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Linear Equations with x on both sides
By Mariana on Sunday, January 30, 2000 - 12:32 pm:

Dear NRICH team,

I would like you to help me on some equations that I don't quite understand.eg,6x+2=3x+5

Could you please explain how to do this.

Mariana

By Peter Robinson (Pmpfr2) on Sunday, January 30, 2000 - 02:06 pm:

Hi Mariana

To solve equations like this, you need to do what's called algebra. I don't know how much you know about this so I'll try and explain as clearly as I can.

The trouble with your equation as it is written is that it has both numbers and x's on both sides of the equals sign. If only it had x 'on its own' then it would be much easier to solve (e.g. x = number; which I think we can both solve!). To get it into this form, you can change your equation by using algebra.

Algebra is like a game - you have a certain amount of freedom to decide what moves to make, but there are rules that you must obey. The most important rule is that you must make sure that whatever you do to the left side you also do to the right. For instance if I want to take away 7 from the left hand side, I must also take away 7 from the right. If I want to divide the right side by 3, I must also divide the left side by 3.

As long as you obey this rule, you shouldn't go wrong - I mean all the algebra you write down should be true. The trick is to pick the 'moves' you make so that the equation can be written in a nice form with x on its own.

I'll start with a nice simple example of my own:
Solve x + 1 = 3

I want to get x on its own, so it would be good to 'get rid of' the 1 on the left hand side. If I want to take away 1 from the left side, I also have to take away 1 from the right hand side because of the rule:

x + 1 - 1 = 3 - 1

Now, I can simplify both sides by actually working out 1 - 1 = 0 and 3 -1 = 2
so
x + 0 = 2
so
x = 2

That's the solution to my simple example. Let's see if we can apply this method to your more complicated one. It's slightly more difficult so it will need more 'moves', but as long as we obey the rule each time it should work:

6x + 2 = 3x + 5

I want to 'get rid of' the 2 on the left hand side, so I could start by taking away 2 from the left hand side. But the rule says that if I want to do that, I also have to take away 2 from the right:

6x + 2 - 2 = 3x + 5 - 2
this is the same as
6x = 3x + 3
by actually doing the sums.

Now, I could 'get rid of' the 3x on the right by taking away 3x. Again, the rule says that I also have to take away 3x from the left side:
6x - 3x = 3x + 3 - 3x
so
3x = 3

(Are you happy that I can say 6x - 3x = 3x and 3x - 3x = 0x = 0 ? If not, think apples instead of x!)

Now, just divide both sides by 3:
3x÷3 = 3÷3
so
x = 1

This is the solution to your equation, but it's much more important that you understand the method used rather than just getting the right answer. Please write back if you haven't understood anything I've said and I'll try to go over it more carefully.

I should say that the rule above is not the only one although it is the most common. Another is that you can NEVER divide by zero. This doesn't come into play as often but you can go badly wrong if you break it.

Peter