Chapter 3 Linear Systems and Matrices - 3.4 Solve Systems of Linear Equations in Three Variables - 3.4 Exercises - Problem Solving - Page 184: 44b

$r=1$ $c=2$ $d=4$

We found: $30c+60d+60r=360 \rightarrow c+2d+2r=12$ $ c+d+r=7$ $d=2c$ Substitute for d: $c+2d+2r=12$ $c+2(2c)+2r=12$ $5c+2r=12$ (1) and $ c+d+r=7$ $ c+2c+r=7$ $3c+r=7$ (2) From (1) and (2): $5c+2r=12$ $3c+r=7$ Multiply both sides of the second equation by $-2$: $5c+2r=12$ $-6c-2r=-14$ ____________________ $-c=-2$ $c=2$ Solve for d: $d=2c=2(2)=4$ Solve for r: $2+4+r=7$ $r=1$