HI Guys,
Can any body solve these simultaneous equations:
2a + 3(a-b) = 7
4b + 3(b-a) = 1
If anybody could show me how to do these that would be great !!
Thanks!
Well start by simplifying:
$\displaystyle 2a + 3(a - b) = 7 \implies 5a - 3b = 7$
and
$\displaystyle 4b + 3(b - a) = 1 \implies -3a + 7b = 1$
The way I typically solve these is by substitution. For example, solve the bottom equation for a:
$\displaystyle a = \frac{7b - 1}{3}$
Then insert this value for a in to the top equation:
$\displaystyle \frac{5}{3} \cdot (7b - 1) - 3b = 7$
Now solve this for b and then put that into the "a" equation above.
-Dan