It's been a while since I've had to do a simultaneous equation and I'm rusty on a few of the particulars.

Say for example I have the following equations:

x + y       = 7
2x + y + 3z = 32
    2y + z  = 13

I know that I need to combine the above 3 equations into 2 other equations, for example, if I combine (counting down) 1 + 2 I'd get

3x + 2y + 3z = 32

And combing 2 with 3 I'd get

2x + 3y + 4z = 45

Which is fine, and I understand. It's the next steps I have trouble understanding. A lot of the examples I've been looking at have a value for each of the x, y z. Looking at this site here I'm not sure what is going on step 2.

I can see that they are multiplying one line by 2. Is that something you always do? Like, with simultaneous equations do you always multiple one of the equations by 2? If not, how do you determine which number to use?

My understanding of simultaneous equations is extremely limited.


Multiply equation (1) by 2 and subtract it from equation (2).

$2x + y + 3z - 2x - 2y = 32 - 14$

$-y +3z = 18$ ......(4)

Now solve equation (3) and (4) to find y and z.

Multiply equation (4) by 2 and add with equation (3).

$2y + z -2y + 6z = 13 - 36$

$7z = -23$

$z = \frac {-23}{7}$

Then from equation (1) find x.

Way to solve.

Try to eliminate one variable and you have two new equations with remaining two variables.

Then solve these two new equations to find both variable.

And when you put value of these 2 variables to find 3rd.

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    $\begingroup$ Still any doubt please ask. $\endgroup$ – Kanwaljit Singh Feb 8 '17 at 11:15
  • $\begingroup$ Thank you for the comment. However, I'm still confused with working out Y and Z. Following from your example I multiplied (4) by 2 so I got -2y + 6z = 36. I then took that away from (3) to give me -5z = 23. Is that correct? It's been over a decade since I've done math like this. $\endgroup$ – RacconLauncher Feb 8 '17 at 11:34
  • $\begingroup$ No. After multiplying equation 4 by 2. As in equation 3 and equation 4 terms containing y are equal and opposite in sign. So add equation 3 and 4. $\endgroup$ – Kanwaljit Singh Feb 8 '17 at 11:37
  • $\begingroup$ Mind if we go into a private chat so I can clarify the points? each steps gives me new questions $\endgroup$ – RacconLauncher Feb 8 '17 at 11:40
  • $\begingroup$ OK. So I have added (3) and (4) to get y + 5z = 31. From there, how do you eliminate a variable? $\endgroup$ – RacconLauncher Feb 8 '17 at 11:50

You already know, that you can add multiple euqations, to get a *new Equation. I put new here, since it actually does not have any new information.

However, instead of adding the first equation to the second, you could do something else. You could substract the second equation from the double of the first. This will eliminate the $x$-part in the new equation, which is a step in the right direction.

You can continue on with more rows, more collumns, just make sure to eliminate some variables on the way.

This is called the Gauß-Algorithm for solving linear systems.


You can "eliminate" one or more of the equations using one of the techniques already listed here. For systems of equations with three variables, a technique my intermediate calculus class uses to find critical points is to solve for one of the variables in (1), substitute that variable into (2) to solve for the unknown, and check validity by subbing both into (3).


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