1. ## equations

Solve the following system of simultaneous equations:
6x1 + 4x2 = 40
2x1 + 3x2 = 20

I'm a little confused, can one help.

2. Originally Posted by sweettea331
Solve the following system of simultaneous equations:
6x1 + 4x2 = 40
2x1 + 3x2 = 20

I'm a little confused, can one help.
I'm a Physicist, so I usually don't do the fancier methods...

Substitution method. Solve one equation for one of the unknowns, then plug it into the other equation. I'll solve the top equation first:
$6x_1+4x_2=40$
Thus $x_2=1/4*(40-6x_1)=10-3/2*x_1$

Now use this value for $x_2$ in the second equation. This leaves us an equation in 1 unknown ( $x_1$).
$2x_1+3(10-3/2*x_1)=20$
$2x_1+30-9/2*x_1=20$
$-5/2*x_1+30=20$

So, solve this equation for $x_1$. Use this value in either of the original equations to get a value for $x_2$. (I get $x_1=4$ and $x_2=4$).

There are other methods to use. One of them is to multiply the first equation by a constant and the second by another constant such that you can subtract the two equations and cancel out one of the variables. In this case, you don't need to do anything to the first equation. Multiply the second equation by 3:
$6x_1+4x_2=40$
$6x_1+9x_2=60$

Now we subtract the two equations from each other:
$6x_1+4x_2=40$
$-6x_1-9x_2=-60$
----------------------------------
$0-5x_2=-20$
This gives a value for $x_2$. Then use this $x_2$ in either of the original equations and solve for $x_1$.

-Dan