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Describing Solutions to Systems of Linear Equations

Describing Solutions to Systems of Linear Equations 1.10 - Solution

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Ron-Jon let dd be the cost of a doughnut and cc the cost of a cupcake. He used the purchases he and Martha-Jon made to write the system of equations. {3d+4c=6.13(I)6d+8c=12.26(II)\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 22 as this will lead to the dd-terms getting the same coefficient.
2(3d+4c)=26.132(3d+4c)=2\cdot 6.13
6d+8c=26.136d+8c=2\cdot 6.13
Now we have the following system. {6d+8c=12.26(I)6d+8c=12.26(II)\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.