# Video: Systems of Linear Equations — Elimination

Learn to use the method of elimination to find the value of two variables in a system of two linear equations. In one example, one of the variables has the same coefficient in both equations, and, in the other, neither variable has the same coefficient.

08:55

### Video Transcript

Systems of Linear Equations — Elimination

So we are going to solve this system of linear equations for 𝑥 and 𝑦 by using the method elimination. So there are steps that we need to follow absolutely every single time to be able to use this method. And the first step we need to do is to label the equations one and two.

And the reason we do this is to help us when we’re trying to work out what we’re actually talking about through the latest steps, so we know exactly we can pinpoint. Now the next thing we need to do is really check that the coefficients are the same in front of either the 𝑥s or the 𝑦s. So in this case, we can see that the coefficients aren’t the same in front of the 𝑥 but they are in front of the 𝑦s, so we check. And then if we check and they aren’t, we’re gonna look at that in the next example. So step two is check the coefficient are the same.

Step three is if the signs are the same, you subtract. So we can see that the signs are the same in front of the 𝑦s, so we’re going to subtract one equation from the other. And in this case, we can see it’s just easier as the second equation is larger than the first equation. We’ve got three 𝑥 and we’ve got nine and seven, so we’re going to do two minus one.

And this is really where our labelling starts to come in handy, so we’ve got three 𝑥 minus two 𝑥. Well three 𝑥 minus two 𝑥 is just 𝑥. 𝑦 minus 𝑦, they’ll obviously cancel out. That’s the reason we we’re doing this subtraction. And then we’ve got nine minus seven, which gives us two. So we now have the answer that 𝑥 is equal to two. Well step four will be put into one. So whichever variable we get here, either the 𝑥 equals something or 𝑦 is equal something or 𝑎 or 𝑏 or whatever variable we get, we substitute it back into equation one. So we can see two multiplied by two plus 𝑦 equals seven.

Well two times two is four. And then subtracting four from both sides, we get seven minus four we know is three, so 𝑦 equals three. Now we’re done there, but it is possible we may have made a mistake somewhere, so we want to make sure that we’ve definitely got it right. And the way that we’ll do that is by substituting in to two, so we’ll say check in two.

So we’re going to substitute our value of 𝑥 equals two and our value of 𝑦 equal to three into equation two to try and find that it works. So we are gonna have three multiplied by two, so that’s six, plus three, that’s the value of 𝑦, and that’s going to be equal to nine. So then is six plus three equals nine? Yes, it is. Give yourself two ticks; you know you’ve done it correct.

We’re gonna follow these same steps every single time with elimination, so we’re gonna look at another system of linear equations following the same steps to find the variables 𝑥 and 𝑦 again.

So now we must solve for 𝑥 and 𝑦 with this system of linear equations, and we may be able to see there’s a couple of things that are different. So first thing we have to do is label the equations as before.

But then this time, we’re checking for the same coefficients. Well there aren’t the same coefficients in front of the 𝑥s and same in front of the 𝑦s. So first of all, we want to see is it possible to multiply just one equation by a constant to get the same coefficients in front of the 𝑥s or the 𝑦s. And the answer to that is no. So we’re gonna have to do is multiply both equations by a constant. Well fi- we could multiply both equations by six and four, respectively, but that’s a larger number. So we wanna pick the smaller numbers always. So we’re going to have to multiply both the first equation by three, because that’s the coefficient in front of the 𝑦 for equation two, and then the second equation by two, because that’s the coefficient in front of the 𝑦 for equation one.

So we’re doing these multiplication to give us the same coefficients in front of the 𝑦s. So by doing each of them, we’ll label three multiplied by equation one. We’re gonna label that equation three. And then two multiplied by all of equation two, we’re gonna label that equation four.

So three now be really, really careful that you make sure to do multiplying by each term, because this is where people usually end up losing marks in the answers, by not doing the multiplication properly. So we’re doing three multiplied by six 𝑥, which gives us eighteen 𝑥; then three multiplied by two 𝑦 gives us six 𝑦; and then fifteen multiplied by three is forty-five.

So now moving on to equation four. We’re gonna multiply each term by two. So first, two multiplied by four 𝑥 is eight 𝑥; then two multiplied by three 𝑦 is six 𝑦; then two multiplied by negative three is negative six.

So now we finally completed step two. We have checked that the coefficients are now the same in front of the 𝑦s in fact because we’ve got six 𝑦 in both equation three and equation four. So we can finally go onto step three, which is if the signs are the same you subtract. Well the signs are the same so we are going to subtract. If it weren’t the same, then we’d have to add. So we can see that three is larger than four, so we’ll do equation three take away equation four.

So we’ve got eighteen 𝑥 take away eight 𝑥 gives us ten 𝑥. Of course the 𝑦s will cancel out because we have six 𝑦 minus six 𝑦, which is zero 𝑦. And then we’ll have forty-five minus minus six, be careful, so we add six. So it’s forty-five add six, which is fifty-one. And that gives us an 𝑥-value of 𝑥 equals fifty-one divided by ten, so five point one.

Then onto step four, put into one. So we look back to equation one and we put our 𝑥-value into it. So we’ve got six multiplied by five point one, which is thirty point six. Then we will add two 𝑦, and that is equal to fifteen. So then subtracting thirty point six from both sides, we get two 𝑦 equals negative fifteen point six. And then dividing that by two to get rid of the two 𝑦, you get that 𝑦 is equal to negative seven point eight.

Hey so moment of truth, we can now check whether this actually works. So we’re gonna check again in two, so we’ve got four multiplied by five point one. We’re adding three multiplied by negative seven point eight, and that should be equal to negative three.

So four multiplied by five point one is twenty point four and then three multiplied by negative seven point eight is negative twenty-three point four. So twenty point four minus twenty-three point four is equal to negative three, so we can give ourselves two ticks because we’ve got the 𝑥 and the 𝑦 correct. Although they seem like quite nasty numbers, actually they work for our system of equations.