Video Transcript
In this video, we’re gonna use the algebraic process of elimination to solve two simultaneous linear equations. We’ll go through to some examples and we’ll built from simple questions to more complicated ones that require a bit more manipulation, and have even fractional or negative or even zero solutions.
So example one: Use algebraic elimination to solve these simultaneous equations. And we’ve got 𝑎 plus three 𝑏 equals seven and 𝑎 plus 𝑏 equals three.
So we’ve got two separate equations and they’re linear equations cause they haven’t got any 𝑎 squareds, or 𝑏 squareds, or 𝑎 cubes, or 𝑏 cubes, or anything like that. And each individual equation has an infinite number of solutions. So for example, if we took the first one 𝑎 plus three 𝑏 equals seven, if I put 𝑎 equal to one and 𝑏 equal to two, I’d have one plus three lots of two, one plus six equals seven, so that would work. I could put 𝑎 equal to four and 𝑏 equal to one and that would work. Or even 𝑏 equal to zero and 𝑎 equal to seven. I could put 𝑎 equal to eight and then 𝑏 equal to negative a third and that would work as well. In fact, I could put any value in for 𝑎 and I could generate a version of 𝑏 which would make that equation work. So as I say, we’ve got an infinite number of solutions to this.
And likewise, I’ve got an infinite number of solutions to the other equation 𝑎 plus 𝑏 equals three. Whatever I- whatever value I put in for 𝑎, I can come up with a value of 𝑏 which would make their sum equal to three.
For example, 𝑎 is a hundred, 𝑏 is negative ninety-seven; add them together you get three. And this is what happens if you’ve got one equation with two unknowns in it, then there are gonna be an infinite number of combinations of those two variables which will make that equation true. Now what we’re looking for here on these simultaneous equations, both of these equations have to be true. So we’re looking for a pair of 𝑎- and 𝑏-values which will be true in this one, and the same pair of 𝑎- and 𝑏-values will be true in this one. And there may be one solution; there may be two solutions, who knows. In this particular case, I think we’ll find there’s only one solution.
So before we go and solve this algebraically, let’s just take a look at the graphs of these and then-then I’ll tell you why I think there was only one solution. So using the 𝑥-axis as our 𝑎-axis and the 𝑦-axis as our 𝑏-axis, for example we’ve got 𝑎 plus 𝑏 equals three here. Now I could put any value I like in for 𝑎. I could put a value of four in for 𝑎. And if I read off the corresponding coordinate, I’ve got a-a 𝑏-value of negative one. I could have an 𝑎-value of negative one and that would generate a 𝑏-value of four. And that works for the other equation. So 𝑎 plus three 𝑏 equals seven. So if I put in an 𝑎-value of five, my 𝑏-value, if I read this off, is about nought point six and a bit. If I put in an 𝑎-value of negative two, reading off my 𝑏-value, that looks like it’s gonna be three.
So I think that you see there’re an infinite number of pairs of coordinates, if you like, along each of those lines. They give us the solutions that match those particular equation, so a whole bunch of different pairs of 𝑎- and 𝑏-values along the blue lines- the blue line, which matches that equation. But there’s one point here where those lines crossover, those straight lines crossover, where the 𝑎 and 𝑏 combinations will work on the blue line and they will also work on the red line. That’s what we’re looking for in this simultaneous equation. Now in this particular case, it looks like we’re gonna get an 𝑎-value of one and a 𝑏-value of two. And that’s probably gonna be our exact answer. But of course it’s not always gonna be exact; it’s not always possible to read exact answers on our graphs. So let’s forget that we saw the graph and now more look at this algebraic method, which is hopefully always gonna give us a hundred percent accurate answer.
Now when I do simultaneous equation questions, I always do like a little curly brace like this, to indicate the fact that they’re simultaneously true. You don’t have to do that, not all places do that. But I quite like that method and I always number my equations so that I can give a running commentary of what I’m doing in my working out, and refer to those numbers. So you’ll see what that looks like as we go through. Now the way these equations are laid out, I’ve got 𝑎 plus three 𝑏 is equal to seven in the first one, and I’ve got 𝑎 plus one 𝑏 is equal to three in the second one. So they’re sort of laid out so that the 𝑎s are above each other, the 𝑏s are above each other, and then the numbers on the other side of the equation are above each other. So what I could do is, I could take equation one and then I could subtract equation two from that, term by term. So let’s have a look at what that would look like.
