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NCERT Solutions for Class 9 Maths Chapter 4 - Linear Equations in Two Variables

Exercise 4.1

Q1. 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)
Ans: Let the cost of a notebook be = ₹ x
Let the cost of a pen be = ₹ y
According to the question,
The cost of a notebook is twice the cost of a pen.
i.e., cost of a notebook = 2×cost of a pen
x = 2 × y
x = 2y
x - 2y = 0
x - 2y = 0 is the linear equation in two variables to represent the statement, ‘The cost of a notebook is twice the cost of a pen.’


Q2. 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= NCERT Solutions for Class 9 Maths Chapter 4 - Linear Equations in Two Variables
Ans: Consider 2x + 3y= NCERT Solutions for Class 9 Maths Chapter 4 - Linear Equations in Two Variables    Equation (1)
⇒ 2x + 3y - NCERT Solutions for Class 9 Maths Chapter 4 - Linear Equations in Two Variables= 0
Comparing this equation with the standard form of the linear equation in two variables, ax + by + c = 0 we have,
a = 2
b = 3
c = -NCERT Solutions for Class 9 Maths Chapter 4 - Linear Equations in Two Variables 


(ii) x – (y/5) – 10 = 0
Ans: The equation x –(y/5)-10 = 0 can be written as,
(1)x+(-1/5)y +(–10) = 0
Now comparing x+(-1/5)y+(–10) = 0 with ax+by+c = 0
We get,
a = 1
b = -(1/5)
c = -10


(iii) –2x + 3y = 6
Ans: –2x+3y = 6
Re-arranging the equation, we get,
–2x+3y–6 = 0
The equation –2x+3y–6 = 0 can be written as,
(–2)x+(3)y+(– 6) = 0
Now, comparing (–2)x+(3)y+(–6) = 0 with ax+by+c = 0
We get, a = –2
b = 3
c =-6


(iv) x = 3y
Ans: x = 3y
Re-arranging the equation, we get,
x-3y = 0
The equation x-3y=0 can be written as,
(1)x+(-3)y+(0)c = 0
Now comparing 1x+(-3)y+(0)c = 0 with ax+by+c = 0
We get a = 1
b = -3
c = 0


(v) 2x = –5y
Ans: 2x = –5y
Re-arranging the equation, we get,
2x+5y = 0
The equation 2x+5y = 0 can be written as,
2x+5y+0 = 0
Now, comparing (2)x+(5)y+0= 0 with ax+by+c = 0
We get a = 2
b = 5
c = 0


(vi) 3x + 2 = 0
Ans: 3x+2 = 0
The equation 3x+2 = 0 can be written as,
3x+0y+2 = 0
Now comparing 3x+0+2= 0 with ax+by+c = 0
We get a = 3
b = 0
c = 2


(vii) y–2 = 0
Ans: y–2 = 0
The equation y–2 = 0 can be written as,
(0)x+(1)y+(–2) = 0
Now comparing (0)x+(1)y+(–2) = 0with ax+by+c = 0
We get a = 0
b = 1
c = –2


(viii) 5 = 2x
Ans: 5 = 2x
Re-arranging the equation, we get,
2x = 5
i.e., 2x–5 = 0
The equation 2x–5 = 0 can be written as,
2x+0y–5 = 0
Now comparing 2x+0y–5 = 0 with ax+by+c = 0
We get a = 2
b = 0
c = -5


Exercise 4.2

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

Ans: Let us substitute different values for x in the linear equation y = 3x + 5,
NCERT Solutions for Class 9 Maths Chapter 4 - Linear Equations in Two Variables
From the table, it is clear that x can have infinite values, and for all the infinite values of x, there are infinite values of y as well.
Hence, (iii) infinitely many solutions is the only option true.


Q2. Write four solutions for each of the following equations:
(i) 2x + y = 7
Ans: To find the four solutions of 2x + y =7 we substitute different values for x and y

Let x = 0
Then,
2x + y = 7
(2)(0) + y = 7
y = 7

(0, 7)
Let x = 1
Then,
2x + y = 7
(2 × 1) + y = 7
2 + y = 7
y = 7 - 2
y = 5
(1, 5)
Let y = 1
Then,
2x + y = 7
(2x) + 1 = 7
2x = 7 - 1
2x = 6
x = 6/2
x = 3
(3, 1)
Let x = 2
Then,
2x + y = 7
(2 × 2) + y = 7
4 + y = 7
y =7 - 4
y = 3
(2, 3)
The solutions are (0, 7), (1, 5), (3, 1), (2, 3)

(ii) πx + y = 9
Ans: To find the four solutions of πx+y = 9 we substitute different values for x and y
Let x = 0
Then,
πx + y = 9
(π)(0) + y = 9
y = 9
(0, 9)
Let x = 1
Then,
πx + y = 9
(π × 1) + y = 9
π + y = 9
y = 9-π
(1, 9-π)
Let y = 0
Then,
πx + y = 9
πx + 0 = 9
πx = 9
x = 9/π
(9/π,0)
Let x = -1
Then,
πx + y = 9
π(-1)+y = 9

