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Video: Solving Linear Equations with Unknowns on Both Sides

Tim Burnham

We solve a series of linear equations with the unknown on both sides, such as 3𝑥 − 5 = 2𝑥 + 3 or 1.5𝑥 + 5 = 3𝑥 − 1. We look at strategies for rearranging the equation in order to make the working as simple as possible.

15:49

Video Transcript

In this video, we’re gonna see how to solve linear equations in which there are terms involving the unknown variable on both sides of the equation. With a bit of careful thought, we can make things easier for ourselves by ensuring we end up with a positive number of the unknown on one side of the equation. Let’s take a look at some examples and find out more.

Number one, solve three 𝑥 minus five equals two 𝑥 plus three. Well as you can see, we’ve got three 𝑥 on the left-hand side and we’ve got two 𝑥 on the right-hand side. So our first task is to try to eliminate 𝑥 from one of the sides. We’ve got the choice of two inverse operations. We can either subtract three 𝑥 from both sides to eliminate 𝑥 from the left-hand side or we can subtract two 𝑥 from both sides to eliminate 𝑥 from the right-hand side. Now if I subtract three 𝑥 from both sides, on the left-hand side I’ve got three 𝑥 take away five take away three 𝑥, and three 𝑥 take away three 𝑥 is nothing. So that’s just gonna leave me with negative five. Then on the right-hand side, I’ve got two 𝑥, and I’m taking away three 𝑥. And two 𝑥 take away three 𝑥 is negative one 𝑥, or just negative 𝑥. And then obviously, I’ve got plus three as well. Well now I’ve got negative five is equal to negative 𝑥 plus three. I’ve got a choice of two things. I can either try to clear off the three from the side that’s got the 𝑥 on it, but then that will still leave me with a negative number of 𝑥s. Or I can add 𝑥 to both sides to get a positive number 𝑥s on one side. And I’m gonna try adding 𝑥 to both sides. So on the right-hand side, I’ve got negative 𝑥 plus 𝑥, which is nothing. So those two to cancel out, just leaving me with three. Now I’ve got negative five plus 𝑥 on the left-hand side is equal to three. So if I add five to both sides, then on the left-hand side I’ve got negative five plus five, which is nothing. So they cancel out, just leaving me with 𝑥. And on the right-hand side, three plus five is eight, so I’ve got 𝑥 equals eight.

Okay, so that works. I’ve got my answer. Let’s rerun that again. But this time instead of taking away three 𝑥 from both sides in the first stage, let’s take away two 𝑥 from both sides. So on the left-hand side, I’ve got three 𝑥 take away two 𝑥 is just 𝑥. And obviously, I’ve still got the minus five term. And on the right-hand side, I’ve got two 𝑥 take away two 𝑥. Well that’s nothing, so they cancel out. So that just leaves me with positive three. Now I can just add five to both sides to get rid of the negative five from the left-hand side. And that means on the left, I’ve got negative five add five, which is nothing. So they cancel each other out, just leaving me with 𝑥. And on the right-hand side, three plus five is equal to eight. So we came up with the same answer in both cases. So I’ve come up with the same answer, 𝑥 equals eight in both questions. But when I eliminated the two 𝑥 from this side and left myself with a positive number of 𝑥s on the other side rather than the negative number of 𝑥s, the whole thing was quicker. I eliminated the need for one of the steps of my calculation. So there’s a top tip; when you’ve got a question with the unknown on both sides of the equation, try to eliminate the one that’s gonna leave you with a positive number of that variable on the other side of the equation, and it would just save you a bit of time.

