AP 9th Class Maths Chapter 4 Linear Equations in Two Variables Important Questions

These AP 9th Class Maths Important Questions 4th Lesson Linear Equations in Two Variables will help students prepare well for the exams.

AP Board Class 9 Maths 4th Lesson Linear Equations in Two Variables Important Questions

9th Class Maths Linear Equations in Two Variables 2 Marks Important Questions

Question 1.
Express the following equation in the form of ax + by + c = 0 and write the values of a, b and c.
\(\frac{x}{2}\) + \(\frac{y}{3}\) + \(\frac{1}{6}\)
Solution:
\(\frac{x}{2}\) + \(\frac{y}{3}\) + \(\frac{1}{6}\) (Given)
⇒ (\(\frac{x}{2}\) × 6) + (\(\frac{y}{3}\) × 6) + \(\frac{1}{6}\) × 6) (Multiply each term by 6)
⇒ 3x + 2y = 1
⇒ 3x + 2y – 1 = 0 (Comparing with ax + by + c = 0)
Now, a = 3, b = 2 and c = – 1.

Question 2.
Give the equations of two lines pass¬ing through (2, 14). How many more such lines are there and why?
Solution:
The equations of two lines passing through (2, 14) can be taken as
x + y = 16 and
7x – y = 0
There are infinitely many such lines because through a point an infinite number of lines can be drawn.

Question 3.
For the graph of the linear equation ax + by + c = 0 to pass through origin which of the three a, b and c is necessarily zero ?
Solution:
The given linear equation is
ax + by + c = 0 …………… (1)
If the graph of (1) passes through origin, then
a(0) + b(0) + c = 0 ⇒ c = 0
Hence, c is necessarily zero.

Question 4.
Write x = -6 as an equation in two variables.
Solution:
x = -6 ⇒ x = 6 = 0
⇒ 1.x + 0.y + 6 = 0
Comparing it with ax + by + c = 0, we get
a = 1, b = 0, c = 6
So, this is in the form of an equation in two variables x and y.

Question 5.
Total number of legs in a herd of goats and hens is 40. Represent this situation in the form of a linear equation in two variables.
Solution:
Let the number of goats and hens be x any y respectively.
(∵ Number of legs of 1 goat = 4)
∴ Number of legs of x goats = 4x
(∵ Number of legs of 1 hen = 2)
∴ Number of legs of y hens = 2y
According to the question,
4x + 2y = 40
⇒ 2x + y = 20
Dividing throughout by 2
which is the required linear equation in two variables.

Question 6.
Find any three solutions of the equation 2x – 5y = 10.
Solution:
2x – 5y = 10
⇒ 5y = 2x – 10
when x = 0,
5y = 2(0) – 10 = 0 – 10 = – 10
⇒ y = -2
when x = 5
5y = 2(5) – 10 = 10 – 10 = 0
⇒ y = 0
when x = – 5,
5y = 2(- 5) – 10 = – 10 – 10 = – 20
⇒ y = – 4
Hence, three solutions are (0, -2), (5, 0) and (-5, -4).

Question 7.
If the point (2, -7) lies on the line 4x + my = 22, then what will be the value of m ?
Solution:
The given line is 4x + my = 22
If it passes through the point (2, -7) then, 4(2) + m(- 7) = 22
⇒ 8 – 7m = 22
⇒ -7m = 22 – 8
⇒ -7m = 14
⇒ m = –\(\frac{14}{7}\) ⇒ m = -2

Question 8.
If x = 1, y = 2 is a solution of the equation kx² + ky = 3, then find the values of k.
Solution:
If x = 1 y = 2 is a solution of the equation k²x + ky = 3, then.
k²(1) + k(2) = 3
k² + 2k = 3
⇒ k² + 2k – 3 = 0
⇒ k² + 3k – k – 3 = 0
⇒ k(k + 3) – 1(k + 3) = 0
⇒ k + 3 = 0 ork – l = 0
⇒ k = -3 or k = 1
∴ k = -3, 1

Question 9.
Express y in terms of x in the equation 2x – 3y = 5. Find two solutions also.
Solution:
2x – 3y = 5 ⇒ 3y = 2x – 5
⇒ y = \(\frac{2 x-5}{3}\)
when x = 4, y = \(\frac{2(4)-5}{3}\) = 1
when x = -2, y = \(\frac{2(-2)-5}{3}\) = -3
So, two solutions are (4, 1) and (-2, -3).

