Linear Equations in Two Variables

Let the cost of a notebook be &rupee; x.
Let the cost of a pen be &rupee; y.
According to the given condition in the question, the cost of a notebook is twice the cost of a pen:
x = 2 × y
Therefore, the linear equation representing this statement is: x - 2y = 0.

(i) 2x + 3y = 9.35̅
Can be written as: 2x + 3y - 9.35̅ = 0
Comparing with ax + by + c = 0, we get: a = 2, b = 3, c = -9.35̅.

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

(iii) -2x + 3y = 6
Can be written as: -2x + 3y - 6 = 0
Comparing, we get: a = -2, b = 3, c = -6.

(iv) x = 3y
Can be written as: x - 3y + 0 = 0
Comparing, we get: a = 1, b = -3, c = 0.

(v) 2x = -5y
Can be written as: 2x + 5y + 0 = 0
Comparing, we get: a = 2, b = 5, c = 0.

(vi) 3x + 2 = 0
Since y is missing, it can be written as: 3x + 0y + 2 = 0
Comparing, we get: a = 3, b = 0, c = 2.

(vii) y - 2 = 0
Since x is missing, it can be written as: 0x + 1y - 2 = 0
Comparing, we get: a = 0, b = 1, c = -2.

(viii) 5 = 2x
Can be written as: 2x + 0y - 5 = 0 (bringing 5 to the RHS side and swapping).
Comparing, we get: a = 2, b = 0, c = -5.

The correct option is (iii) Infinitely many solutions.

Reason: For any given value of x, there will exist a corresponding unique value of y, and vice versa. Since we can choose an infinite number of real values for x, we will get an infinite number of corresponding y values. Thus, a linear equation in two variables always has infinitely many solutions.

(i) 2x + y = 7 ⇒ y = 7 - 2x
When x = 0, y = 7 - 2(0) = 7 ⇒ (0, 7)
When x = 1, y = 7 - 2(1) = 5 ⇒ (1, 5)
When x = 2, y = 7 - 2(2) = 3 ⇒ (2, 3)
When x = 3, y = 7 - 2(3) = 1 ⇒ (3, 1)
Four solutions: (0, 7), (1, 5), (2, 3), (3, 1).

(ii) πx + y = 9 ⇒ y = 9 - πx
When x = 0, y = 9 - π(0) = 9 ⇒ (0, 9)
When x = 1, y = 9 - π(1) = 9 - π ⇒ (1, 9 - π)
When x = 2, y = 9 - π(2) = 9 - 2π ⇒ (2, 9 - 2π)
When x = -1, y = 9 - π(-1) = 9 + π ⇒ (-1, 9 + π)
Four solutions: (0, 9), (1, 9 - π), (2, 9 - 2π), (-1, 9 + π).

(iii) x = 4y
When y = 0, x = 4(0) = 0 ⇒ (0, 0)
When y = 1, x = 4(1) = 4 ⇒ (4, 1)
When y = 2, x = 4(2) = 8 ⇒ (8, 2)
When y = -1, x = 4(-1) = -4 ⇒ (-4, -1)
Four solutions: (0, 0), (4, 1), (8, 2), (-4, -1).

Equation is x - 2y = 4. We substitute x and y from each coordinate to see if LHS = RHS (4).

(i) (0, 2): LHS = 0 - 2(2) = -4. RHS = 4. Since LHS ≠ RHS, (0, 2) is not a solution.

(ii) (2, 0): LHS = 2 - 2(0) = 2. RHS = 4. Since LHS ≠ RHS, (2, 0) is not a solution.

(iii) (4, 0): LHS = 4 - 2(0) = 4. RHS = 4. Since LHS = RHS, (4, 0) is a solution.

(iv) (√2, 4√2): LHS = √2 - 2(4√2) = √2 - 8√2 = -7√2. RHS = 4. Since LHS ≠ RHS, it's not a solution.

(v) (1, 1): LHS = 1 - 2(1) = 1 - 2 = -1. RHS = 4. Since LHS ≠ RHS, (1, 1) is not a solution.

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

The equation 5y = 2 can be rewritten as:
0.x + 5y - 2 = 0
Comparing this with ax + by + c = 0:
a = 0, b = 5, and c = -2.

Let the tens digit be x and the units digit be y.
Original Number = 10x + y
Number after reversing digits = 10y + x
Sum: (10x + y) + (10y + x) = 110
11x + 11y = 110
x + y = 10.
Therefore, the linear equation is x + y - 10 = 0.

Equation: x + 2y = 6
To find the intersection with the x-axis: Put y = 0.
x + 2(0) = 6 ⇒ x = 6. Point is (6, 0).
To find the intersection with the y-axis: Put x = 0.
0 + 2y = 6 ⇒ y = 3. Point is (0, 3).
To graph, plot (6, 0) and (0, 3) on the Cartesian plane and draw a straight line through them.

