Welcome back, students! Today, we will explore Exercise 4.2 of Chapter 4 (Linear Equations in Two Variables). This exercise focuses on finding solutions to linear equations and checking if a given pair of values satisfies an equation. This is a crucial concept not just for your school exams but also for competitive examinations like NTSE and Olympiads. Let's solve these step-by-step!
Exercise 4.2 Solutions
Q1. Which one of the following options is true, and why?
y = 3x + 5 has:
(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions.
Answer: (iii) infinitely many solutions
Reason: The given equation y = 3x + 5 is a linear equation in two variables. For any value of x we choose, we will get a corresponding value of y. Since there are infinitely many real numbers that we can substitute for x, there will be infinitely many corresponding values of y. Thus, the equation has infinitely many solutions.
Q2. Write four solutions for each of the following equations:
(i) 2x + y = 7
We can rewrite the equation as: y = 7 - 2x
- Let
x = 0⇒y = 7 - 2(0) = 7. Solution: (0, 7) - Let
x = 1⇒y = 7 - 2(1) = 5. Solution: (1, 5) - Let
x = 2⇒y = 7 - 2(2) = 3. Solution: (2, 3) - Let
x = 3⇒y = 7 - 2(3) = 1. Solution: (3, 1)
Four solutions are: (0, 7), (1, 5), (2, 3), and (3, 1).
(ii) πx + y = 9
We can rewrite the equation as: y = 9 - πx
- Let
x = 0⇒y = 9 - π(0) = 9. Solution: (0, 9) - Let
x = 1⇒y = 9 - π(1) = 9 - π. Solution: (1, 9 - π) - Let
x = 2⇒y = 9 - π(2) = 9 - 2π. Solution: (2, 9 - 2π) - Let
x = -1⇒y = 9 - π(-1) = 9 + π. Solution: (-1, 9 + π)
Four solutions are: (0, 9), (1, 9 - π), (2, 9 - 2π), and (-1, 9 + π).
(iii) x = 4y
- Let
y = 0⇒x = 4(0) = 0. Solution: (0, 0) - Let
y = 1⇒x = 4(1) = 4. Solution: (4, 1) - Let
y = 2⇒x = 4(2) = 8. Solution: (8, 2) - Let
y = -1⇒x = 4(-1) = -4. Solution: (-4, -1)
Four solutions are: (0, 0), (4, 1), (8, 2), and (-4, -1).
Q3. Check which of the following are solutions of the equation x - 2y = 4 and which are not:
Our test equation is: LHS = x - 2y and RHS = 4
- (0, 2): Substitute x = 0, y = 2.
LHS = 0 - 2(2) = -4 ≠ RHS. Not a solution. - (2, 0): Substitute x = 2, y = 0.
LHS = 2 - 2(0) = 2 ≠ RHS. Not a solution. - (4, 0): Substitute x = 4, y = 0.
LHS = 4 - 2(0) = 4 = RHS. Is a solution. - (√2, 4√2): Substitute x = √2, y = 4√2.
LHS = √2 - 2(4√2) = √2 - 8√2 = -7√2 ≠ RHS. Not a solution. - (1, 1): Substitute x = 1, y = 1.
LHS = 1 - 2(1) = -1 ≠ RHS. Not a solution.
Q4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
Since (2, 1) is a solution of the equation, substituting x = 2 and y = 1 must satisfy the equation:
2x + 3y = k
⇒ 2(2) + 3(1) = k
⇒ 4 + 3 = k
⇒ k = 7
Hence, the value of k is 7.
ЁЯТб Aptitude Corner (NTSE / Olympiad Special)
Concept: Finding Number of Positive Integral Solutions
In competitive exams, a common question is: "Find the number of positive integer solutions (where x > 0, y > 0 are integers) for an equation like ax + by = c."
Example: Find the number of positive integer solutions for 2x + y = 7.
- Rewrite as:
y = 7 - 2x - Since y must be positive (y > 0):
7 - 2x > 0⇒2x < 7⇒x < 3.5. - Since x must be a positive integer,
xcan only be1, 2, or 3. - This gives exactly 3 positive integer solutions: (1, 5), (2, 3), and (3, 1).
I hope these solutions made Exercise 4.2 easy to understand! Remember, practicing finding coordinates is the first step toward masterfully plotting graphs of linear equations. If you have any doubts, feel free to ask in class or write down your queries. Keep practicing, and happy learning!
Academic Discussion & Doubts