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!

 

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

 

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.

 

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.

Correct. If in doubt, a liberal use of brackets should help you break the calculation down.

Dave

 

Thanks mate.

 

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.

 

If you would like I can write a tutorial on these methods, or you can just read about them on wikipedia.

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