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Ron-Jon let $d$ be the cost of a doughnut and $c$ the cost of a cupcake. He used the purchases he and Martha-Jon made to write the system of equations.
$\begin{cases}3d+4c=6.13 & \, \text {(I)}\\ 6d+8c=12.26 & \text {(II)}\end{cases}$
If Ron-Jon is right we should be able to solve this system. Let's solve it using the elimination method. First we will multiply Equation (I) by $2$ as this will lead to the $d$-terms getting the same coefficient.
Now we have the following system.
$\begin{cases}6d+8c=12.26 & \, \text {(I)}\\ 6d+8c=12.26 & \text {(II)}\end{cases}$
We can see that Ron-Jon's system consists of two identical equations. This means that there exist infinitely many solutions. With infinitely many solutions there is no way to tell the exact cost of neither a doughnut nor a cupcake. Therefore, Martha-Jon is right.

$3d+4c=6.13$

$2(3d+4c)=2\cdot 6.13$

DistrDistribute $2$

$6d+8c=2\cdot 6.13$

MultiplyMultiply

$6d+8c=12.26$