NCERT Solutions: Linear Equations in Two Variables (Exercise 4.2)

# Linear Equations in Two Variables (Exercise 4.2) NCERT Solutions - Mathematics (Maths) Class 9

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, 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 Linear Equations in Two Variables (Exercise 4.2) NCERT Solutions | Mathematics (Maths) Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

## FAQs on Linear Equations in Two Variables (Exercise 4.2) NCERT Solutions - Mathematics (Maths) Class 9

 1. What are linear equations in two variables? Ans. Linear equations in two variables are equations that involve two variables and can be expressed in the form of ax + by + c = 0, where a, b, and c are constants.
 2. How do you graph a linear equation in two variables? Ans. To graph a linear equation in two variables, we first rewrite the equation in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. Then, we plot the y-intercept on the graph and use the slope to find additional points and draw a straight line through these points.
 3. What is the solution to a system of linear equations in two variables? Ans. The solution to a system of linear equations in two variables is the common point that satisfies all the equations in the system. It is the point where the graphs of the equations intersect.
 4. How do you solve a system of linear equations in two variables algebraically? Ans. To solve a system of linear equations in two variables algebraically, we can use the substitution method or the elimination method. In the substitution method, we solve one equation for one variable and substitute it into the other equation. In the elimination method, we eliminate one variable by adding or subtracting the equations.
 5. What are the applications of linear equations in two variables? Ans. Linear equations in two variables have various applications in real-life situations. They can be used to solve problems related to distance, speed, and time, as well as problems involving cost and revenue, such as profit maximization and break-even analysis. They are also used in fields like engineering, physics, and economics to model and solve real-world problems.

## Mathematics (Maths) Class 9

62 videos|426 docs|102 tests

## Mathematics (Maths) Class 9

62 videos|426 docs|102 tests
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Track your progress, build streaks, highlight & save important lessons and more! (Scan QR code)
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;