Video: Solving Simultaneous Using Elimination, Where Both of the Equations Needs to be Multiplied

Using elimination, solve the simultaneous equations 4𝑥 + 6𝑦 = 20, 3𝑥 − 4𝑦 = −2.


Video Transcript

Using elimination, solve the simultaneous equations for 𝑥 plus six 𝑦 equals 20 and three 𝑥 minus four 𝑦 equals negative two.

So when we look into this question, the key thing is this: it tells us to use elimination. So it doesn’t matter if you know any other methods, you’ve gotta use elimination to actually solve this problem. So in order to actually use elimination, what we need to have is either the same number of 𝑥s and same number of 𝑦s, so same coefficient of 𝑥 and same coefficient of 𝑦.

So to enable this to happen, what we’re actually gonna do is actually multiply the first equation by three and the second equation by four. So when I multiply the first question, I get 12𝑥 plus 18𝑦 equals 60. And what we’ve got to do here is to make sure that we actually multiply every term, a common mistake is actually to just multiply the 𝑥 term that we’re actually looking at.

Okay! And I’ve also labeled that equation one. And then what I’ve done is I’ve actually multiplied the second equation by four, like I said, and I get 12𝑥 minus 16 𝑦 equals negative eight. Okay, so now we have two equations and they’re actually labeled one and two. So now what I’m gonna do is actually subtract equation two from equation one. And I’m gonna do that so that I can actually eliminate the 𝑥 terms.

And I know that actually it’s going to be a subtraction because the 𝑥 terms or the coefficient of 𝑥 have the same sign; they’re both positive. And if it’s the same signs, we subtract. However, if they had different signs, so let’s say for instance in this case we’re dealing with coefficients of 𝑦, then we’ll actually add because you’d have to add to actually make them eliminate each other.

Okay, so great! Let’s get on and actually subtract equation two form equation one. Well first of all, we’re just gonna have zero because 12𝑥 minus 12𝑥 is actually zero. And then we’ve got positive 34𝑦, and that’s because positive 18𝑦 minus negative 16𝑦 is gonna give us positive 34𝑦. And then this is gonna be equal to 68 because you’ve got 60 minus negative eight. And again, if you minus a negative, it turns it positive. So we get 68. And then what we’re gonna do is actually divide each side of the equation by 34.

And when we do that, we’re gonna find our 𝑦 value. So we get 𝑦 is equal two. So great we actually now know what 𝑦 is. So now we move on to the next stage. We need to find 𝑥. Okay, so what we’ll do now is actually substitute 𝑦 equals two into equation three. I’ve actually just labeled one of our original questions equation three for doing this. You can actually substitute it into any of them, so equation one two or the other one that we haven’t used from the beginning.

I’ve just chosen this one because it’s gonna be the easiest to work out, and with less multiplying by higher factors. Okay, so when we do this, what we get is that there three 𝑥 minus four multiplied by two, because this is where our 𝑦 was equal to two, is equal to negative two. So you get three 𝑥 minus eight is equal to negative two. And then we add eight to each side of the equation, and we get three 𝑥 is equal to six. That’s cause negative two plus eight is six.

And then we actually divide each side of the equation by three. And then we’re left with 𝑥 is equal to two. So therefore, we could say that using elimination to solve the simultaneous equations four 𝑥 plus six 𝑦 equals 20 and three 𝑥 minus four 𝑦 equals negative two, you get the answers 𝑥 is equal to two and 𝑦 is equal to two.

And what we can do is that actually just double check that our answer is correct. And we can do that just by substituting the values into the first equation that we were given. So if we do that, we get four multiplied by two plus six multiplied by two is equal to 20, which would give us eight plus 12 is equal to 20. Well, this is correct. So great, yea. We know the values that 𝑥 equal two and 𝑦 equal two work.

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