LINEAR EQUATIONS IN TWO VARIABLES — NCERT Solutions | Class 9 Mathematics | Academia Aeternum
Ch 4  ·  Q–
0%
Class 9 Mathematics Exercise-4.2 NCERT Solutions Olympiad Board Exam
Chapter 4

LINEAR EQUATIONS IN TWO VARIABLES

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

4 Questions
10–15 min Ideal time
Q1 Now at
Begin Q1 ↑ Progress bar
Q1
NUMERIC2 marks
Which one of the following options is true, and why?
y = 3x + 5 has
  1. a unique solution
  2. only two solutions
  3. infinitely many solutions
📘 Concept & Theory Theory / Concept ›

A linear equation in two variables is generally written as:

\[ ax+by+c=0 \]

Such equations represent a straight line on a Cartesian plane.

Every point lying on the line satisfies the equation.

Since a straight line contains infinitely many points, a linear equation in two variables usually has infinitely many solutions.

A solution of the equation means an ordered pair \((x,y)\) that satisfies the equation.

Value of \(x\) Equation Value of \(y\) Ordered Pair
\(0\) \(y=3(0)+5\) \(5\) \((0,5)\)
\(1\) \(y=3(1)+5\) \(8\) \((1,8)\)
\(2\) \(y=3(2)+5\) \(11\) \((2,11)\)
\(-1\) \(y=3(-1)+5\) \(2\) \((-1,2)\)
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Identify the type of equation.

  2. Understand how values of \(x\) determine values of \(y\).

  3. Observe that infinitely many ordered pairs satisfy the equation.

  4. Conclude the correct option

    .
📊 Graph / Figure Graph / Figure ›
x y (0,5) (1,8) (2,11) y = 3x + 5
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  20 steps
  1. The given equation is
  2. \[y=3x+5\]
  3. This equation contains two variables \(x\) and \(y\).
  4. Therefore, it is a linear equation in two variables.
  5. For different values of \(x\), we obtain different values of \(y\).
  6. Let us verify this
  7. If \[x=0\]
  8. \[ \begin{align} y&=3(0)+5\\ y&=5 \end{align} \]
  9. So, one solution is
  10. \[(0,5)\]
  11. Again, if \[x=1\]
  12. \[ \begin{align} y&=3(1)+5\\ y&=8 \end{align} \]
  13. Another solution is
  14. \[(1,8)\]
  15. Again, if \[x=2\]
  16. \[ \begin{align} y&=3(2)+5\\ y&=11 \end{align} \]
  17. Another solution is
  18. \[(2,11)\]
  19. In this way, we can choose infinitely many values of \(x\), and each value gives a corresponding value of \(y\).
  20. Hence, the equation has infinitely many solutions.
💡 Answer Final Answer ›
Correct Option:

\[ \boxed{\text{(iii) infinitely many solutions}} \]

🎯 Exam Significance Exam Significance ›
  • Helps students understand the concept of solutions of linear equations.
  • Frequently asked in CBSE Board examinations as conceptual reasoning questions.
  • Forms the foundation of graph plotting and coordinate geometry.
  • Important for NTSE, Olympiads and Polytechnic entrance examinations where graphical interpretation is tested.
  • Strengthens understanding of straight line equations used in higher mathematics.
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1 / 4  ·  25%
Q2 →
Q2
NUMERIC3 marks
Write four solutions for each of the following equations:
(i) \(2x + y = 7\)
(ii) \(πx + y = 9\)
(iii) \(x = 4y\)
📘 Concept & Theory Theory / Concept ›

A solution of a linear equation in two variables is an ordered pair \((x,y)\) which satisfies the equation.

To find solutions:

  1. Choose any convenient value of one variable.
  2. Substitute the chosen value into the equation.
  3. Calculate the corresponding value of the other variable.
  4. Write the ordered pair in the form \((x,y)\).

Since linear equations represent straight lines, they have infinitely many solutions.

🗺️ Solution Roadmap Step-by-step Plan ›
  1. Take suitable values of one variable.
  2. Put those values into the equation.
  3. Solve step-by-step for the second variable.

