If \(3x+6y+9z = \dfrac{20}{3}, 6x+9y + 3z = \dfrac{17}{3}\) and \(18x+ 27y - z = \dfrac{113}{9}\) , then what is the value of \(75x+113y \ ?\)
163/3
Step 1 — Add E1 and E2: \(9x+15y+12z=\tfrac{37}{3}\).
Step 2 — Eliminate z using E3: Multiply this by 1 and combine with \(12\cdot E3\); solving the system gives \(x=\tfrac{2}{9},\ y=\tfrac{1}{3}\) (verified in E1).
Step 3 — Compute:
\[75x+113y=75\cdot\tfrac{2}{9}+113\cdot\tfrac{1}{3}=\tfrac{50}{3}+\tfrac{113}{3}=\dfrac{163}{3}\]
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