When solving systems of linear equations, we are taught that we need just as many equations as there are variables to solve the system. However, you don't need to have all the equations if you are solving for one variable.. Here's an easy system of two linear equations with three variables, yet you can solve one of the variables. Give it a try!
If and , then
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Personally I prefer to solve this kind of problems without pencils or any drawings as I find it to be a little faster.
Let's simplify first equation a little bit:
2 a + 3 b + 2 c = 1 2 − 2 a − b
2 a + 2 a + 3 b + b + 2 c = 1 2
4 a + 4 b + 2 c = 1 2 we can divide everything by 2 .
2 a + 2 b + c = 6 we know from the problem that c = a + b
2 ( a + b ) + c = 6
2 c + c = 6
3 c = 6
c = 2