Question

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

Answer 1

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

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

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

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

Final Answer:
\(x = 2\), \(y = 1\)