# simultaneous equations

Discussion in 'Math' started by Thinker, Jan 29, 2007.

1. ### Thinker Thread Starter Active Member

Jan 9, 2007
61
0
Can someone solve this simultaneous equations for me please?

3x - 2y = -18
2x - 5y = -23

I can get the formula for the first bit, but when i try use the same formula for the second i get stuck!

2. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
Rearranage equation 2 to get the equation as a subject of either x or y, for example as a subject of x:

x = (-23 + 5y)/2

Sub into equation 1:

3/2(-23 + 5y) - 2y = -18

Solve for y:

y = 3

Sub into 2:

2x - (5x3) = -23

Solve for x:

x = -4

Check in both your original equations. Any problems?

Dave

3. ### Thinker Thread Starter Active Member

Jan 9, 2007
61
0
Okay let me check dave.

(3 * -4) - (2 * 3) = -18
= (-12) - (6)

(2 * -4) - (5 * 3) = -23
= (-8) - (15)

This is what i got, i think where i was going wrong is i treated the minus numbers like positive numbers.

Thanks again dave.

4. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
Correct. If in doubt, a liberal use of brackets should help you break the calculation down.

Dave

Jan 9, 2007
61
0
Thanks mate.

6. ### DrNick Active Member

Dec 13, 2006
110
2
You can solve this in various other ways that are much quicker (involving linear aglebra). You may want to look in to cramer's rule, or finding inverse matricies. If you would like I can write a tutorial on these methods, or you can just read about them on wikipedia. Just to get you started that perticular linear system can be re-writen as:

y=A*x

where y is a vector contining [-18, -23],
A is the matrix:

[3 -2]
[2 -5]

x = [x, y]

Hope this helps out.

7. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
Please feel free to post up a tutorial here in this thread. We can then look at against the existing Simultaneous Equations section in the on-line e-book and if all parties agree we could add it to the e-book, obviously acknowledging your contribution to this body of work.

Dave