GSEB Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Exercise 4.2

Question 1. Which of the following options is true and why?
(i) a unique solution
(ii) only two solutions
(iii) infinitely many solutions
Answer: Option (iii) is correct. The equation \( y = 3x + 5 \) has many solutions. For every real value of \( x \), you will always find a matching value for \( y \), and the opposite is also true.
In simple words: This equation has countless solutions. You can pick any number for \( x \), and you'll get a valid number for \( y \), and vice versa, meaning there are endless pairs that work.

Exam Tip: Remember that linear equations in two variables always have infinitely many solutions, as there are endless points on a line.

Question 2. Write four solutions for each of the following equations.
(1) \( 2x + y = 7 \)
(2) \( \pi x + y = 9 \)
(3) \( x = 4y \)
Answer:
(1) Given equation: \( 2x + y = 7 \)
Rearranging, we get \( y = 7 - 2x \).
Let's take a few values for \( x \):
If \( x = 0 \), then \( y = 7 - 2(0) = 7 \). So, \( (0, 7) \) is a solution.
If \( x = 1 \), then \( y = 7 - 2(1) = 7 - 2 = 5 \). So, \( (1, 5) \) is a solution.
If \( x = 2 \), then \( y = 7 - 2(2) = 7 - 4 = 3 \). So, \( (2, 3) \) is a solution.
If \( x = 3 \), then \( y = 7 - 2(3) = 7 - 6 = 1 \). So, \( (3, 1) \) is a solution.
Thus, four solutions are \( (0, 7), (1, 5), (2, 3), \) and \( (3, 1) \).
In simple words: To find solutions, rearrange the equation to solve for \( y \). Then, pick some easy numbers for \( x \) (like 0, 1, 2, 3) and calculate the matching \( y \) value for each. Each pair \( (x, y) \) is a solution.

Exam Tip: Always show at least two steps for each calculation to demonstrate your method. Make sure to present your final solutions as ordered pairs \( (x, y) \).

(2) Given equation: \( \pi x + y = 9 \)
Rearranging, we get \( y = 9 - \pi x \).
Let's substitute a few values for \( x \):
If \( x = 0 \), then \( y = 9 - \pi (0) = 9 \). So, \( (0, 9) \) is a solution.
If \( x = 1 \), then \( y = 9 - \pi (1) = 9 - \pi \). So, \( (1, 9 - \pi) \) is a solution.
If \( x = 2 \), then \( y = 9 - \pi (2) = 9 - 2\pi \). So, \( (2, 9 - 2\pi) \) is a solution.
If \( x = 3 \), then \( y = 9 - \pi (3) = 9 - 3\pi \). So, \( (3, 9 - 3\pi) \) is a solution.
Thus, four solutions are \( (0, 9), (1, 9 - \pi), (2, 9 - 2\pi), \) and \( (3, 9 - 3\pi) \).
In simple words: Just like before, rearrange to find \( y \). Pick values for \( x \) (0, 1, 2, 3) and figure out what \( y \) should be. Even if \( \pi \) is in the equation, the method stays the same.

Exam Tip: When dealing with constants like \( \pi \), treat them as numbers. Do not approximate \( \pi \) unless specifically asked to do so, leave it in its exact form.

(3) Given equation: \( x = 4y \)
Here, the equation is already solved for \( x \). Let's pick values for \( y \):
If \( y = 0 \), then \( x = 4(0) = 0 \). So, \( (0, 0) \) is a solution.
If \( y = 1 \), then \( x = 4(1) = 4 \). So, \( (4, 1) \) is a solution.
If \( y = 2 \), then \( x = 4(2) = 8 \). So, \( (8, 2) \) is a solution.
If \( y = 3 \), then \( x = 4(3) = 12 \). So, \( (12, 3) \) is a solution.
Thus, four solutions are \( (0, 0), (4, 1), (8, 2), \) and \( (12, 3) \).
In simple words: This time, the equation gives \( x \) using \( y \). So, choose numbers for \( y \) (like 0, 1, 2, 3) and then find the matching \( x \) values. List the \( (x, y) \) pairs you get.

Exam Tip: You can choose values for either \( x \) or \( y \) depending on which variable is easier to isolate or already isolated in the equation.

