Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$2 = A_0 + A_1 $

$0 = -A_0 + X_1A_1$

$2/3 = A_0 + X_1^2A_1$

I'm trying to find values for $A_0, A_1$, and $X_1$. If I knew $X_1$ this would be a pretty simple process (I'd probably just plug it into a matrix or set $A_0 = 2 - A_1$ and solve), but with that being an unknown I'm sort of confused as to how I should be going about this.

Any help would be appreciated!

I also don't know what tag to give this. I was looking for one for System of Equations but there doesn't seem to be one. I assumed Matrices would be an appropriate tag since I'd guess you could solve this with a matrix.

share|improve this question
    
$X_1$ is an unknown. You have three equations in three unknowns-is the problem that it appears squared? The substitution method still works, though sometimes it increases the degree and sometimes it gets you roots you don't like. mythealias takes you through this one nicely, my comment hopes to have you not scared of these in the future. –  Ross Millikan Nov 11 '12 at 3:59

1 Answer 1

up vote 2 down vote accepted

Add first two equations to get $$ 2 = A_1(1 + X_1) $$ Add last two equations to get $$ 2/3 = A_1 X_1 (1 + X_1)$$ From this it follows, $X_1 = 1/3$. substitute back to get $A_1 = 3/2$ and $A_0 = 1/2$

The general approach to solving such problem it to try and eliminate variables (while being careful to not divide by zero). You could have used $A_0 = 2 - A_1$ and then try to find $A_1$ in terms of $X_1$ from second equation. I say $A_1$ because when you go to third equation you have $X_1^2$ but only $A_1$. This means if you had to substitute for $X_1$, you will be required to square a term, which will only increase the work. Which way to proceed usually comes from trial and error and experience.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.