So equation one take away equation two, so I’m gonna have 𝑎 take away 𝑎 in the first column; so 𝑎 take away 𝑎. And then I’m gonna have positive three 𝑏 take away 𝑏 in the second column; so positive three 𝑏 take away 𝑏. And on the right-hand side, I’m gonna have seven take away three. Now 𝑎 take away 𝑎 is nothing, so that process eliminated 𝑎. So we’ve created a new equation here and we could call it equation three, which has eliminated 𝑎. Because we had the same number of 𝑎s in each, if I take one away from the other I’ve got no 𝑎s left. Three 𝑏 minus 𝑏 is two 𝑏, and seven take away three is four.
So doing equation one and subtracting equation two from it has eliminated 𝑎 and left us with a new equation, which we’ll call equation number three; two 𝑏 is equal to four. Now if I divide both sides of that equation by two, a half of two 𝑏 is one 𝑏, so just 𝑏 and a half of four is two. So we now know that 𝑏 is equal to two.
So now I could substitute that value of 𝑏 into either equation two or equation one, to find out what the corresponding 𝑎-value would be. Now I’m gonna substitute it into equation two because the numbers look a bit smaller. It’s just 𝑎 plus 𝑏 equals three. So it’s just a small advantage, but it ma- it’s gonna make the math a little bit easier. So substituting the value 𝑏 equals two into equation two. Equation two begins with 𝑎 plus 𝑏. Well we know that 𝑏 equals two so instead of writing two, I’m just gonna- instead of writing 𝑏, I’m just gonna write two. And that’s equal to three, so 𝑎 plus two equals three. Now I’ve got an equation 𝑎 plus two equals three. I can take away two from both sides of that equation. So on the left-hand side, 𝑎 plus two take away two is just 𝑎 and on the right-hand side, three take away two is one. So I’ve got 𝑎 equals one.
Well that’s great cause that tailored with the graph that we saw earlier, but you gotta remember we wouldn’t normally have seen the graph by now. So we need to have a way of just checking that our answer is correct. And we used equation two to do some substituting of 𝑏, so we’ve got equation one that we haven’t really played around with yet. So I’m gonna substitute the values 𝑎 and 𝑏, so 𝑎 is one, 𝑏 is two, back into equation one now just to do a check. So the value of 𝑎 is one and then it’s plus three 𝑏, so that’s plus three times the value of 𝑏 and the value of 𝑏 is two. And we’re hoping that that’s gonna equal seven. Well one and three times two is six, one plus six does in fact equal seven. So yep, we’re happy that that’s the right answer. So all that remains is to put that answer there in a nice box and make it lovely and clear. Okay. Let’s go on to the next question.
And now we’ve gotta use algebraic elimination again to solve the simultaneous equations three 𝑎 plus two 𝑏 equals seventeen and four 𝑎 minus two 𝑏 is equal to four.
So I’m gonna put my braces around them to show they’re simultaneous and I’m gonna number them one and two. Now if I do equation one take away equation two, I’m gonna have three 𝑎 take away four 𝑎, which is gonna give me negative 𝑎. So I’m not gonna eliminate the 𝑎s. And if I did two 𝑏 take away negative two 𝑏, that’s the same as two 𝑏 add two 𝑏, that’ll be four 𝑏. So I’m not gonna eliminate 𝑏. So I’m just gonna come up with another equation with two variables in it. So that’s not gonna work. But looking at these equations, I’ve got the same number of 𝑏s. There are two 𝑏s in each, but in one case it’s positive two 𝑏, the other case it’s negative two 𝑏. So if I add the two equations together, I’m gonna get two 𝑏 plus negative two 𝑏. That’s gonna be no 𝑏. So that’s gonna cancel out the 𝑏s. So that’s what I’m gonna do.