Question 10.
Write three solutions for the equation 2x + y = 7.
Solution:
The given equation is
2x + y = 7
⇒ y = 7 – 2x ……….. (1)
When x = 0, then, y = 7 – 2(0) = 7 – 0 = 7
When x = 1, then, y = 7 – 2(1) = 7 – 2 = 5
When x = 2, then, y = 7 – 2(2) = 7 – 4 = 3
Hence, three solutions for the equation (1) are (0, 7), (1, 5) and (2, 3).

Question 11.
Write the coordinates of the point where the line 2x – 7y = 14 intersects x-axis.
Solution:
The given line is
2x- 7y = 14 …………. (1)
It will intersect x – axis at the point for which y = 0. So, putting y = 0 in equation (1), we get
2x – 7(0) = 14
⇒ 2x = 14
⇒ x = 7
Hence, the required point is (7, 0).

Question 12.
Rehaan is concerned about his health. He balances his intake of calories by doing some physical activity. He wants to burn 250 Kcal. He chooses running up the stairs and jogging as physical activities. Write a linear equation for the same and give two solutions.
Solution:
Let Rehaan burn x K cal by running up the stairs and y K cal by jogging.
Then, according to the question,
x + y = 250 ………. (1)
which is the required linear equation.
From(1), y = 250 – x
when x = 0, y = 250
when x = 250, y = 0
Hence, two solutions are (0, 250) and (250, 0).

Question 13.
Shahid wants to burn 200 kilo calories in a day doing physical activity. He spend 2 minutes for running and 3 minutes for running up stairs for burning kilo calories in a day.
Write a linear equation for the above situation.
Solution:
Let he spends for physical activity
= 2 mins = 2x
For running up stairs = 3 mins = 3y
He wants to burn 200 kilo calories for a day.
∴ 2x + 3y – 200

9th Class Maths Linear Equations in Two Variables 4 Marks Important Questions

Question 1.
The taxi fare in a city is as follows: For the first kilometre, the fare is ₹ 8 and for the subsequent distance it is ₹ 5 per km. Taking the distance covered as x km and total fare as ₹ y, write a linear equation for this information and draw its graph.
Solution:
Total distance covered = x km
Toted fare = ₹ y
Fare for the first kilometre = ₹ 8
Subsequent distance = (x – 1) km
Fare for the subsequent distance = ₹ 5(x – 1)
According to the question,
y = 8 + 5(x – 1)
⇒ y = 8 + 5x – 5
⇒ y = 5x + 3
Table of solution

x	0	1
y	3	8

We plot the Points (0, 3) and (1, 8) on the graph paper and join the same by a ruler to get the line which is the graph of the equation y 5x + 3.

Question 2.
If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
i) 2 units
ii) 0 unit.
Solution:
Let the work done by the constant force be y units and the distance travelled by the body be x units.
Constant force = 5 units
We know that,
Work done = Force × Displacement
y = 5x
Table of solutions

x	0	1
y	0	5

We plot the points (0, 0) and (1, 5) on the graph paper and join the same by a ruler to get the line which is the graph of the equation y = 5x.

i) Let A → (2, 0). Through A, draw a line parallel to OY to intersect the graph of the equation y = 5x at B. Through B, draw a line parallel to OX to intersect OY at C. Then,
C → (0, 10)
∴ Work done when the distance travelled by the body is 2 units = 10 units.
ii) Clearly y = 0 when x = 0. So, the work done when the distance travelled by the body is 0 units’is 0 units.

Question 3.
Give the geometric representations of y = 3 as an equation
i) in one variable
ii) in two variables.
Solution:
The given equation is y = 3
i) In one variable : The representation of y = 3 on the number line is as shown below :

ii) In two variables :
y = 3
⇒ 0.x + 1.y = 3

It is a linear equation in two variables x and y. This is represented by a line. AH the values of x are permissible because 0.x is always 0. However, y must satisfy the relation y = 3. Hence, two solutions of the given equation are x = 0, y = 3 and x = 2, y = 3.
Thus the graph AB is a line parallel to the x – axis at a distance of 3 units above it.

