Lesson Video: Solving Systems of Linear Equations Using Substitution Mathematics • 8th Grade

In this video, we will learn how to solve systems of linear equations using substitution.


Video Transcript

Here in this video, we’re gonna use the process of algebraic substitution to solve some linear simultaneous equations. It’s a process that isn’t always the most suitable method, but it can be really useful. So we’ll see some examples of when it works well and one or two where it actually makes life harder for us.

Now remember that every point on a straight line graph represents a unique and different solution to the equation that represents that line. So for example, this point here I’ve got an 𝑥-value. So if I put an 𝑥-value of four into that equation, the corresponding 𝑦-value will be one. If I put an 𝑥-value of seven into the equation, the corresponding 𝑦-value that makes that true will be zero. And as we said every single point on that line represents a unique combination of 𝑥- and 𝑦-values that when I do the 𝑥-value and I add three times the 𝑦-value, I’m gonna get the answer seven. And for the other line 𝑥 plus 𝑦 equals three, again every point on that line represents another unique combination of 𝑥- and 𝑦-values that will sum to three.

So 𝑥 is negative one, 𝑦 is four; they sum to three. 𝑥 is two, 𝑦 is one; they sum to three and so on. But what’s so special about simultaneous equations is there in this particular case when we’ve got two lines, there is one pair of 𝑥- and 𝑦-values that give- that match both of those equations. They give three when you add them together. But if you add three times the 𝑦-coordinate to the 𝑥-coordinate. They also give seven. That’s this point here on the graph where the two lines intersect. So solving simultaneous equations is all about finding where those lines intersect.

So here’s a question.

Use algebraic substitution to solve the simultaneous equations 𝑦 equals three 𝑥 minus two and 𝑥 equals three 𝑦 minus ten. So they’ve told us the specific method we’ve got to use and they’ve told us that these two equations are simultaneous. Now it’s always good when you do number your equations so you can refer to those numbers when you’re explaining your working out. But I like to put a little brace to indicate that those two things are simultaneously true as well. And not everybody does that, but it’s I just find it’s a good idea to communicate this idea or simultaneousness of the equations. Now the thing about substitution is in the first equation, we’ve got the fact that 𝑦 is equal to this bunch of stuff here three 𝑥 plus [minus] two. So substitution is all about saying- well in the second equation, we’ll replace 𝑦 with all that bunch of stuff because those two things are true at the same time. So whilst 𝑦 is represented in the second equation, we also want it to be equal to three 𝑥 minus two. So in this case we’re gonna take the value that 𝑦 has in equation one and substitute 𝑦 in the second equation.

And we could have done that the other way around if we wanted to. Equation two says that 𝑥 is equal to three 𝑦 minus ten. So we could’ve replaced 𝑥 in the first equation with three 𝑦 minus ten. So it doesn’t matter which way around you do it as long as you’re completely replacing the letter that you’re trying to get rid of. So now we’ve got an equation here. Let’s call that equation number three, which is purely in terms of 𝑥. So we’re gonna be able to find a unique solution to this for 𝑥. So using the distributive law of multiplication, 𝑥 is equal to three lots of three 𝑥; so that’s nine 𝑥 and three times negative two which is negative six. And then we’ve got to take away ten as well. So 𝑥 is equal to nine 𝑥 minus six minus another ten; that’s nine 𝑥 minus sixteen. Now I want to get all the 𝑥s onto one side. And it’s generally speaking a good idea to have a positive number of 𝑥s when we do that. So I’m gonna just subtract one 𝑥 from both sides leaving me with eight 𝑥 on the right-hand side and no 𝑥s on the left-hand side. So we’re subtracting 𝑥 on both sides and as we said 𝑥 minus 𝑥 on the left-hand side, it gives us zero. And nine 𝑥 take away 𝑥 on the right-hand side leaves us with eight 𝑥. Now I can add sixteen to both sides to get rid of that number from the right-hand side. And on the left-hand side, zero plus sixteen is just sixteen. And on the right-hand side, negative sixteen plus sixteen is zero. So we’re left with sixteen is equal to eight 𝑥. Now I wanna know what one 𝑥 is. So if I divide both sides by eight, on the left-hand side sixteen divided by eight is two. And on the right-hand side the eights cancel out to leave me with just 𝑥. So I know that 𝑥 is equal to two. Now I want to know what 𝑦 is equal to. But remember equation one has got an equation which is 𝑦 is equal to three 𝑥 minus two. So if I substitute that value of 𝑥 back into equation one, it’ll tell me straight away what 𝑦 is. So 𝑦 is three times two take away two. So that’s six take away two, which means that 𝑦 is four.

