Q1
Solve the following pair of linear equations by the elimination method and the substitution
Concept Foundation: Pair of Linear Equations
A pair of linear equations in two variables represents two straight lines. The solution of the pair is the point where both lines intersect.
- If lines intersect → Unique solution
- If lines are parallel → No solution
- If lines coincide → Infinite solutions
Standard form:
\[a_1x + b_1y + c_1 = 0\] \[a_2x + b_2y + c_2 = 0\]In this exercise, we solve using:
- Elimination Method → Remove one variable
- Substitution Method → Replace one variable
Solution Roadmap (Exam Strategy)
- Write equations clearly with numbering
- Choose elimination or substitution wisely
- Make coefficients equal (if elimination)
- Eliminate one variable
- Substitute back to get second variable
- Always verify using second method
Geometrical Meaning (Visual Insight)
The intersection point represents the solution of both equations.
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\(x + y = 5\), \(2x - 3y = 4\)
Elimination Method
\[x + y = 5 \tag{1}\] \[2x - 3y = 4 \tag{2}\] Multiply (1) by 2: \[2x + 2y = 10 \tag{3}\] Subtract (3) from (2): \[ (2x - 3y) - (2x + 2y) = 4 - 10 \] \[2x - 3y - 2x - 2y = -6\] \[-5y = -6\] \[y = \dfrac{6}{5}\] Substitute in (1): \[x + \dfrac{6}{5} = 5\] \[x = \dfrac{25 - 6}{5} = \dfrac{19}{5}\]Substitution Method
\[x = 5 - y\] Substitute: \[2(5 - y) - 3y = 4\] \[10 - 2y - 3y = 4\] \[10 - 5y = 4\] \[y = \dfrac{6}{5}, \quad x = \dfrac{19}{5}\]Final Answer: \(x=\frac{19}{5}, y=\frac{6}{5}\)
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\(3x + 4y = 10\), \(2x - 2y = 2\)
Elimination Method
\[3x + 4y = 10 \tag{1}\] \[2x - 2y = 2 \tag{2}\] Multiply (2) by 2: \[4x - 4y = 4 \tag{3}\] Add (1) and (3): \[7x = 14\] \[x = 2\] Substitute in (1): \[3(2) + 4y = 10\] \[6 + 4y = 10\] \[y = 1\]Substitution Method
\[x = \dfrac{10 - 4y}{3}\] Substitute: \[2\left(\dfrac{10 - 4y}{3}\right) - 2y = 2\] Multiply by 3: \[20 - 8y - 6y = 6\] \[-14y = -14\] \[y = 1, \quad x = 2\]Final Answer: \(x=2, y=1\)
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\(3x - 5y - 4 = 0\), \(9x = 2y + 7\)
Convert:
\[3x - 5y = 4 \tag{1}\]
\[9x - 2y = 7 \tag{2}\]
Multiply (1) by 3:
\[9x - 15y = 12 \tag{3}\]
Subtract (2) from (3):
\[ (9x - 15y) - (9x - 2y) = 12 - 7 \]
\[-13y = 5\]
\[y = -\dfrac{5}{13}\]
Substitute:
\[3x - 5\left(-\dfrac{5}{13}\right) = 4\]
\[3x + \dfrac{25}{13} = 4\]
Multiply by 13:
\[39x + 25 = 52\]
\[39x = 27\]
\[x = \dfrac{9}{13}\]
Final Answer: \(x=\frac{9}{13}, y=-\frac{5}{13}\)
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\(\frac{x}{2} + \frac{2y}{3} = -1\), \(x - \frac{y}{3} = 3\)
Multiply first equation by 6:
\[3x + 4y = -6 \tag{1}\]
Multiply second by 3:
\[3x - y = 9 \tag{2}\]
Subtract (1) from (2):
\[ (3x - y) - (3x + 4y) = 9 - (-6) \]
\[-5y = 15\]
\[y = -3\]
Substitute:
\[3x + 4(-3) = -6\]
\[3x - 12 = -6\]
\[3x = 6\]
\[x = 2\]
Final Answer: \(x=2, y=-3\)
Why This Exercise is Important
- Directly asked in CBSE Board Exams (2–4 marks guaranteed)
- Builds foundation for Coordinate Geometry
- Used in Physics (motion equations)
- Forms base for Linear Programming in higher classes
Competitive Exam Relevance
- Asked in NTSE, Olympiads
- Important for SSC, Banking Algebra
- Used in Data Interpretation simplifications
Exam Tips
- Always number equations
- Choose easiest elimination path
- Avoid sign mistakes
- Verify answer (very important for full marks)