Question 4.
Give the geometric representations of 2x + 9 = 0 as an equation
i) in one variable
ii) in two variables.
Solution:
The given equation is 2x + 9 = 0
i) In one variable :
2x + 9 = 0 ⇒ 2x = -9 ⇒ x = –\(\frac{9}{2}\)
The representation of 2x + 9 = 0 on the number line is as shown below :

(ii) In two variables:
2x + 9 = 0
⇒ 2x + 0y + 9 = 0
It is a linear equation in two variables x and y. This is represented by a line.
All the values of y are permissible because 0y is always 0. However x must satisfy the relation 2x + 9 = 0, i.e.,
x = –\(\frac{9}{2}\)
Hence, two solutions of the given equation are x = –\(\frac{9}{2}\), y = 0 and x = – \(\frac{9}{2}\), y = 2.
The graph AB is a line parallel to the y – axis and a distance of \(\frac{9}{2}\) units to the left of origin O.

Question 5.
Find the value of k, if x = 1, y = 0 is n solution of the equation 3x – 8y – k.
solution:
Given equation, 3x – 8y = k
x = 1 and y = 0 is a solution of given equation,
⇒ 3(1) – 8(0) = k
3 – 0 = k
∴ k = 3

Question 6.
Observe the graphs of lines l₁, l₂, l₃ and l₄, answer the questions that follow.


i) Equation of l₁ is
A) x + y = 4
B) x – y + 4 = 0
C) x + y + 4 = 0
D) x – y – 4 = 0
Answer:
A) x + y = 4

ii) Equation of l₂ is
A) x + y = 4
B) x – y + 4 = 0
C) x + y + 4 = 0
D) x – y – 4 = 0
Answer:

iii) Equation of l₃ is
A) x + y = 4
B) x – y + 4 = 0
C) x + y + 4 = 0
D) x – y – 4 = 0
Answer:
C) x + y + 4 = 0

iv) Equation of l₄ is ( D )
A) x + y = 4
B) x – y + 4 = 0
C) x + y + 4 = 0
D) x – y – 4 = 0
Answer:
D) x – y – 4 = 0

v) Equation of the line through B and D
A) x = 0
B) y = 0
C) x = 4
D) y = 4

9th Class Maths Linear Equations in Two Variables 8 Marks Important Questions

Question 1.
Angles of quadrilateral are 5x + 50°, 4x + 60°, 6y + 60° and 3y + 100°. Write a linear equation which satisfies this data. Also draw the graph for the same.
Solution:
We know that the sum of the angles of a quadrilateral is 360°.
∴ (5x + 50°) + (4x + 60°) + (6y + 60°) + (3y + 100°) = 360°
⇒ 9x + 9y + 270° = 360°
⇒ 9x + 9y = 360° – 270° = 90°
⇒ x + y = 10° (Dividing throughout by 9 which is the required linear equation)
x + y = 10 ⇒ y = 10 – x
Table of solutions

x	0	10
y	10	0

We plot the points (0, 10) and (10, 0) on a graph paper and join the same by a ruler to get the line which is the graph of the equation x + y = 10.

Question 2.
Cost of a pencil is 3 times the cost of an eraser. Represent this in the form of a linear equation in two variables. Also, find four solutions of the above equation.
Solution:
Let the cost of a pencil and cost of an eraser be ₹ x and ₹ y respectively.
Then, according to the question,
x = 3y ⇒ x – 3y = 0 …………. (1)
which is the required linear equation in two variables.
From (1), 3y = x ⇒ y = \(\frac{x}{3}\)
when x = 0, y = \(\frac{0}{3}\) = 0
when x = 3, y = \(\frac{3}{3}\) = 1
when x = 6, y = \(\frac{6}{3}\) = 2
when x = 9, y = \(\frac{9}{3}\) = 3
Hence, four solutions of equation (1) are (0, 0), (3, 1), (6, 2) and (9, 3).

