First you must solve system of equations.
Multiply the first equation by 4 and second one by 3.
You result with,
[tex]
16x-12y=4 \\
15x+12y=27
[/tex]
Add the equations so y terms cancel out.
[tex]31x=31\Longrightarrow x=1[/tex]
Insert x that was found in either one of the equations. I'll pick first one.
[tex]4\cdot1-3y=1[/tex]
Solve for y.
[tex]
-3y=-3 \\
y=1
[/tex]
The solutions to the system of equations are,
[tex]\boxed{x=1},\boxed{y=1}[/tex]
Therefore the number missing is 1.
Hope this helps.
r3t40