  4. Form ordered pairs \((x,y)\).

✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. (i) \[2x+y=7\]
  2. We choose convenient values of \(x\) and calculate corresponding values of \(y\).
  3. From the equation:
  4. \[ \begin{align} 2x+y&=7\\ y&=7-2x \end{align} \]
  5. Chosen value of \(\small x\) Calculation of \(\small y\) Value of \(\small y\) Solution
    \(0\) \(y=7-2(0)\) \(7\) \((0,7)\)
    \(1\) \(y=7-2(1)\) \(5\) \((1,5)\)
    \(2\) \(y=7-2(2)\) \(3\) \((2,3)\)
    \(3\) \(y=7-2(3)\) \(1\) \((3,1)\)

    Therefore, four solutions are:

    \[ (0,7),\ (1,5),\ (2,3),\ (3,1) \]

💡 Answer Final Answer ›
Final Answer: Solutions are \((0,7),\quad (1,5),\quad (2,3),\quad (3,1)\)
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. (ii) \[\pi x+y=9\]
  2. We choose convenient values of \(x\) and calculate corresponding values of \(y\).
  3. From the equation:
  4. \[ \begin{align} \pi x+y&=9\\ y&=9-\pi x \end{align} \]
  5. Chosen value of \(\small x\) Calculation of \(\small y\) Value of \(\small y\) Solution
    \(0\) \(y=9-\pi(0)\) \(9\) \((0,9)\)
    \(\dfrac{1}{\pi}\) \(y=9-\pi\left(\dfrac{1}{\pi}\right)\) \(8\) \(\left(\dfrac{1}{\pi},8\right)\)
    \(1\) \(y=9-\pi(1)\) \(9-\pi\) \((1,9-\pi)\)
    \(\dfrac{9}{\pi}\) \(y=9-\pi\left(\dfrac{9}{\pi}\right)\) \(0\) \(\left(\dfrac{9}{\pi},0\right)\)

    Therefore, four solutions are:

    \[ (0,9),\ \left(\frac{1}{\pi},8\right),\ (1,9-\pi),\ \left(\frac{9}{\pi},0\right) \]

💡 Answer Final Answer ›
Final Answer: Solutions are \((0,9),\quad \left(\frac{1}{\pi},8\right),\quad (1,9-\pi),\quad \left(\frac{9}{\pi},0\right)\)
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. (iii) \[x=4y\]
  2. Here, we choose convenient values of \(y\) and calculate corresponding values of \(x\).
  3. From the equation
  4. \[x=4y\]
  5. Chosen value of \(\small y\) Calculation of \(\small x\) Value of \(\small x\) Solution
    \(-1\) \(x=4(-1)\) \(-4\) \((-4,-1)\)
    \(0\) \(x=4(0)\) \(0\) \((0,0)\)
    \(1\) \(x=4(1)\) \(4\) \((4,1)\)
    \(2\) \(x=4(2)\) \(8\) \((8,2)\)

    Therefore, four solutions are:

    \[ (-4,-1),\ (0,0),\ (4,1),\ (8,2) \]

💡 Answer Final Answer ›
Final Answer: Solutions are \((-4,-1),\quad (0,0),\quad (4,1),\quad (8,2)\)
🎯 Exam Significance Exam Significance ›
  • Helps students understand how ordered pairs satisfy linear equations.
  • Important for plotting graphs of straight lines in Coordinate Geometry.
  • Frequently asked in CBSE Board examinations in table-based and graph-based questions.
  • Forms the foundation for solving simultaneous linear equations in higher classes.
  • Useful for NTSE, Olympiads and entrance examinations where analytical reasoning and graph interpretation are tested.
← Q1
2 / 4  ·  50%
Q3 →
Q3
NUMERIC3 marks

Check which of the following are solutions of the equation \[ x-2y=4 \] and which are not:

  1. \(\small (0,2)\)
  2. \(\small (2,0)\)
  3. \(\small (4,0)\)
  4. \(\small (\sqrt{2},4\sqrt{2})\)
  5. \(\small (1,1)\)
📘 Concept & Theory Theory / Concept ›

A solution of a linear equation in two variables is an ordered pair \((x,y)\) that satisfies the equation.

To verify whether an ordered pair is a solution:

  1. Substitute the given value of \(x\) and \(y\) into the equation.
  2. Simplify the left-hand side step-by-step.
  3. Compare the obtained value with the right-hand side.
  4. If both sides are equal, the ordered pair is a solution.
  5. If both sides are not equal, the ordered pair is not a solution.
Ordered Pair \((x,y)\) Substitute In Equation Check LHS = RHS
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Take one ordered pair at a time.