Question 3.
Write the equations of the lines drawn in following graph:

Also, find the area enclosed between these lines.
Solution:
From graph, we see that

1. Line a is a line parallel to y – axis and is at a distance of 2 units from origin to the right of y-axis. Therefore, its equation is x = 2
2. Line b is a line parallel to x – axis and is at a distance of 1 unit from origin above x – axis. Therefore, its equation is y = 1.
3. Line c is a line parallel to x – axis and is at a distance of 3 units from origin above x – axis. Therefore, its equation is y = 3.
4. Line d is a line parallel to y – axis and is at a distance of 1 unit from origin to the left of y-axis.
   Therefore, its equation is x = -1
   Again, area enclosed between these lines = 3 × 2 = 6 square units.

Question 4.
Shade the trapezium formed by the graphs of the equation x + y = 3 and x + y = 1 and the y – axis and the x – axis. Write the coordinates of the vertices of the trapezium so formed. Also, find its area.
Solution:
First line is
x + y = 3 …………. (1)
⇒ y = 3 – x
Table of solutions

x	0	3
y	3	0

We plot the points (0, 3) and (3, 0) on a graph paper and join the same by a ruler to get the line which is the graph of the equation (1).
Second line is
x + y = 1 ………… (2)
⇒ y = 1 – x
Table of solutions

x	1	0
y	0	1

We plot the points (1, 0) and (0, 1) on the same graph paper and join the same by a ruler to get the line which is the graph of the equation (2).
Third line is y-axis.
Fourth line is x-axis.
From graph, we see that the coordinates of the vertices of the trapezium ABCD so formed are A(3, 0), B(0, 3), C(0, 1) and D(1, 0).
Also, area of the trapezium formed = Area of ΔOAB – Area of ΔOCD
= \(\frac{3 \times 3}{2}\) – \(\frac{1 \times 1}{2}\) 4 square units.

Question 5.
On her birthday, Anisha donates 2 toffees to each child of an orphanage and 15 chocolates to adults working there. Taking the total items distributed as x and the number of children as y write a linear equation in 2 variables for the above situation.
a) Write the equation in standard form.
b) How many children are there if 61 items were distributed ?
Solution:
Total items distributed = x
Number of children = y
Number of toffees given to each child = 2
∴ Number of toffees given to y children = 2y
Number of chocolates given to adults = 15
∴ x = 15 + 2y ………….. (1)
This is the required linear equation in 2 variables.
a) Equation (1) in standard form is
x – 2y – 15 = 0
as it is of the form
ax + by + c = 0

b) If x = 61, then from (1)
61 = 15 + 2y
⇒ 2y = 61 – 15 = 46 ⇒ y = \(\frac{46}{2}\) = 23
Hence, there are 23 children.

Question 6.
The parking charges (per hour) for vehicles in a parking in ₹ 20 for the first two hours and ₹ 10 per hour for subsequent hours (where x ≥ 2) and the total parking charges is ₹ y. Write a linear equation to express the above relation and draw its graph.
Solution:
Total hours = x
Total parking charges = ₹ y
Charges for the first two hours = ₹ 20
Subsequent hours = (x – 2)
Charges for the subsequent hours = ₹ (x – 2)10
According to the questions
y = 20 + 10(x – 2)
⇒ y = 20 + 10x – 20 ⇒ y = 10x ……….. (1)
which is the required linear equation.
Table of solutions

x	0	10
y	10	0

We plot the points 0(0, 0) and A(1, 10) on a graph paper and join the same by a ruler to get the line which is the graph of equation (1).

Question 7.
Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and x-axis. Find the area of the shaded region.
Solution:
2x + y = 6 ……….. (1)
⇒ y = 6 – 2x
Table of solutions

x	0	3
y	6	0

We plot the points A(0, 6) and B(3, 0) on a graph paper and join the same by a ruler to get the line which is the graph of equation (1).
2x – y + 2 = 0 ……….. (2)
⇒ y = 2x + 2
Table of solutions

x	-1	0
y	0	2

We plot the points C(- 1, 0) and D(0, 2) on the same graph paper and join the same by a ruler to get the line which is the graph of equation (2).

From graph, we see that the two lines intersect at the point E(1, 4).
The region Δ ECB bounded by these lines and x-axis has been shaded.
Area of the shaded region (Δ ECB)
= \(\frac{\text { Base } \times \text { Height }}{2}\) = \(\frac{4 \times 4}{2}\) = 8 square units.