Okay, let’s move on to another question. Solve four 𝑥 minus three equals eleven minus three 𝑥. Well I’m gonna use inverse operations either to eliminate four 𝑥 from this side by subtracting four 𝑥 from both sides or eliminating negative three 𝑥 from this side by adding three 𝑥 to both sides. So I just need to think carefully which one am I gonna do. If I subtract four 𝑥 from both sides, yes I will eliminate four 𝑥 from the left-hand side; but on the right hand side, I’m gonna have negative three 𝑥 take away another four 𝑥. I’m gonna leave myself with negative seven 𝑥, so I’m not gonna do that. If I add three 𝑥 to both sides, that will eliminate negative three 𝑥 on the right-hand side. But on the left-hand side, I’m gonna have four 𝑥 plus three 𝑥, which will leave me with positive seven 𝑥. So this is the inverse operation that I’m gonna do to both sides. So I’ve got my original equation, but I’m adding three 𝑥 to both sides. And four 𝑥 plus three 𝑥 is seven 𝑥. And then I’ve got to subtract three, so still got this term here. On the right-hand side, I’ve got negative three 𝑥 add three 𝑥. Well that becomes zero, so they cancel out. So I’ve just got eleven. Now on the left-hand side, I’ve got seven 𝑥 take away three. So the inverse operation of taking away three is adding three. So I’m gonna add three to both sides of the equation. So on the left-hand side, I had my seven 𝑥 minus three and I’m adding three to that. And on the right-hand side, I had eleven and I’m adding three to that. Now over on the left-hand side, negative three add three is nothing. That was the whole point of adding three to both sides, so I’m just left with seven 𝑥. And on the right-hand side, eleven plus three is fourteen. For now I’ve got seven 𝑥 is equal to fourteen. Well this means seven times 𝑥. And the inverse operation to times-ing by seven is dividing by seven. So I’m going to divide both sides of my equation by seven. And that means on the left-hand side, I’ve got seven on the top and seven on the bottom. So divide seven by seven, I get one. Divide seven by seven, I get one. So I’ve just got 𝑥 on the left-hand side. And fourteen divided by seven is just two on the right-hand side, so my answer is 𝑥 equals two. Now I’m just gonna check my answer as well just to make sure that’s right. So on the left-hand side of my original equation, it said four 𝑥 minus three. Well we now know that 𝑥 is equal to two. So we’re gonna replace the 𝑥 with two. So that becomes four times two minus three. And four times two is eight, so eight minus three and eight minus three is equal to five. So the left-hand side is equal to five. So let’s see what the right-hand side would be when we substitute 𝑥 equals two into that. Well eleven minus three 𝑥, if 𝑥 is two, that is the same as eleven minus three times two. And three times two is six. So eleven minus six is equal to five. And that is the same as the left-hand side, so they are equal. So it looks like we were right. So if you’re doing these sorts of questions in a test or in an exam, then if you’ve got a bit of time left at the end, always do the check cause then you know whether you got the right answer or not. If you did make a slip somewhere and write the wrong number down, at least you’ve got time to correct that.

Okay number three then. Ooh! Look at this! We’ve got one point five 𝑥 plus five is equal to three 𝑥 minus one, so the numbers aren’t always as nice as we’d like. But basically we could, looking at the inverse operations to eliminate 𝑥 from one side, we could either subtract one point five 𝑥 from both sides or we could subtract three 𝑥 from both sides. If we subtract one point five 𝑥 from the left-hand side, that will eliminate 𝑥. And if we subtract one point five 𝑥 on the right-hand side, we’ll have three 𝑥 minus one point five 𝑥. We’ll have positive one point five 𝑥 on the right-hand side. So that looks like that’s gonna be a good plan. Let’s just check it the other way around. If we subtract three 𝑥 from the right-hand side, we’ll have no 𝑥s; three 𝑥 minus three 𝑥 is nothing. If we subtract three 𝑥 from the left-hand side, one point five 𝑥 take away three 𝑥 is negative one point five 𝑥, which will be a bad thing, so I’m not going to do that. So subtracting one point five 𝑥 from the left-hand side, we started off with one point five 𝑥 and we’re taking away one point five 𝑥. So we’ve got no 𝑥s. So we’re just left with five. And on the right-hand side, we started off with three 𝑥s. We’re taking away one point five 𝑥s, we’re leaving ourselves with one point five 𝑥 and the negative one. So now I’ve got one point five 𝑥 minus one. Well the inverse operation of taking away one is adding one. So with the interest of just eliminating everything else and leaving the 𝑥 term on its own, I’m gonna add one to both sides of that equation. Then on the right-hand side, I’ve got negative one add one is nothing. So it just leaves me with one point five 𝑥. And on the left-hand side, five plus one is six. So now I’m left with one point five 𝑥 is equal to six. Well the inverse operation to multiplying by one point five, which is what we’re doing here, is dividing by one point five. Now if I’m comfortable dividing both sides by one point five, how many one point five’s go into six? If we can spot that, then I would recommend dividing by one point five at this time. And in fact, I happen to know one point five plus one point five is three. So one point five goes into six four times. So I could do that one in my head. But if you don’t feel comfortable with that, another top tip is to just try to get rid of those decimals at this stage by maybe doubling both sides of our equation. So two, lots of six on the left-hand side are twelve. And two, lots of one point five 𝑥 on the right-hand side make three 𝑥. Now I can do the inverse operation of multiplying by three, which is dividing by three. And that’s a nice whole number; it’s a bit easier to work with. So on the right-hand side, I’ve got three 𝑥 divided by three, which is just one 𝑥. And on the right-hand side, twelve divided by three is four. So our answer is four is equal to 𝑥 or 𝑥 is equal to four. And now if I have time, I can do the left-hand-right-hand check again if I wanted to. But in fact, I’m just gonna to take a moment to show you another slightly shorter way of doing all the working out. And I’ve been writing it out, all the equations again and adding one and subtracting one and-and showing all of that very clearly. But there’s a bit of a shortcut you can take when you’re used to doing that and you know what you’re doing. So let’s have a look at that. So in stage one, we said we were gonna to subtract one point five 𝑥 from both sides of our equation. Now I can recognise that on the left-hand side, I’ve got one point five 𝑥. I’m taking away one point five 𝑥. I don’t really need to write it all out like this. I can just do that straight away and write down the answer five. And then on the right-hand side, three 𝑥 take away one point five 𝑥 is just one point five 𝑥. And I’ve also got minus one. Now I’m gonna to add one to both sides. So that’s the inverse operation of subtracting one, so I can just do that calculation straight away: five plus one is six. So I can just write that down. And on the right-hand side, one point five 𝑥 minus one plus one is just one point five 𝑥. Now I’m going to double both sides like I did before. And again, I can just write those answers down. And now I’m gonna divide both sides by three. And again, I can just write the answers down; twelve divided by three is four; three 𝑥 divided by three is just 𝑥. So you see that we end up with a much more efficient-looking solution with a lot less writing. We’ve still got pretty clear commentary on each stage of the process, but we just have less writing to wade through in order to get to our answer. So we’re gonna use that method then for our final example, number four.

