Question

Solve the following pair of equations: \[ \frac{5}{x - 1} + \frac{1}{y - 2} = \frac{7}{4}, \] \[ \frac{6}{x - 1} - \frac{2}{y - 2} = \frac{1}{2}. \]

Hint:

To solve a pair of equations with fractions, first eliminate the fractions by multiplying through by suitable factors, then solve for the variables.

Answer 1

Let us assume: \[ a = x - 1 \quad \text{and} \quad b = y - 2. \] The given system of equations becomes: \[ \frac{5}{a} + \frac{1}{b} = \frac{7}{4} \quad \text{(1)}, \] \[ \frac{6}{a} - \frac{2}{b} = \frac{1}{2} \quad \text{(2)}. \]
Step 1: Multiply equation (1) by 4 and equation (2) by 2 to eliminate the fractions: From equation (1): \[ 4 \left(\frac{5}{a} + \frac{1}{b}\right) = 7 \quad \Rightarrow \quad \frac{20}{a} + \frac{4}{b} = 7 \quad \text{(3)}. \] From equation (2): \[ 2 \left(\frac{6}{a} - \frac{2}{b}\right) = 1 \quad \Rightarrow \quad \frac{12}{a} - \frac{4}{b} = 1 \quad \text{(4)}. \]
Step 2: Add equations (3) and (4): \[ \left(\frac{20}{a} + \frac{4}{b}\right) + \left(\frac{12}{a} - \frac{4}{b}\right) = 7 + 1. \] Simplifying: \[ \frac{32}{a} = 8 \quad \Rightarrow \quad a = 4. \]
Step 3: Substitute \( a = 4 \) into equation (3): \[ \frac{20}{4} + \frac{4}{b} = 7 \quad \Rightarrow \quad 5 + \frac{4}{b} = 7 \quad \Rightarrow \quad \frac{4}{b} = 2 \quad \Rightarrow \quad b = 2. \]
Step 4: Now, substitute \( a = 4 \) and \( b = 2 \) into the relations \( a = x - 1 \) and \( b = y - 2 \): \[ x - 1 = 4 \quad \Rightarrow \quad x = 5, \] \[ y - 2 = 2 \quad \Rightarrow \quad y = 4. \]
Conclusion: The solution is \( x = 5 \) and \( y = 4 \).