### Video Transcript

Solve the simultaneous equations two 𝑥 plus three 𝑦 plus two 𝑧 equals 21 over five, negative six 𝑥 minus two 𝑦 plus seven 𝑧 equals negative 67 over five, and 𝑥 plus five 𝑦 plus three 𝑧 is equal to 22 over five.

So, the first thing to look at is the fact that we’ve got three different equations. And in these three equations, we have three variables, 𝑥, 𝑦, and 𝑧. And the first thing I’m gonna do is label our equations so we can follow what we’re doing, so equation one, equation two, and equation three. Now, to be able to solve these simultaneous equations, what we first want to do is work out how we can eliminate one of our variables.

Well, what we need to look at is for where we’ve got some coefficients that are the same for some variables, for instance if we got the same amount of 𝑥s, 𝑦s, or 𝑧s in two of our equations. Well, in fact, we don’t. But what we can do is we can multiply an equation to make one of our variables equal. So, for example, if I look at equation one, we’ve got two 𝑥s. If I look at equation three, we’ve got one 𝑥. So therefore, if I multiply all of equation three by two, then I’m gonna have two 𝑥s, and I can look to eliminate 𝑥s.

So, when I multiply equation three by two, I get two 𝑥 plus 10𝑦 plus six 𝑧 is equal to 44 over five. And I’ve called this equation four. Make sure though you do multiply the right-hand side of the equation by, in this case, two as well cause often this is forgotten but needs to be every single term in our equation was multiplied by two. So, now, what we can see is that we’ve got two 𝑥s in both equation one and equation four. So, what I’m gonna do is use the elimination method. And I’m gonna subtract equation one from equation four. And I’m gonna do this because this means we’re gonna be able to eliminate our 𝑥s.

Well, if I subtract equation one from equation four, I’m gonna have two 𝑥 minus two 𝑥, which is just zero. Then, 10𝑦 minus three 𝑦, which is seven 𝑦. Then, we’ve got six 𝑧 minus two 𝑧, which is four 𝑧. Then, finally, we have 44 over five minus 21 over five. Well, if we’re gonna subtract two fractions with the same denominator, we just subtract the numerators. So, 44 minus 21 gives us 23. So therefore, we’ve now got 23 over five, so we can say that seven 𝑦 plus four 𝑧 is equal to 23 over five.

So, great, we’ve now got an equation that hasn’t got one of our variables. So, we’ve only got 𝑦 and 𝑧. However, this isn’t very much use at the moment because we need another equation that only has 𝑦 and 𝑧 in before we can solve to find our variables. So, once again, what I’m gonna do is take a look at our 𝑥 terms. I could have looked at 𝑦s or 𝑧s, but I’m just concentrating with the 𝑥s because it seems the easiest at the moment because we’ve got two, which is a multiple of six.

So, now, I can see that in equation one, I’ve got two 𝑥. In equation two, I’ve got negative six 𝑥. So, what I’m gonna do is multiply equation one by three. And when I do that, I’m gonna get six 𝑥, cause three multiplied by two 𝑥 is six 𝑥, plus nine 𝑦 plus six 𝑧 is equal to 63 over five. And I got the 63 over five because I had 21 over five multiplied by three gives us 63 over five. Okay, great, this is now equation five. So, now, what I’m gonna do to eliminate our 𝑥 term is add equation five and equation two.

And I’m gonna do that because our coefficients of 𝑥 are different signs; we’ve got negative six 𝑥 and positive six 𝑥. And if we add negative six 𝑥 and positive six 𝑥, we’re gonna get zero. So, we’re gonna eliminate our 𝑥 term. So, we’re gonna get seven 𝑦. And that’s because if we have nine 𝑦 add negative two 𝑦, that gives us seven 𝑦. Then, we’ve got add 13𝑧, and that’s cause six 𝑧 add seven 𝑧 is 13𝑧, is equal to negative four over five. And that’s cause we had negative 67 over five. And if we add on 63 over five, we get to negative four over five.