So we’ve gotta solve two lots of two times three 𝑥 minus five plus four times 𝑥 plus one is equal to seven times two 𝑥 minus four. So we’ve got some parentheses to multiply out here, and then we’ve got to gather some like terms in order to get a sort of simpler version of our equation with our unknown on both sides. So I’m gonna use the distributive law and calculate two times three 𝑥 and two times negative five. And then I’m gonna do positive four times 𝑥 and positive four times positive one. And then I’m gonna do seven times two 𝑥 and seven times negative four. So two times three 𝑥 is six 𝑥 and two times negative five is negative ten. Positive four times 𝑥 is positive four 𝑥, and positive four times positive one is positive four. So that’s the left-hand side. And on the right-hand side, seven times two 𝑥 is fourteen 𝑥 and seven times negative four is negative twenty-eight. So now I’ve got to gather like terms on the left-hand side. Well I’ve got six 𝑥 and I’ve got another four 𝑥. So six 𝑥 and four 𝑥 is ten 𝑥. And then I’ve got negative ten add four is negative six. And on the right-hand side, I’ve still got fourteen 𝑥 take away twenty-eight. So now I’ve got to make a decision. Am I going to subtract fourteen 𝑥 from both sides or am I going to subtract ten 𝑥 from both sides? Well if I subtract fourteen 𝑥 from both sides, I’d end up with ten 𝑥 take away fourteen 𝑥 and then we’d end up with negative four 𝑥 on the left-hand side. But if I take away ten 𝑥 from both sides, then on the right-hand side I’m gonna have fourteen 𝑥 minus ten 𝑥. I’ll have positive four 𝑥. So on the left-hand side, ten 𝑥 take away ten 𝑥 is nothing, so I’m just left with my negative six. Be very careful to remember your negative sign in front of the six. Fourteen 𝑥 take away ten 𝑥 is just four 𝑥. And I’ve still got my negative twenty-eight. So now four 𝑥 minus twenty-eight on the right-hand side, I want to get that 𝑥 term on its own. So I’m gonna have to do the inverse of subtracting twenty-eight from both sides. So I’m going to add twenty-eight to both sides. So on left-hand side, if I start off at negative six and I add twenty-eight, I’m gonna get positive twenty-two. And on the right-hand side, four 𝑥 take away twenty-eight plus twenty-eight is just four 𝑥. And then lastly, I’ve got to do the inverse operation of multiplying by four to undo this four times 𝑥 to give me one 𝑥. So I’m going to divide both sides by four. And dividing the right-hand side by four, I’ve got just 𝑥, and the left hand side is twenty-two over four. But look, I can simplify that down. Look, the top and the bottom are both divisible by two. So our simpler version of that fraction is eleven over two. So my answer is 𝑥 is equal to eleven over two. Or if you prefer that as a mixed number, 𝑥 is equal to five and a half.

So let’s quickly summarise the process that we’ve been through then in solving these linear equations with unknowns on both sides. First, if we had parentheses, we needed to multiply those out. Second, we needed to gather any like terms. Third, we needed to make a decision about what we were gonna do to both sides to eliminate 𝑥 from one side. Now remember, the ideal rule there is to try to leave yourself with a positive number of 𝑥s on one side of your equation. We then needed to do more inverse operations to get the 𝑥 term on its own. And finally, we had to decide which inverse operation to do to leave ourselves with just one 𝑥 on its own and just a number on the other side. Then if we have time, for example, we have time at the end of our exam or our test, then we can substitute our answer back into the left-hand side of the equation and the right-hand side of the equation and just check that they are in fact still equal and we’ve got the right answer.