\left\{\begin{array}{ccc}c+3d=8\\c=4d-6\end{array}\right\\\\substitute:\\\\(4d-6)+3d=8\\4d+3d-6=8\ \ \ \ /+6\\7d=14\ \ \ \ /:7\\d=2\\\\c=4\cdot2-6=8-6=2\\\\\left\{\begin{array}{ccc}c=2\\d=2\end{array}\right
The solution to the system of equations c + 3 d = 8 and c = 4 d − 6 is c = 2 and d = 2 .
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