Math 181 - Quiz 1A, linear systems solutions
Problem 1 Solve the system
2a + b − 2c = 3
4a + 3b + 6c = 13
a − b + 5c = 12
Solution. There are many, many ways to do this! as an example, we
decide to eliminate b from I, II using III. So we write
I + III
II + 3III
3a + 3c = 15
7a + 21c = 49
We could have used any equation to eliminate any of the variables from the
two others, the important thing is just to make a decision and stick with it.
Also, some choices are better than others – the best choice is an equation
where your chosen variable has coefficient 1 or -1 (so you don’t ever need
fractions, and don’t need to cross-multiply). We don’t have this luxury right
now, but we spot that we can dvide our first equation by 3 and our 2nd
equation by 7 to get
a+c = 5
a + 3c = 7
which is much nicer! Then we subtract these from each other to get 2c = 2,
hence c = 1. Backwards substitution gives a = 4 and finally b = −3.