Now we can check our answers. We just used substitution in equation one to work out that 𝑦 was equal to four. So I’m gonna use the other equation — equation two — just to check that my 𝑥 and 𝑦 match up. So I’ve substituted in two for 𝑥 and four for 𝑦. And I’ve got two is equal to three times four take away ten. So two is equal to twelve minus ten which is two. That’s correct; so we’re happy with our answer. And so we can just put a nice box around our answer to make it lovely and clear. So that’s an example where you’ve got a ready-made equation that says 𝑦 is equal to something; 𝑥 is equal to something. And it’s very easy to substitute either 𝑥 or 𝑦 into the other equation.

So let’s move on to a second example.

Use algebraic substitution to solve the simultaneous equations 𝑥 is equal to two 𝑦 plus eight and two 𝑥 plus three 𝑦 equals twenty-three. Now looking at that first equation, we’ve got 𝑥 is equal to something not involving 𝑥, so two 𝑦 plus eight. So there’s an obvious thing that we can use to substitute into the second equation. In order to find out what 𝑦 is equal to in the second equation or even what 𝑥 is equal to to substitute it back into the first, we’d have to do quite a bit of work to rearrange it. So the obvious substitution is to take 𝑥 from equation one and substitute it into equation two.

So substitution is just a matter of saying in the first equation 𝑥 is equal to all this stuff here. So where we see 𝑥 in our second equation, we’ll replace 𝑥 with all that stuff. Now we’re going to multiply out the parentheses and solve that equation to find out the value of 𝑦. So two lots of two 𝑦 is four 𝑦; two lots of eight equals sixteen. We’ve still got our three 𝑦 and that’s all equal to twenty-three. Now we’ve got four 𝑦 and three 𝑦; so that’s seven 𝑦. So that seven 𝑦 plus sixteen equals twenty-three. Now if I subtract sixteen from both sides, I’ll be just left with seven 𝑦 on the left. So seven 𝑦 plus sixteen take away sixteen. Well sixteen take away sixteen is nothing; so that wipes that out. We’ve just got seven 𝑦 and twenty-three take away sixteen is just seven. So seven 𝑦 is equal to seven. Now if I divide both sides by seven, I’ll find that 𝑦 is equal to one.

And luckily our first equation told us the value of 𝑥 in terms of 𝑦. So I just need to substitute that value of 𝑦 back into that equation and I’ll find out what 𝑥 is. So if 𝑥 was two times 𝑦 plus eight and we know that 𝑦 is equal to one, that means that 𝑥 is equal to two times one plus eight. So 𝑥 is ten. Always a good idea to check your answers, so we’ll use the other equation, equation two, just to substitute 𝑥 is ten and 𝑦 is one and just check that it all works out. So if all is good, we’ve got two times 𝑥 is ten plus three times 𝑦 is one and that will be equal to twenty-three. Well two times ten is twenty and three times one is three. So yep that’s true; looks like we’ve got the right answer. So that’s pretty much the substitution method.

So let’s look at number three then.

