Question

Solve the following pair of equations algebraically:
\[ 101x + 102y = 304 \quad \text{(i)}\\ 102x + 101y = 305 \quad \text{(ii)} \]

Hint:

Use elimination method by cross-multiplying equations to eliminate one variable quickly.

Answer 1

Given:
We are given the system of equations:
\[ (1)\quad 101x + 102y = 304\\ (2)\quad 102x + 101y = 305 \]
We will solve this system algebraically using the method of elimination.

Step 1: Subtract equation (1) from equation (2)
Subtracting (1) from (2): \[ (102x + 101y) - (101x + 102y) = 305 - 304\\ (102x - 101x) + (101y - 102y) = 1\\ x - y = 1 \quad \text{...(3)} \]
Step 2: Express one variable in terms of the other
From equation (3): \( x = y + 1 \)

Step 3: Substitute into equation (1)
Substitute \( x = y + 1 \) into equation (1):
\[ 101(y + 1) + 102y = 304\\ 101y + 101 + 102y = 304\\ (101y + 102y) + 101 = 304\\ 203y + 101 = 304 \]
Step 4: Solve for \( y \)
\[ 203y = 304 - 101 = 203\\ y = \frac{203}{203} = 1 \]
Step 5: Find \( x \)
From equation (3): \( x = y + 1 = 1 + 1 = 2 \)

Final Answer:
\[ \boxed{x = 2,\quad y = 1} \]