If the system of equations has infinitely many solutions, we know they must coincide each other. If two equations intersect at a point, we know they have a single solution at the intersection. If they are parallel with different vertical shifts, we know they have no solutions for their intersection. But if they are the same lines, they must have infinitely many solutions (or intersections)
We can write these two equations as a set of linear equations
\(Ax+By=12 [1]\)
\(2x+8y=60[2]\)
If we divide Equation 2 by 12 (the right hand side of equation 1), we find the solution is 5. Therefore, we need to divide equation 2 by 5 to get the values in relation to equation 1.
\(\frac{2x}{5} + \frac{8y}{5} = \frac{60}{5}\)
\(\frac{2x}{5} + \frac{8y}{5} = 12\)
This is in the same form as the first equation. This means we can see the values for A and B will be 2/5 and 8/5 respectively.
Now all we need to do is divide A/B to find the solution to your problem.
\(\frac{A}{B}=\frac{\frac{2}{5}}{\frac{8}{5}}\)
[tex]\frac{2}{8}=\frac{1}{4}[\tex]
