Solve the following system of equations algebraically : $30x + 44y = 10$; $40x + 55y = 13$
Generated by claude-sonnet-4-6 · 2026-06-15 10:34 · grounding rag
Model Answer
Elimination Method:
Given: $30x + 44y = 10$ … (1) and $40x + 55y = 13$ … (2)
Multiply (1) by 4 and (2) by 3:
$120x + 176y = 40$ … (3)
$120x + 165y = 39$ … (4)
Subtract (4) from (3):
$11y = 1 \Rightarrow y = \dfrac{1}{11}$
Substitute in (1): $30x + 44 \times \dfrac{1}{11} = 10$
$30x + 4 = 10 \Rightarrow 30x = 6 \Rightarrow x = \dfrac{1}{5}$
$\therefore x = \dfrac{1}{5},\quad y = \dfrac{1}{11}$
Source: Chapter 3, Section 3.3.2 (Elimination Method)
Explanation
- Make coefficients of one variable equal by multiplying both equations by suitable numbers, then subtract to eliminate that variable.
- After finding one variable, substitute back into either original equation.
- Examiners award marks for: correct multiplication step, correct subtraction, correct value of each variable.