# Stuck on solving 2 simultaneous equations with variables. [Serge Lang - Basic Mathematics]

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.

• khanacademy.org/math/algebra-home/alg-system-of-equations/… – Kenta S Jan 15 at 0:17
• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Jan 15 at 0:20
• 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 – J. W. Tanner Jan 15 at 0:31
• 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. – Saeed Jan 15 at 4:30

## 1 Answer

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 $$y$$.