-π+y=9
y = 9 + π
(-1, 9+π)
The solutions are (0, 9), (1, 9-π), (9/π, 0), (-1, 9+π)
(iii) x = 4y
Ans:
To find the four solutions of x = 4y we substitute different values for x and y
Let x = 0
Then,
x = 4y
0 = 4y
4y = 0
y = 0/4
y = 0
(0, 0)
Let x = 1
Then,
x = 4y
1 = 4y
4y = 1
y = 1/4
(1, 1/4)
Let y = 4
Then,
x = 4y
x= 4 × 4
x = 16
(16, 4)
Let y =
Then,
x = 4y
x = 4×1
x = 4
(4,1)
The solutions are (0,0), (1,1/4), (16,4), (4,1)


Q3. Check which of the following are solutions of the equation x–2y = 4 and which are not:
(i) (0, 2)
Ans: (x, y) = (0, 2)
Here, x = 0 and y = 2
Substituting the values of x and y in the equation x – 2y = 4, we get,
x – 2y = 4
⟹ 0 – (2 × 2) = 4
But, -4 ≠4
Therefore, (0, 2) is not a solution of the equation x – 2y = 4


(ii) (2, 0)
Ans: (x,y) = (2, 0)
Here, x = 2 and y = 0
Substituting the values of x and y in the equation x - 2y = 4, we get,
x - 2y = 4
⟹ 2-(2 × 0) = 4
⟹ 2 - 0 = 4
But, 2 ≠ 4
(2, 0) is not a solution of the equation x - 2y = 4


(iii) (4, 0)
Ans: (x, y) = (4, 0)
Here, x = 4 and y = 0
Substituting the values of x and y in the equation x - 2y = 4, we get,
x – 2y = 4
⟹ 4 – 2 × 0 = 4
⟹ 4-0 = 4
⟹ 4 = 4
(4, 0) is a solution of the equation x – 2y = 4


(iv) (√2, 4√2)
Ans: (x, y) = (√2, 4√2)
Here, x = √2 and y = 4√2
Substituting the values of x and y in the equation x – 2y = 4, we get,
x –2y = 4
⟹ √2-(2×4√2) = 4
√2-8√2 = 4
But, -7√2 ≠ 4
(√2, 4√2) is not a solution of the equation x – 2y = 4


(v) (1, 1)
Ans: (x, y) = (1, 1)
Here, x = 1 and y = 1
Substituting the values of x and y in the equation x – 2y = 4, we get,
x – 2y = 4
⟹ 1 -(2 × 1) = 4
⟹ 1 - 2 = 4
But, -1 ≠ 4
(1, 1) is not a solution of the equation x–2y = 4


Q4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k.
Ans: The given equation is
2x + 3y = k
According to the question, x = 2 and y = 1.
Now, Substituting the values of x and y in the equation 2x + 3y = k,
We get,
(2 × 2)+(3 × 1) = k
⟹ 4 + 3 = k
⟹ 7 = k
k = 7
The value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k, is 7.

The document NCERT Solutions for Class 9 Maths Chapter 4 - Linear Equations in Two Variables is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on NCERT Solutions for Class 9 Maths Chapter 4 - Linear Equations in Two Variables

1. What are linear equations in two variables?
Ans. Linear equations in two variables are equations that involve two variables, usually represented by x and y, and their coefficients are constants. These equations can be written in the form ax + by = c, where a, b, and c are real numbers.
2. How do you solve a linear equation in two variables?
Ans. To solve a linear equation in two variables, you need to find the values of the variables that satisfy the equation. This can be done by using various methods such as substitution, elimination, or graphing. By substituting the value of one variable in terms of the other in the given equation, you can find the value of one variable and then substitute it back to find the other variable.
3. What is the importance of linear equations in two variables?
Ans. Linear equations in two variables have several real-life applications. They are used to solve problems involving quantities that vary with respect to each other, such as distance and time, cost and quantity, or speed and time. They provide a mathematical representation of relationships between variables and help in making predictions and solving optimization problems.
4. Can linear equations in two variables have more than one solution?
Ans. Yes, linear equations in two variables can have more than one solution. When the equation represents a line, it can intersect with another line at a single point, infinitely many points, or not intersect at all. The number of solutions depends on the relationship between the coefficients of the variables in the equation.
5. How can linear equations in two variables be graphically represented?
Ans. Linear equations in two variables can be graphically represented as straight lines on a coordinate plane. The x and y values of the points on the line satisfy the equation. To graph a linear equation, you can find any two points that satisfy the equation and plot them on the coordinate plane. Then, you can draw a straight line passing through these points to represent the equation.
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