go back up one at a time to solve for Z and then to X. So we're going to go back up to Z.
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Okay, and let's call back substitute, put in the one half for Y. So now I solve for Z is 8, Y, which is 8 times a half, right?
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minus 1, which is going to be 4 minus 1 or 3. So now I know Z, okay? So keep in mind, I knew that Y was a half and now I know that Z equals 3.
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And now I go back up to here, put both of those in for X and I'll find out what X is. Now I already did that previously. We plug that in and we found out that X was negative 1.
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Okay, so just notice we're going to get the same three answers for X, Y, and Z. So the order pair is still going to be negative 1, 1 half, 3. The check is the same. So here we go. Our final answer is this order pair, negative 1, 1 half, 3.
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Please visit my website at yourmathgal.com where you can view all of my videos which are organized by topic.