Use algebraic substitution to solve the simultaneous equations three 𝑥 plus three 𝑦 equals twenty-seven and two 𝑥 plus five 𝑦 equals thirty-six. Now just looking at this-this pair of equations, I- my gut reaction would be — if they didn’t tell me which method to use I’d probably be leaning towards elimination. I’d be perhaps doing two times equation one and three times equation two and then subtracting one from the other and eliminating 𝑥 from our enquiries, working out what 𝑦 is and then substituting that back in. But that’s not the case; we’ve been told specifically we’ve gotta use algebraic substitution. So we need to rearrange one of those equations to get either 𝑥 or 𝑦 on its own and then substitute that into the other equation. Now looking at both of those equations, I think I’m gonna mess about with equation one because the coefficients of 𝑥 and 𝑦 are both three. And twenty-seven is a multiple of three. So if I get 𝑥 or 𝑦 on its own, I can divide everything through by three. And I won’t have any fractions involved in this day. So let’s try that.

So I’m gonna try to make 𝑦 the subject. So first of all I need to get rid of the three 𝑥 term. So I’m gonna subtract three 𝑥 from both sides of the equation, so subtracting three 𝑥 from this side and subtracting three 𝑥 from this side. Well subtracting three 𝑥 from the left-hand side just leaves me with three 𝑦 and subtracting three 𝑥 from the right-hand side. I’m gonna write negative three 𝑥 plus twenty-seven. You could say twenty-seven minus three 𝑥. It wouldn’t make any difference in the long run, but I’m just gonna do this for now. Now that’s what three 𝑦 is. So I need to divide everything by three in order to get what one 𝑦 is. So a third of three 𝑦 is just 𝑦, a third of negative three 𝑥 is just negative 𝑥, and a third of twenty-seven is positive nine.

So from equation one, we know that 𝑦 is equal to negative 𝑥 plus nine; it’s equal to all that stuff. Now we can substitute that version of 𝑦 negative 𝑥 plus nine into the second equation. So remember our second equation was two 𝑥 plus five 𝑦 equals thirty-six. So two 𝑥 plus five 𝑦; now we’re saying 𝑦 is equal to all this stuff. So we’re gonna put this stuff instead of 𝑦 five times that and that’s equal to thirty-six. So now we’re gonna multiply out the parentheses here, which gives us two 𝑥 minus five 𝑥 plus forty-five equals thirty-six and two 𝑥 minus five 𝑥 is negative three 𝑥. So I would advise adding three 𝑥 to both sides to give me a positive number of 𝑥s somewhere. So adding three 𝑥 to the left-hand side just leaves me with forty-five. And adding three 𝑥 to the right-hand side, I can say three 𝑥 plus thirty-six or thirty-six plus three 𝑥. I’m gonna do it this way round this time just for a change. And then I’m gonna subtract thirty-six from both sides, just to leave the 𝑥 term on its own on the right-hand side. So subtracting thirty-six on the right-hand side obviously just leaves me with three 𝑥. And subtracting thirty-six on the left-hand side, so forty-five minus thirty-six is nine. So three 𝑥 is equal to nine. So now I can divide both sides by three to tell me what one 𝑥 is. Third of three 𝑥 is one 𝑥 and third of nine is three; so 𝑥 is equal to three.

Now down here we had said that 𝑦 is equal to negative 𝑥 plus nine. Let’s call that equation three, shall we? So using the equation number three, 𝑦 is equal to negative 𝑥 plus nine. We can now substitute 𝑥 is three into that equation to find out what 𝑦 is. So 𝑦 is the negative of the 𝑥-value; so negative of three plus nine which is equal to six. And now I’m gonna check those values back in one of my original equations just to see that everything adds up properly. So I’m gonna go for equation two. That looks more interesting. And equation two said two times the 𝑥-value plus five times the 𝑦-value equals thirty-six. That’s two times three plus five times six, which is six plus thirty. And yup that is thirty-six. So our answer is 𝑥 is three and 𝑦 is six. So when we had you know complicated equations at the beginning that didn’t have a nice simple 𝑦 equals or 𝑥 equals that we can substitute into the other one, then we had quite a bit of work to do before we could actually get on to do the substitution. So just bear that in mind when using this method.

