$3x+11y$ has remainder $2$ and $9x+5y$ has remainder $3$ when divided by $7$ then the remainder when $x-y$ is divided by $7$

Question

If $x > y$ are positive integers such that

• $3x + 11y$ leaves a remainder $2$ when divided by $7$ and

• $9x + 5y$ leaves a remainder $3$ when divided by $7$,

then the remainder when $x – y$ is divided by $7$, equals?

The answer to this question is $6$ (provided in the answer key) but I am getting something different

$\begin{align} 3x+11y &=7m+2\tag{i}\\ 9x\ +5y\ &=7n+3\tag{ii}\\ \hline 3\times(i)-(ii)\Rightarrow\ 28y &=7(3m-n)-3 \end{align}$

But how is this possible as the LHS is divisible by $7$ but the RHS is $3.$

Answer 1

Use congruences mod. $7$:

• $3x+11y\equiv 3x-3y=3(x-y)$, and $3^{-1}\equiv 5\mod 7$ so $$3x+11y\equiv 2\iff x-y\equiv 5\cdot2=10\equiv 3\mod 7.$$
• $9x+5y\equiv 2(x-y)$ and $2{-1}\equiv 4\mod 7$, so $$9x+5y\equiv 3\iff x-y\equiv 4\cdot 3=12\equiv 5 \mod 7.$$ Therefore the two congruences are inconsistent (indeed the determinant of the system of congruences is equal to $0$).

Answer 2

Given: $$\begin{cases}3x+11y\equiv 2 \pmod{7} \\ 9x+5y\equiv 3 \pmod{7}\end{cases}$$ Note that: $$3x+11y\equiv 3x+4y\equiv 2 \pmod{7} \Rightarrow \\ 3(3x+4y)\equiv 3\cdot 2 \pmod{7} \Rightarrow 9x+12y\equiv 9x+5y\equiv 6 \pmod{7}.$$ So: $$\begin{cases}9x+5y\equiv 6 \pmod{7} \\ 9x+5y\equiv 3 \pmod{7}\end{cases} \Rightarrow \emptyset.$$

Answer 3

Correct: $\bmod 7\!:\,\ 0\equiv \underbrace{\overbrace{9x\!+\!5y}^{\Large\color{#c00} 3}-3(\overbrace{3x\!+\!11y}^{\Large\color{#c00} 2})}_{\Large{\rm eliminate}\ x\ }\equiv \color{#c00}{-3},\,$ contradiction, so the system is inconsistent