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Let $ad - bc \neq 0$.

The system is: $$ ax + by = 1 \\ cx + dy = 2$$

I couldn't get far with elimination. I heard you could do this with row reduction, but i'd like an algebraic answer to this. Using substitution, solving for y in the first equation i get $y = \frac{1 - ax}{b}$. Plugging this into the second yields: $cx + \frac{d(1 - ax)}{b} = 2$.

Now, when solving for x, i can get as far as $cx * b = 2b - d(1 - ax)$. But once i get to this point i'm quite stuck as to what to do next.

Is there a better way to solve this? Am i on the right track? A walk through would be really nice--i have similar to answer.

user121947
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  • You're on the right track. You have $bcx = 2b - ad + adx$, and you can separate the $x$ and express it in terms of $a,b,c,d$. Substituting it back to the $y = \frac{1-ax}{b}$ equation you got earlier should give you the $y$. – 2012ssohn Jan 17 '14 at 01:56
  • Gaussian Elimination = Row reduction, no? – user121947 Jan 17 '14 at 01:59
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    @user121947 Have you studied matrices yet? – David H Jan 17 '14 at 02:04
  • @DavidH I have not. It was expected to use algebraic techniques only; I do not mind taking the next step to learn. But i'd like to know how to do it both ways. – user121947 Jan 17 '14 at 02:07

2 Answers2

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Take one of the equations and express $x$ in terms of $y$. Here's the first one:

$$ax+by=1 \to x=\frac{1-by}{a}.$$

Now substitute this expression for $x$ into the second:

$$c \frac{1-by}{a} + dy = 2 \\ \frac{c}{a} - \frac{bcy}{a} + dy = 2 \\ \left(d - \frac{bc}{a}\right)y = 2 - \frac{c}{a} \\ y = \frac{2 - \frac{c}{a}}{d - \frac{bc}{a}} = \frac{2a - c}{da - bc}.$$

Then

$$x = \frac{1 - b{\frac{2a - c}{da - bc}}}{a} = \frac{(da-bc) - b}{(da-bc)a}.$$

That's the straightforward way to solve it.

John
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    Or to get $x$, multiply first equation by $d$, second by $b$, and substract. It is the same with $y$. (I do not consider this comment worthy of being a full answer) – chubakueno Jan 17 '14 at 02:09
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The simplest way is probably via Cramer's Rule.

Your solution is: $$ \begin{align} x &= \dfrac{ \det \left[ \begin{array}{cc} 1 & b\\ 2 & d \end{array} \right] }{ad-bc} = \dfrac{d-2b}{ad-bc}\\ y &= \dfrac{ \det \left[ \begin{array}{cc} a & 1\\ c & 2 \end{array} \right] }{ad-bc} = \dfrac{2a-c}{ad-bc} \end{align} $$

baudolino
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