Now number four, we’ve got to use algebraic substitution to solve the simultaneous equations four 𝑥 plus three 𝑦 equals three and five 𝑥 plus four 𝑦 equals three and eleven-twelfths.

So not-not only have we not got an obvious 𝑦 equals or 𝑥 equals, but there’s not even a very nice way to rearrange these to get nice numbers without fractions in them. So you know if I wanted to make 𝑦 the subject in either of those, I’m gonna have some fraction of 𝑥 and then fractional numbers; it’s all very horrible. So this is probably a classic example of where you would not use algebraic substitution to solve the simultaneous equation. I’m gonna show you it quickly anyway so that you can see how horrible it turns and just so that you can avoid it in future yourself if you come across ones like these and if they don’t tell you that you’ve got to use algebraic substitution that is.

So I’m gonna rearrange equation one to make 𝑦 the subject. So I’ve taken away four 𝑥 from both sides giving me three 𝑦 is three minus four 𝑥 and then divided every term on both sides by three. So I’ve got 𝑦 is equal to this stuff here: one minus four-thirds of 𝑥. And I’ve called that equation three. So I’m gonna substitute that version, that value of 𝑦, into equation two. So we’ve taken that value of 𝑦 and we’ve replaced 𝑦 in the equation with that value. The other thing that I’ve done here is I converted the mixed number three and eleven-twelfths into forty-seven over twelve. It’s generally easier to do your calculations with top heavy fractions rather than messing about with mixed numbers. So now we’re gonna multiply out the brackets four times one and four times negative four-thirds 𝑥. And that gives us a five 𝑥 plus four minus sixteen-thirds 𝑥 equals forty-seven over twelve. So we’ve got five 𝑥 and we’re gonna take away sixteen-thirds of 𝑥. So really we want to express five 𝑥 in a top heavy fraction with a denominator of three to end this easiest; so fifteen over three 𝑥. So five 𝑥 is the same as fifteen over three 𝑥. We’ve now got fifteen over three 𝑥 subtract sixteen over three 𝑥. So that’s negative one over three 𝑥, so negative a third 𝑥. If I add a third 𝑥 to both sides, then I’m gonna end up with a positive number of 𝑥s over on the right-hand side.

And then I’m gonna subtract forty-seven over twelve from both sides. But at the same time, I’m gonna convert four into a top heavy fraction involving twelve. So that’s gonna be forty-eight over twelve, which is the same as four. So four becomes forty-eight over twelve with subtracting forty-seven over twelve. And when I subtract forty-seven over twelve from the right-hand side, it gets rid of that; it just leaves me with a third of 𝑥. So forty-eight-twelfths minus forty-seven-twelfths is just one-twelfth. Now to find out what 𝑥 is equal to, I’m gonna multiply both sides by three. So 𝑥 is equal to three-twelfths, which is obviously the same as a quarter. Now I can substitute that value of 𝑥 back into this equation we had up here: 𝑦 equals one minus four-thirds of 𝑥 and that will tell me what 𝑦 is equal to. So 𝑦 is one minus four-thirds times a quarter. Well the four is gonna cancel here; so that’s one-third. So 𝑦 is equal to one minus a third which is two-thirds.

Now I’ll leave it to you to go back and check that in one of the original equations. But take it from me once you start doing algebraic substitution on unsuitable pairs of equations, it does get quite tricky; there’s lots of fractions and lots of negative numbers. So it all gets a bit messy. Yep so overall the algebraic substitution method is okay in some cases for linear equations. But it really comes into its own when you’ve got a nonlinear equation and a linear equation. And you can do a nice simple substitution that way. So why not check out that particular video solving simultaneous nonlinear equations using algebraic substitution. Okay thanks for now.

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