The system of equations have no solution.
Explanation:
The equations are [tex]4r+\frac{1}{2}s=12[/tex] and [tex]2r+\frac{1}{4} s=8[/tex]
To find the two numbers, let us solve the equations using substitution method.
From the equation [tex]2r+\frac{1}{4} s=8[/tex], let us determine the value of r,
[tex]$\begin{aligned} 2 r+\frac{1}{4} s &=8 \\ 2 r &=8-\frac{1}{4} s \\ r &=\frac{1}{2}\left(8-\frac{1}{4} s\right) \\ r &=4-\frac{1}{8} s \end{aligned}$[/tex]
Let us substitute the value of r in [tex]4r+\frac{1}{2}s=12[/tex], we get,
[tex]$\begin{aligned} 4\left(4-\frac{1}{8} s\right)+\frac{1}{2} s &=12 \\ 16-\frac{1}{2} s+\frac{1}{2} s &=12 \\ 16 &=12 \end{aligned}$[/tex]
which is not possible.
This means that the system has no solution.
