Solve the system 8(x + y) = 3520 and 6x + 12y = 3480.

Step-by-step solution

1. 1

   Divide each equation by its common factor before anything else. The numbers are large but both equations reduce dramatically:

   $8(x+y)=3520\ \Longrightarrow\ x+y=440$

   $6x+12y=3480\ \Longrightarrow\ x+2y=580$

   This single move avoids all four-digit arithmetic later.

2. 2

   Notice the two reduced equations are already aligned. Both start with a coefficient of $1$ on $x$, so subtracting them eliminates $x$ with no multiplication needed:

   $\begin{cases}x+y=440\\ x+2y=580\end{cases}$

3. 3

   Subtract the first from the second.

   $(x+2y)-(x+y)=580-440\ \Longrightarrow\ y=140$

4. 4

   Back-substitute to get $x$. Using $x+y=440$:

   $x+140=440\ \Longrightarrow\ x=300$

5. 5

   Check in the original (unreduced) equations. $8(300+140)=8\times 440=3520\ \checkmark$ and $6(300)+12(140)=1800+1680=3480\ \checkmark$. Checking against the originals, not the simplified copies, is what catches an arithmetic slip made during the reduction step.