So, what we’ve got now are two equations. And I’ve labeled them six and seven, just again so we can see what we’re doing. And these two equations only have two variables, 𝑦 and 𝑧. And so, what we’re gonna do is we’re gonna use the elimination method once more to find out what our variables are. So, if we take a look at equation six and seven, we can see that they both got positive seven 𝑦. So therefore, what I’m gonna do is subtract one from the other. And I’m gonna subtract equation six from equation seven.

So, if I do that, I’m gonna have seven 𝑦 minus seven 𝑦, which is zero. Then, I’ve got 13𝑧 minus four 𝑧, which is nine 𝑧. And then, I’m gonna have negative four over five minus 23 over five, which is gonna be negative 27 over five. So, great, we’ve now got nine 𝑧 equals negative 27 over five. So therefore, if I wanna find out what 𝑧 is, what I’m gonna do is divide each side of our equation by nine. And when I do that, I’m gonna get 𝑧 is equal to negative three over five. So, we can think about it as having negative 27 over five. Well, divide 27 by nine to get three, so we’re left with negative three over five.

Okay, great, we found 𝑧. Now, let’s see if we can use this to find 𝑦. So, to find 𝑦, what I’m gonna do is substitute 𝑧 equals negative three over five into equation six. I could also have done it into equation seven, but I’ve just chosen to do it into equation six. So, when I do that, we’re gonna get seven 𝑦 plus four multiplied by negative three over five equals 23 over five. And when we do that, we’re gonna get seven 𝑦 minus 12 over five is equal to 23 over five.

So, then, if we add 12 over five to both sides of the equation, we’re gonna be left with seven 𝑦 is equal to 35 over five. And now, we divide both sides of the equation by seven. And then, when we do that, what we’re gonna get is 𝑦 is equal to five over five. So, 𝑦 is just equal to one. We could’ve written in a step above seven 𝑦 is equal to seven. And that’s because 35 over five is seven. But either way we end up with the same answer, which is 𝑦 is equal to one.

So, great, we found 𝑧, we found 𝑦, now let’s get on and find 𝑥. So, now, I’ve cleared a bit of space, but we’ve still got that 𝑦 is equal to one and 𝑧 is equal to negative three over five. So, now, to find 𝑥, what we’re gonna do is substitute both these values back into equation one. We could substitute them into any of our equations. I’ve just chosen to do it into equation one. So, when we do that, what we’re going to get is two 𝑥 plus three multiplied by one plus two multiplied by negative three over five is equal to 21 over five.

So, this is gonna give us two 𝑥 plus three minus six over five equals 21 over five. So, what I’ve done is I’ve converted our three into fifths, and that’s 15 over five because three multiplied by five is 15. So, what I’ve got is two 𝑥 plus 15 over five minus six over five is equal to 21 over five. So therefore, if I simplify, I’m gonna get two 𝑥 plus nine over five is equal to 21 over five. So, then, if I subtract nine over five from each side of the equation, I get two 𝑥 is equal to 12 over five. So, then, divide both sides of the equation by two, and I get 𝑥 is equal to six over five.

So, great, I’ve now found 𝑥, 𝑦, and 𝑧. Because I can say that the solutions to our simultaneous equations are 𝑥 equals six over five, 𝑦 equals one, 𝑧 equals negative three over five. Well, now, what I can do is run a little check. And the check I can run is by substituting our values into another one of our equations. So, I’m gonna sub it into three. So, when I do that, I get six over five plus five multiplied by one plus three multiplied by negative three over five is equal to 22 over five.

So, that’s gonna give me six over five plus five minus nine over five equals 22 over five. Well, now, if I convert the five into fifths, we’re gonna have 25 over five. So, we got six over five plus 25 over five, which is 31 over five, minus nine over five is 22 over five. So therefore, the three values do work, and we could say that we are correct, and that the solution to the equations is 𝑥 equals six over five, 𝑦 equals one, and 𝑧 equals negative three over five.