So three 𝑎 plus four 𝑎 is seven 𝑎, two 𝑏 add negative two 𝑏 is nothing, and seventeen plus four is twenty-one. So seven 𝑎 is equal to twenty-one. I want to know what one 𝑎 is, so if I divide both sides by seven, I get 𝑎 is equal to twenty-one divided by seven is three; 𝑎 is equal to three. So I’m now gonna substitute that back into either equation, one or two, in order to work out the value of 𝑏. So I’m gonna go for equation one. Just for the sake of argument, it doesn’t really matter which way around you go. So 𝑎 is three. So that first term there is three times three plus two 𝑏 is equal to seventeen, so that’s nine plus two 𝑏 is equal to seventeen. So if I take away nine from each side of that equation, I’ll just leave the 𝑏 term on it’s own on the left. So as I say nine take away nine leaves me no number, so that’s just the two 𝑏. And seventeen take away nine leaves eight. So now I can just divide both sides by two, and I’ve got 𝑏 equals four. So there’re my answers: 𝑎 is three and 𝑏 is four.
Now I used equation one to do my substitution for 𝑎, so I’m gonna use equation two to do my check now. So all being well, four 𝑎, so that’s four times three minus two 𝑏; so that’s two times four is equal to four. Well four times three is twelve and two times four is eight. And twelve minus eight does in fact equal four. So we’re pretty confident we’ve got the right answer.
Number three then: Use algebraic elimination to solve the simultaneous equations two 𝑎 plus four 𝑏 equals ten and four 𝑎 minus 𝑏 equals eleven.
So let’s label up our equations, one and two. Now when we look at the terms, I’ve got two 𝑎 in the first and four 𝑎 in the second; I’ve got four 𝑏 in the first and negative 𝑏 in the second. If I just add or subtract equations one or two, I’m not gonna eliminate either of those variables. I’m gonna, you know, end up with some 𝑎s and some 𝑏s in my equation. So what I need to do is generate a new equation that’s either got the same number of 𝑎s or the same number of 𝑏s, as one of the other equations. Now if multiply everything in the first equation by two, I’ll have four 𝑎 plus eight 𝑏 equals twenty. But I’ll have four 𝑎, so I’ll have the same number of 𝑎s as I did in equation two.
So let’s call that equation three. Now you’ll also notice that I could’ve multiplied everything in equation two by four and that would’ve given me sixteen 𝑎, which is different. But it would’ve given me negative four 𝑏, so that would’ve worked just as well. But I’d rather multiply everything by two than multiply everything by four, just to get the numbers smaller. So that’s what we’ve done.
So equations two and three have now got the same number of 𝑎s; they both got four 𝑎s. So I’m gonna take one equation away from the other to eliminate the 𝑎s. Now I could either do equation two take away equation three, or I could do equation three take away equation two. Both of those would eliminate the 𝑎s. But if I do three minus two, then I’m gonna have eight take away negative 𝑏, so I’m gonna have nine 𝑏. If I did it the other way around, I’d have negative 𝑏 take away eight, so I’d have negative nine 𝑏. And I’d rather end up with a positive number of 𝑏s on one side of my equation, cause that just makes subsequent steps a little bit easier. So that’s what we’re gonna do. Equation three take away equation two.
So as we say four 𝑎 minus four 𝑎 is nothing, eight 𝑏 take away negative 𝑏 is positive nine 𝑏, and twenty take away eleven is nine; so we’ve got nine 𝑏 is equal to nine. Now I can divide both sides by nine, so that I get the result of 𝑏 is equal to one. And I can substitute that back into equation one, two, or three in order to find out the next bit of the answer, to find out the value of 𝑎. Now I’m gonna substitute it into equation two, which is gonna give me four 𝑎 take away one is equal to eleven. And if I add one to both sides of that equation, four 𝑎 take away one plus one is just four 𝑎 and on the right-hand side, eleven plus one is twelve. Now dividing both sides by four, I get 𝑎 equals three. And I used equation two so I’m gonna check my answers in equation one. So substituting in three for 𝑎 and one for 𝑏, I’m hoping that two times three plus four times one equals ten. Well two times three is six and four times one is four and six plus four is ten. So hopefully you’ve got the right answer.
