# Linear Algebra General Solution

1. Jan 22, 2010

### Precursor

The problem statement, all variables and given/known data
Find the general solution to the system:
$$ax+ by= 1$$
$$cx+ dy= 2$$

Consider the case when
$$ad- bc \neq 0$$

The attempt at a solution
Like in my other post, I multiplied the first equation by "c" and the second equation by "a", and then I subtracted the two equations. I just seem to be stuck.

I got the following matrix:
$$0.....ad- bc.....2a- c$$
$$ac.....ad.....2a$$

Therefore, $$(ad- bc)y= 2a- c$$
But what relation can I deduce from this equation to help my determine a general solution for the system? The previous question I posted was much more obvious, where I was able to solve y= 0. But this isn't the case. I appreciate any help.

2. Jan 22, 2010

### Staff: Mentor

Solve for y by dividing both sides by ad - bc. You know that it is legitimate to do this because you are told that ad - bc != 0.

Just as before, the general solution is going to be just a single point - a pair of numbers. The reason that a unique solution exists, although it might not seem obvious to you at this point in your studies, is that ad - bc != 0.

3. Jan 22, 2010

### Precursor

When I divide both sides by y, I get y= (2a- c)/(ad- bc)
If I substitute this back into either one of the original equations, I get an even more complicated equation. Is there some way to simply it?

4. Jan 22, 2010

### Staff: Mentor

What you get for x will be no more complicated than what you got for y, at least when it is simplified. Also, you can check your answers by substituting them into your original system of equations. It should be true that ax + by = 1 and cx + dy = 2.

5. Jan 22, 2010

### Precursor

When I substitute what I got for y into the first original equation, I get ax= 1- b((2a-c)/(ad-bc))
Can I simplify this?

6. Jan 22, 2010

### Staff: Mentor

Yes. Get a common denominator and things simplify.

Also, as in the other problem you posted, a and c can't both be zero, due to the restriction that ad - bc != 0.