  2. Substitute the values of \(x\) and \(y\).

  3. Simplify carefully step-by-step.
  4. Decide whether the equality is true or false.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. (i) Verify \((0,2)\)
  2. Substitute
  3. \[ x=0,\qquad y=2 \]
  4. \[ \begin{aligned} x-2y&=4\\ 0-2(2)&=4\\ 0-4&=4\\ -4&=4 \end{aligned} \]
  5. Since \[ -4\ne4 \] the statement is false.
  6. Hence, \((0,2)\) is not a solution.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. (ii) Verify \((2,0)\)
  2. Substitute
  3. \[ x=2,\qquad y=0 \]
  4. \[ \begin{aligned} x-2y&=4\\ 2-2(0)&=4\\ 2-0&=4\\ 2&=4 \end{aligned} \]
  5. Since \[ 2\ne4 \] the statement is false.
  6. Hence, \((2,0)\) is not a solution.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  7 steps
  1. (iii) Verify \((4,0)\)
  2. Substitute
  3. \[ x=4,\qquad y=0 \]
  4. \[ \begin{aligned} x-2y&=4\\ 4-2(0)&=4\\ 4-0&=4\\ 4&=4 \end{aligned} \]
  5. Since both sides are equal,
  6. \[ 4=4 \]
  7. Hence, \((4,0)\) is a solution.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. (iv) Verify \((\sqrt{2},4\sqrt{2})\)
  2. Substitute
  3. \[ x=\sqrt{2},\qquad y=4\sqrt{2} \]
  4. \[ \begin{aligned} x-2y&=4\\ \sqrt{2}-2(4\sqrt{2})&=4\\ \sqrt{2}-8\sqrt{2}&=4\\ (1-8)\sqrt{2}&=4\\ -7\sqrt{2}&=4 \end{aligned} \]
  5. Since \[ -7\sqrt{2}\ne4 \] the statement is false.
  6. Hence, \((\sqrt{2},4\sqrt{2})\) is not a solution.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. (v) Verify \((1,1)\)
  2. Substitute
  3. \[ x=1,\qquad y=1 \]
  4. \[ \begin{aligned} x-2y&=4\\ 1-2(1)&=4\\ 1-2&=4\\ -1&=4 \end{aligned} \]
  5. Since \[ -1\ne4 \] the statement is false.
  6. Hence, \((1,1)\) is not a solution.
🎯 Exam Significance Exam Significance ›
  • Helps students understand how to verify ordered pairs in linear equations.
  • Frequently asked in CBSE Board examinations as concept-based algebra questions.
  • Builds a strong foundation for graph plotting and coordinate geometry.
  • Important for NTSE, Olympiads and entrance examinations involving logical verification.
  • Strengthens algebraic substitution and simplification skills.
← Q2
3 / 4  ·  75%
Q4 →
Q4
NUMERIC3 marks

Find the value of \(k\) if\(x=2,\quad y=1\)is a solution of the equation \(2x+3y=k\)

📘 Concept & Theory Theory / Concept ›

A pair of values \((x,y)\) is called a solution of a linear equation if it satisfies the equation after substitution.

To find an unknown constant:

  1. Substitute the given values of variables into the equation.
  2. Simplify the left-hand side carefully.
  3. The resulting value equals the constant term.

Since \((2,1)\) is a solution, it must satisfy the equation exactly.

🗺️ Solution Roadmap Step-by-step Plan ›
  1. Write the given equation clearly.

  2. Substitute the values of \(x\) and \(y\).

  3. Simplify step-by-step.

  4. Obtain the value of \(k\).

✏️ Solution Complete Solution ›
Step-by-step Solution  ·  12 steps
  1. Given equation
  2. \[2x+3y=k\]
  3. It is given that
  4. \[x=2,\qquad y=1\]
  5. Since \((2,1)\) is a solution of the equation, it must satisfy the equation.
  6. Substitute the values of \(x\) and \(y\):
  7. \[ \begin{aligned} 2x+3y&=k\\ 2(2)+3(1)&=k \end{aligned} \]
  8. Multiply the terms
  9. \[ \begin{aligned} 4+3&=k \end{aligned} \]
  10. Add the numbers:
  11. \[ \begin{aligned} 7&=k \end{aligned} \]
  12. Therefore,

    \[ \boxed{k=7} \]

📊 Graph / Figure Graph / Figure ›
(2,1) 2x + 3y = 7 2 1 Point satisfies equation
🎯 Exam Significance Exam Significance ›
  • Helps students understand substitution in linear equations.
  • Frequently asked in CBSE Board examinations as short-answer algebra questions.
  • Builds strong foundations for solving simultaneous linear equations in higher classes.
  • Important for NTSE, Olympiads and entrance examinations involving algebraic reasoning.
  • Strengthens arithmetic simplification and equation verification skills.
← Q3
4 / 4  ·  100%
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NCERT Class 9 Maths Exercise 4.2 Solutions
NCERT Class 9 Maths Exercise 4.2 Solutions — Complete Notes & Solutions · academia-aeternum.com
Chapter 4 of NCERT Class 9 Mathematics, Linear Equations in Two Variables, introduces students to the world of equations containing two variables. Through practical examples and exercises, learners discover how such equations graphically represent straight lines, the method to find infinite solutions, and how to solve contextual problems using algebraic and graphical approaches. The chapter covers the formation, solution, and real-life application of linear equations in two variables, fostering…
🎓 Class 9 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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