Question 3. Check which of the following are solutions of the equation \( x - 2y = 4 \) and which are not?
(1) \( (0, 2) \)
(2) \( (2, 0) \)
(3) \( (4, 0) \)
(4) \( (\sqrt{2}, 4\sqrt{2}) \)
(5) \( (1, 1) \)
Answer:
We need to substitute the given points into the equation \( x - 2y = 4 \) to see if they satisfy it.
(1) For the point \( (0, 2) \):
Substitute \( x = 0 \) and \( y = 2 \) into the left-hand side (LHS) of the equation:
LHS \( = x - 2y = 0 - 2(2) = 0 - 4 = -4 \)
The right-hand side (RHS) of the equation is 4.
Since LHS \( = -4 \) and RHS \( = 4 \), we have LHS \( \neq \) RHS.
So, \( (0, 2) \) is not a solution to the given equation.
In simple words: Put the \( x \) and \( y \) numbers from the point into the left side of the equation. If the answer matches the right side, it's a solution. If not, it isn't. For \( (0,2) \), it doesn't match.

Exam Tip: When checking if a point is a solution, always clearly show the substitution into the LHS and compare it with the RHS.

(2) For the point \( (2, 0) \):
Substitute \( x = 2 \) and \( y = 0 \) into the left-hand side (LHS) of the equation:
LHS \( = x - 2y = 2 - 2(0) = 2 - 0 = 2 \)
The right-hand side (RHS) of the equation is 4.
Since LHS \( = 2 \) and RHS \( = 4 \), we have LHS \( \neq \) RHS.
So, \( (2, 0) \) is not a solution to the given equation.
In simple words: For \( (2,0) \), putting the numbers into the equation gives 2 on the left side, which is not 4. So this point is not a solution either.

Exam Tip: Be careful with calculations, especially with negative signs. Even a small error can lead to an incorrect conclusion.

(3) For the point \( (4, 0) \):
Substitute \( x = 4 \) and \( y = 0 \) into the left-hand side (LHS) of the equation:
LHS \( = x - 2y = 4 - 2(0) = 4 - 0 = 4 \)
The right-hand side (RHS) of the equation is 4.
Since LHS \( = 4 \) and RHS \( = 4 \), we have LHS \( = \) RHS.
So, \( (4, 0) \) is a solution to the given equation.
In simple words: For \( (4,0) \), putting the numbers into the equation makes both sides equal to 4. This means \( (4,0) \) is a solution.

Exam Tip: A point is a solution only if it makes the equation true, meaning LHS equals RHS after substitution.

(4) For the point \( (\sqrt{2}, 4\sqrt{2}) \):
Substitute \( x = \sqrt{2} \) and \( y = 4\sqrt{2} \) into the left-hand side (LHS) of the equation:
LHS \( = x - 2y = \sqrt{2} - 2(4\sqrt{2}) = \sqrt{2} - 8\sqrt{2} = -7\sqrt{2} \)
The right-hand side (RHS) of the equation is 4.
Since LHS \( = -7\sqrt{2} \) and RHS \( = 4 \), we have LHS \( \neq \) RHS.
So, \( (\sqrt{2}, 4\sqrt{2}) \) is not a solution to the given equation.
In simple words: When we use the square root numbers for \( x \) and \( y \), the left side of the equation becomes \( -7\sqrt{2} \), which is not 4. So, this point is not a solution.

Exam Tip: Be careful when simplifying expressions involving square roots. Remember that \( a\sqrt{b} - c\sqrt{b} = (a-c)\sqrt{b} \).

(5) For the point \( (1, 1) \):
Substitute \( x = 1 \) and \( y = 1 \) into the left-hand side (LHS) of the equation:
LHS \( = x - 2y = 1 - 2(1) = 1 - 2 = -1 \)
The right-hand side (RHS) of the equation is 4.
Since LHS \( = -1 \) and RHS \( = 4 \), we have LHS \( \neq \) RHS.
So, \( (1, 1) \) is not a solution to the given equation.
In simple words: For the point \( (1,1) \), putting the numbers into the equation results in -1 on the left side, which does not match the 4 on the right. Therefore, this point is also not a solution.

Exam Tip: Double-check your arithmetic, especially with subtraction, as simple mistakes can lead to incorrect conclusions.

Question 4. Find the value of \( k \), if \( x = 2, y = 1 \) is a solution of the equation \( 2x + 3y = k \).
Answer: The given equation is \( 2x + 3y = k \).
Since \( (x, y) = (2, 1) \) is a solution to this equation, it means that these values must satisfy the equation when substituted.
Substitute \( x = 2 \) and \( y = 1 \) into the equation:
\( 2(2) + 3(1) = k \)
\( 4 + 3 = k \)
\( 7 = k \)
Therefore, the value of \( k \) is 7.
In simple words: If a point is a solution, its \( x \) and \( y \) values work in the equation. Just put \( x=2 \) and \( y=1 \) into \( 2x+3y=k \), calculate the left side, and that number will be \( k \).

Exam Tip: When a point is given as a solution, always substitute its coordinates into the equation to find any unknown constants.