Answer :
[tex] \left \{ {{x+3y=12} \atop {x-y=8}} \right. \\\\ \left \{ {{x=12-3y} \atop {x=8+y}} \right. \\\\
12-3y=8+y\ \ \ | subtract\ y\\\\
-4y+12=8\ \ \ | subtract\ 12\\\\
-4y=-4\ \ \ | divide\ by\ -4\\\\
y=1\\\\
x=8+y=8+1=9[/tex]
First, we need to find any value for x or y (I will do x)
So, taking x-y=8, if you add y, you get x=8+y. Even though this is not a real integer, we can still substitute it into the other equation. So, because x = 8+y, then, 8+y + 3y = 12. Then solve it by adding like terms, 8+4y=12, then subtract 8, 4y=4, and last, divide by four. y=1. However, we are not done yet, by substituting the value of y into the second equation, we get x-1=8. Add one and x=9.
So, taking x-y=8, if you add y, you get x=8+y. Even though this is not a real integer, we can still substitute it into the other equation. So, because x = 8+y, then, 8+y + 3y = 12. Then solve it by adding like terms, 8+4y=12, then subtract 8, 4y=4, and last, divide by four. y=1. However, we are not done yet, by substituting the value of y into the second equation, we get x-1=8. Add one and x=9.
