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LINEAR EQUATIONS IN TWO VARIABLES - Q&A

Exercise 4.1

1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be ₹ x and that of a pen to be ₹ y).

Let the cost of a notebook be = ₹ x
Let the cost of a pen be = ₹ y
According to the problem, the cost of a notebook is twice the cost of a pen.
So, x = 2 × y
x = 2y
Subtracting 2y from both sides, we get:
x - 2y = 0
This is the required linear equation in two variables.

2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

(i) 2x + 3y = 9.3̅5

We need to bring all terms to the left side to match the form ax + by + c = 0.
2x + 3y - 9.3̅5 = 0
Comparing this with ax + by + c = 0, we get:
a = 2
b = 3
c = -9.3̅5

(ii) x - y/5 - 10 = 0

This is already in the form ax + by + c = 0.
x - (1/5)y - 10 = 0
Comparing with ax + by + c = 0:
a = 1
b = -1/5
c = -10

(iii) -2x + 3y = 6

Subtract 6 from both sides:
-2x + 3y - 6 = 0
Comparing with ax + by + c = 0:
a = -2
b = 3
c = -6

(iv) x = 3y

Subtract 3y from both sides:
x - 3y = 0
This can be written as x - 3y + 0 = 0.
Comparing with ax + by + c = 0:
a = 1
b = -3
c = 0

(v) 2x = -5y

Add 5y to both sides:
2x + 5y = 0
This can be written as 2x + 5y + 0 = 0.
Comparing with ax + by + c = 0:
a = 2
b = 5
c = 0

(vi) 3x + 2 = 0

This equation has no y term, so we can write it as:
3x + 0y + 2 = 0
Comparing with ax + by + c = 0:
a = 3
b = 0
c = 2

(vii) y - 2 = 0

This equation has no x term, so we can write it as:
0x + y - 2 = 0
Comparing with ax + by + c = 0:
a = 0
b = 1
c = -2

(viii) 5 = 2x

Subtract 5 from both sides (or move 2x to the left):
2x - 5 = 0
This can be written as:
2x + 0y - 5 = 0
Comparing with ax + by + c = 0:
a = 2
b = 0
c = -5


Exercise 4.2

1. 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

Option (iii) infinitely many solutions is true.
Reason: A linear equation in two variables (like y = 3x + 5) represents a straight line. A line consists of infinite points, and every point on the line is a solution to the equation. For every value we choose for x, we get a corresponding unique value for y. Since we can choose infinite values for x, there are infinite corresponding values for y.

2. Write four solutions for each of the following equations:

(i) 2x + y = 7

We can express y in terms of x: y = 7 - 2x
1. Let x = 0, then y = 7 - 2(0) = 7. Solution: (0, 7)
2. Let x = 1, then y = 7 - 2(1) = 5. Solution: (1, 5)
3. Let x = 2, then y = 7 - 2(2) = 3. Solution: (2, 3)
4. Let x = 3, then y = 7 - 2(3) = 1. Solution: (3, 1)
The four solutions are (0, 7), (1, 5), (2, 3), and (3, 1).

(ii) πx + y = 9

Express y in terms of x: y = 9 - πx
1. Let x = 0, then y = 9 - π(0) = 9. Solution: (0, 9)
2. Let x = 1, then y = 9 - π(1) = 9 - π. Solution: (1, 9 - π)
3. Let x = -1, then y = 9 - π(-1) = 9 + π. Solution: (-1, 9 + π)
4. Let x = 2, then y = 9 - π(2) = 9 - 2π. Solution: (2, 9 - 2π)
The four solutions are (0, 9), (1, 9 - π), (-1, 9 + π), and (2, 9 - 2π).

(iii) x = 4y

Here it is easier to choose values for y and find x.
1. Let y = 0, then x = 4(0) = 0. Solution: (0, 0)
2. Let y = 1, then x = 4(1) = 4. Solution: (4, 1)
3. Let y = -1, then x = 4(-1) = -4. Solution: (-4, -1)
4. Let y = 2, then x = 4(2) = 8. Solution: (8, 2)
The four solutions are (0, 0), (4, 1), (-4, -1), and (8, 2).

3. Check which of the following are solutions of the equation x - 2y = 4 and which are not:

(i) (0, 2)

Substitute x = 0 and y = 2 in LHS:
LHS = 0 - 2(2) = -4
RHS = 4
Since LHS ≠ RHS, (0, 2) is not a solution.

(ii) (2, 0)

Substitute x = 2 and y = 0 in LHS:
LHS = 2 - 2(0) = 2
RHS = 4
Since LHS ≠ RHS, (2, 0) is not a solution.

(iii) (4, 0)

Substitute x = 4 and y = 0 in LHS:
LHS = 4 - 2(0) = 4
RHS = 4
Since LHS = RHS, (4, 0) is a solution.

(iv) (√2, 4√2)

Substitute x = √2 and y = 4√2 in LHS:
LHS = √2 - 2(4√2)
LHS = √2 - 8√2 = -7√2
RHS = 4
Since LHS ≠ RHS, (√2, 4√2) is not a solution.

(v) (1, 1)

Substitute x = 1 and y = 1 in LHS:
LHS = 1 - 2(1) = 1 - 2 = -1
RHS = 4
Since LHS ≠ RHS, (1, 1) is not a solution.

4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Since x = 2 and y = 1 is a solution, these values must satisfy the equation.
Substitute x = 2 and y = 1 into the equation:
2(2) + 3(1) = k
4 + 3 = k
k = 7
Therefore, the value of k is 7.

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