So number four: Use algebraic elimination again to solve the simultaneous equations three 𝑎 minus five 𝑏 equals fifteen and two 𝑎 plus seven 𝑏 equals forty-one.
Well I can’t simply subtract one from two or two from one to eliminate 𝑎 or 𝑏 and I’m- neither of them are simple multiples of the other. So I can’t just multiply one equation by something to get the same number of 𝑎s or 𝑏s as the other one, well not whole numbers anyway. So what if I multiply the first equation by two and the second equation by three? Well doubling everything in the first equation three 𝑎 times two is six 𝑎, minus five 𝑏 times two is minus ten 𝑏, and fifteen times two is thirty. And three times the second equation gives us six 𝑎 plus twenty-one 𝑏 is equal to a hundred and twenty-three.
Now let’s call those equations three and four. So now I’ve got the same number of 𝑎s in each, and I’m gonna be able to subtract one of those equations three and four from the other, in order to eliminate the number of 𝑎s. Now just before we go on, just to bear in mind, we could’ve multiplied equation one by seven and equation two by five, and then we we’d have had thirty-five 𝑏s in each case. And we could’ve eliminated 𝑏. But I thought it was easier to multiply it by two and three, than it is to multiply it by five and seven cause it will leave us with slightly smaller numbers. So are we gonna do equation four subtract equation three? Or are we gonna do equation three subtract equation four? Well my top tip will be four minus three. Because that means we’re gonna do six 𝑎 take away six 𝑎 is nothing, and twenty-one 𝑏 take away negative ten 𝑏 is the same as adding ten 𝑏 so that’s gonna be thirty-one 𝑏, and a hundred and twenty-three minus thirty is ninety-three. So we’ve ended up with a positive number of 𝑏s and a positive number on the right-hand side, so that’s gonna make life a little bit easier. Now I just need to divide both sides by thirty-one. And for those of you who are not so comfortable with your thirty-one times table, it- very often you’ll find in these questions that the numbers workout nice and easily. Three times thirty-one is ninety-three, so we end up with 𝑏 equals three. Now I’m gonna substitute that value for 𝑏 into equation one.
Now I could’ve substituted it into equation two just as happily, but the numbers looked a bit smaller in equation one; so I went for that one. So we’ve got three 𝑎 minus five lots of 𝑏 is three, so five lots of three is equal to fifteen. So just evaluating that five times three is fifteen so three 𝑎 minus fifteen is equal to fifteen. If I add fifteen to both sides, I’ll eliminate the negative fifteen from the left-hand side. So that’s gives us three 𝑎 is equal to thirty. And then dividing both sides by three, I get 𝑎 equals ten.
So let’s just quickly do a check. We used equation one to substitute, remember. So let’s use equation two to do our check. And it’s two 𝑎, so that’s two times ten plus seven 𝑏, so that’s seven times three. And we’re hoping that equals forty-one. And indeed twenty plus twenty-one does equal forty-one. So we’ve got our answer 𝑎 equals ten and 𝑏 equals three.
Now number five, I’m not gonna go through in great detail. We’ve got four 𝑎 minus two 𝑏 equals eight and six 𝑎 plus four 𝑏 equals negative nine.
So pause the video and have a good look, if you want to, to see all the working out. But I just wanted to highlight the fact that sometimes the answers can be fractions and sometimes the answers can be negative. And sometimes the answer can even be zero, as in this question here. And that’s something that often catches people out.
So just one more thing, sometimes you need to do a little bit of manipulation to get your 𝑎s and 𝑏s to line up properly in the first place, so like in this example. So I’ve done lots of rearranging just to get two equations with the same number of 𝑎s, and all the 𝑎s and the 𝑏s and the other numbers lined up properly. So we can then subtract some equations to eliminate 𝑎 and go on and solve the problem.
