Question

Solve the following pair of equations algebraically: \[ \begin{aligned} 101x + 102y &= 304 \\ 102x + 101y &= 305 \end{aligned} \]

Answer 1

System of equations:
\[\begin{cases}101x + 102y = 304 \\102x + 101y = 305\end{cases}\]

Step 1: Eliminate \(x\)
Multiply the first equation by 102 and the second by 101: \[102 \times (101x + 102y) = 102 \times 304 \implies 10302x + 10404y = 31008\]
\[101 \times (102x + 101y) = 101 \times 305 \implies 10302x + 10201y = 30805\]

Step 2: Subtract the second equation from the first
\[(10302x + 10404y) - (10302x + 10201y) = 31008 - 30805\]
\[(10404y - 10201y) = 203\Rightarrow 203y = 203\Rightarrow y = 1\]

Step 3: Substitute \(y=1\) into the first original equation
\[101x + 102(1) = 304\Rightarrow 101x + 102 = 304\Rightarrow 101x = 202\Rightarrow x = 2\]

Solution:
\(x = 2\), \(y = 1\)