Since (1, 1) is a solution, substitute x = 1, y = 1 in 3x - ay = 6:
3(1) - a(1) = 6
3 - a = 6
-a = 6 - 3 = 3
a = -3.

We need linear equations where LHS = RHS for x = 2, y = 14.
Let y = 7x (Because 14 = 7 × 2). Let's write it as 7x - y = 0.
Let x + y = 2 + 14 = 16. Let's write it as x + y - 16 = 0.
There are infinitely many lines passing through this point because through a single point on a plane, infinite distinct lines can be drawn.

Equation is 4y = 12 - 3x ⇒ y = (12 - 3x)/4.
When x = 0, y = 12/4 = 3 ⇒ (0, 3)
When x = 4, y = (12 - 12)/4 = 0 ⇒ (4, 0)
When x = -4, y = (12 + 12)/4 = 24/4 = 6 ⇒ (-4, 6)
When x = 8, y = (12 - 24)/4 = -12/4 = -3 ⇒ (8, -3)

Substitute x = 3, y = 4 in the equation:
3(4) = a(3) + 7
12 = 3a + 7
12 - 7 = 3a
5 = 3a ⇒ a = 5/3.

Let the total distance covered be x km.
Let the total fare be &rupee; y.
Fare for the 1st km = &rupee; 10.
Remaining distance = (x - 1) km.
Fare for remaining distance = 4(x - 1).
Total fare y = 10 + 4(x - 1) = 10 + 4x - 4
y = 4x + 6 (or 4x - y + 6 = 0).

A line meets the x-axis where y = 0.
Substitute y = 0 into 2x + 3y = 6:
2x + 3(0) = 6 ⇒ 2x = 6 ⇒ x = 3.
The point is (3, 0).

A line parallel to the x-axis has an equation in the form y = a.
Since it is 4 units below the origin, its constant y-coordinate everywhere is -4.
Therefore, the equation is y = -4 or y + 4 = 0.

Equation is y = 2x.
If x = 0, y = 0. So, (0, 0) is a solution. Yes, the origin is a solution, thereby the line passes through the origin.
Other points: (1, 2), (-1, -2). Plotting these gives a straight slanting line passing strictly through the origin (0, 0).

Let daily wages be y. Let the number of hours worked be x.
Equation: y = 2x (or 2x - y = 0).
To graph:
For x = 4 hours, y = 8.
For x = 6 hours, y = 12.
Plot (4, 8) and (6, 12). Join them to form a line passing through the origin (0, 0).

The line cuts the y-axis when the x-coordinate is 0.
Substituting x = 0 in 3x - y = 3:
3(0) - y = 3 ⇒ -y = 3 ⇒ y = -3.
The point is (0, -3).

LCM of 5 and 4 is 20:
[4(2x - 3) + 5(x + 3)] / 20 = (4x + 1) / 7
(8x - 12 + 5x + 15) / 20 = (4x + 1) / 7
(13x + 3) / 20 = (4x + 1) / 7
Cross-multiply:
7(13x + 3) = 20(4x + 1)
91x + 21 = 80x + 20
91x - 80x = 20 - 21
11x = -1 ⇒ x = -1/11.

A line parallel to the y-axis has the equation x = a.
Lying on the left means taking negative coordinates.
So, x = -3, which implies x + 3 = 0.

Substitute x = m and y = 2m - 1 in the equation:
4(m) - 3(2m - 1) + 5 = 0
4m - 6m + 3 + 5 = 0
-2m + 8 = 0
2m = 8 ⇒ m = 4.

Let father's age be y. Let son's age be x.
Equation: x + y = 45.
Calculation: When y = 35:
x + 35 = 45 ⇒ x = 10.
The son's age is 10 years.

2y - 5y = 6 + 3
-3y = 9 ⇒ y = -3.
(i) On a number line, this represents a single point at -3 on the y-axis.
(ii) On the Cartesian plane, treating y = -3 structurally as a linear equation in two variables 0.x + y = -3, it represents a line parallel to the x-axis, intersecting the y-axis at (0, -3).

The equation y = -x passes through the origin. Solutions: (1, -1), (2, -2), (-1, 1).
The line bisects Quadrants II and IV exactly in half.
Therefore, it makes an angle of 45° with the negative x-axis (or 135° with the positive x-axis).

The point of intersection of x = 2 and y = 3 is firmly the coordinate (2, 3).
We substitute (2, 3) back into the equation 3x + 2y = 12:
LHS = 3(2) + 2(3) = 6 + 6 = 12.
RHS = 12.
Since LHS = RHS, the statement is true. The line does pass through the point of intersection.