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This problem from Serge Lang's Basic Mathematics in Chapter 2, question 9a.

Let $a,b,c,d$ be numbers such that $ad-bc \neq 0$. Solve the following systems of equations for $x$ and $y$ in terms of $a,b,c,d$. a) \begin{align*} ax + by & = 1\\ cx + dy & = 2 \end{align*}

I'm fine with solving these sorts of equations with numbers in place of $a,b,c,d$ but trying to solve it with just variables has been a problem. I assume that the information "$ad-bc \neq 0$" is some sort of hint, possibly that I'm able to divide by $ad-bc$ at some point but I don't know how to apply it.

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  • $\begingroup$ khanacademy.org/math/algebra-home/alg-system-of-equations/… $\endgroup$ – Kenta S Jan 15 at 0:17
  • $\begingroup$ Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Jan 15 at 0:20
  • $\begingroup$ with matrices, the system is $\pmatrix{a&b\\c&d}\pmatrix{x\\y}=\pmatrix{1\\2},\\$ and $\pmatrix{a&b\\c&d}$ can be inverted to solve for $\pmatrix{x\\y}$ if its determinant is non-zero $\endgroup$ – J. W. Tanner Jan 15 at 0:31
  • $\begingroup$ Note that $a$ , $b$, $c$ and $d$ are not variables. Consider them constants, just like numbers. We call them parameters. Try solving the system of equations just the same way you solve a system with numbers, and show us what you get. $\endgroup$ – Saeed Jan 15 at 4:30
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Hint

Multiply the 1st eqn by $c$ and the 2nd eqn by $a$. Then the two altered eqn's will each have, as their first term, $(ac)x$. Therefore, by subtracting the 2nd altered eqn from the 1st altered eqn, you will obtain an eqn that has only the single variable $y$.

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