Question 8.
Solve for x :
\(\frac{3}{x-1}\) + \(\frac{1}{x+1}\) = \(\frac{4}{x}\) (where x ≠ 0, 1, -1)
Solution:
The given equation is
\(\frac{3}{x-1}\) + \(\frac{1}{x+1}\) = \(\frac{4}{x}\)
⇒ \(\frac{3(x+1)+1(x-1)}{(x-1)(x+1)}\) = \(\frac{4}{x}\)
⇒ \(\frac{3 x+3+x-1}{x^2-1}\) = \(\frac{4}{x}\) ⇒ \(\frac{4 x+2}{x^2-1}\) = \(\frac{4}{x}\)
⇒ (4x + 2)x = 4(x² – 1)
⇒ 4x² + 2x = 4x² – 4
⇒ 2x = -4 ⇒ x = -2

Question 9.
Find four different solutions of 10x – y = 2.
Solution:
Given equation 10x – y = 2
Take, x = 0
⇒ 10(0) -y = 2 ⇒ y = -2
∴ Solution (0,-2)
Take, x = 1
⇒ 10(1) – y = 2 ⇒ 10 – 2 = y ⇒ 8 = y
∴ Solution (1, 8)
Take, y = 0
⇒ 10x – (0) = 2 ⇒ x = \(\frac{2}{10}\) = \(\frac{1}{5}\)
∴ Solution (\(\frac{1}{5}\), 0) = (0.2, -2)
Take, y = 1
⇒ 10x – 1 = 2 ⇒ 10x = 2 + 1
x = \(\frac{3}{10}\) = 0.3
Solution (\(\frac{3}{10}\), 1) = (0.3, 1).
∴ Four different solutions of given linear equation in two variables are (0, -2), (1, 8), (0.2 -2) and (0.3, 1).

AP 9th Class Maths Important Questions Chapter 4 Linear Equation in Two Variables

Question 1.
Express the following statement as a linear equation in two variables. “Neeraja and Girija of class IX students together contributed Rs. 300/- towards the C.M. relief fund”.
Solution:
Contribution of Neeraja = ₹x
Contribution of Girija = ₹y
Sum of contribution of Neeraja and Girija is 300/-
∴ x + y = 300.

Question 2.
Re state the following statement with appropriate conditions to make it true statement.
“For every real number x, x2 ≥ x”.
Solution:
If x ≤ 0 or x ≥ 1 then x² ≥ x.
For every integer x, x² ≥ x.

Question 3.
Write the statement given below as a linear equation two in variables. “The sum of two numbers x and y is 75”.
Solution:
The sum of two numbers x and y is 75.
∴ x + y = 75

Question 4.
If x = 2 – α and y = 2 + α in a solution of 5x + 3y – 7 = 0 and x = 2β + 1 and y = β – 1 in a solution of 3x – 2y + 6 = 0 then find the value of α + β.
Solution:
Given equation = 5x + 3y – 7 = 0
x = 2 – α and y = 2 + α
∴ 5(2 – α) + 3(2 + α) – 7 = 0
⇒ 10 – 5α+ 6 + 3α – 7 = 0
⇒ 9 – 2α = 0
⇒ 2α = 9 ⇒ α = \(\frac{9}{2}\) = 4.5
Given equation 3x – 2y + 6 = 0
x = 2β + 1 and y = β – 1
∴ 3(2β + 1)- 2(β – 1) + 6 = 0
⇒ 6β + 3 – 2β + 2 + 6 = 0
⇒ 4β + 11 = 0 ⇒ β = \(\frac{-11}{4}\)
∴ α + β = \(\frac{9}{2}\) – \(\frac{11}{4}\) = \(\frac{18-11}{4}\) = \(\frac{7}{4}\)

Question 5.
Draw the graph of the equation 2x + 3y = 6. Find the coordinates of the points where the graph cuts the coordinate axes.
Solution:
Given equation 2x + 3y = 6


Coordinates of the points are (3, 0) (0, 2)

Question 6.
Draw the graph of the linear equations 2x + 3y = 12. At what points, the graph of the equation cuts the X-axis and Y-axis.
Solution:
2x + 3y = 12


The graph of the equation cuts the x – axis = (6, 0)
y – axis = (0, 4)