NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2.
- Linear Equations in Two Variables Class 9 Ex 4.1
- Linear Equations in Two Variables Class 9 Ex 4.2
- Linear Equations in Two Variables Class 9 Ex 4.3
- Linear Equations in Two Variables Class 9 Ex 4.4
Board | CBSE |
Textbook | NCERT |
Class | Class 9 |
Subject | Maths |
Chapter | Chapter 4 |
Chapter Name | Linear Equations in Two Variables |
Exercise | Ex 4.2 |
Number of Questions Solved | 4 |
Category | NCERT Solutions |
NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2
Ex 4.2 Class 9 Maths Question 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 ,
Solution:
(iii) We know that, a linear equation in two variables has infinitely many solutions. Here, given equation is a linear equation in two variables x and y, so it has infinitely many solutions.
Ex 4.2 Class 9 Maths Question 2.
Write four solutions for each of the following equations :
(i) 2x + y = 7,
(ii) πx + y = 9
(iii) x = 4y Sol.
Solution:
(i) Given equation, 2x + y=7 …(i)
Now, putting x = 0, in eq. (i), we get 2(0) + y=7
⇒ 0 + y = 7 => y=7
So, x = 0 and y = 7 is a solution of given equation.
On putting x = 1 in eq (i), we get
2(1) + y =7 ⇒ y=7-2 ⇒ y=5
So, x = 1 and y = 5 is a solution of given equation
On putting x = 2, in eq. (i) we get, 2(2) + y = 7 ⇒ y= 7 – 4 ⇒ y= 3
So, x = 2 and y = 3 is a solution of given equation.
On putting x = 3 in eq(i), we get 2(3) + y =7
⇒ y= 7-6 ⇒ y = 1
So, x = 3 and y = 1 is a solution of given equation.
Hence, four out of the infinitely many solutions of the given equation are (0, 7), (1, 5) (2, 3) and (3,1).
(ii) Given equation πx + y =9 – – – (i)
On putting x = 0 in eq. (i) we get, π(0) + y=9
⇒ y = 9 – 0 =>y = 9
So, x = 0 and y = 9 is a solution of given equation.
On putting x = 1 in eq. (i), we get π(1) + y= 9
y-9-π
So, x = 1, and y = (9 -π) is a solution of given equation.
On putting x = 2 in eq (i), we get π(2) + y=9
y = 9 – 2π
So x = 2 and y = (9 – 2π) is a solution of given equation.
On putting x = -1 in eq (i), we get π(-9) + y=9
y = 9 + π
So x = -1 and y = (9 + π) is a solution of given equation.
Hence, four out of the infinitely many solutions of the given equation are
(0, 9), {1,(9 – π)}, {2,(9 – 2π)}, {-1, (9 + π)}
(iii) Given equation x = 4y …(i)
On putting x = 0, in eq. (i), we get 4y=0 ⇒ y=0
So, x – 0 and y = 0 is a solution of given equation.
On putting x = 1 in eq. (i), we get 4y=1 ⇒ y= \(\frac { 1 }{ 4 } \)
So, x -1 and y = \(\frac { 1 }{ 4 } \) is a solution of given equation.
On putting x – 4, in eq (i), we get
4y= 4
⇒ y= – \(\frac { 4 }{ 4 } \) =>y = 1
So, x = 4 and y = 1 is a solution of given equation, on putting x = – 4 in eq. (i), we get
4y= -4 ⇒ y= \(\frac { -4 }{ 4 } \) = -1
So, x = – 4 and y = -1 is a solution of given equation.
Hence, four out of the infinitely many solutions of given equation are
(0, 0), (1, \(\frac { 1 }{ 4 } \)), (4,1) and (-4, -1).
Ex 4.2 Class 9 Maths Question 3.
Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0, 2)
(ii) (2, 0)
(iii) (4, 0)
(iv) (\( \sqrt { 2 }\) ,4\( \sqrt { 2 }\) )
(v) (1, 1) .
Solution:
(i) Given equation is x – 2y = 4.
(0, 2) means x = 0 and y = 2
On putting x = 0 and y = 2 in L.H.S., get
L.H.S. = x – 2y = 0 – 2(2) = – 4
But R.H.S. = 4
L.H.S ≠ R.H.S
Hence (0, 2) is not a solution of x – 2y – 4.
(ii) Given equation is x – 2y = 4 (2, 0) means x – 2 and y = 0
On putting x = 2 and y = 0 in L.H.S., we get
L.H.S. = x – 2y = 2 – 2(0) = 2-0 = 2
But – R.H.S. = 4
∴ L.H.S = R.H.S.
Hence (2, 0) is not a solution of x – 2y – 4.
(iii) Given equation is x – 2y = 4 (4, 0) means x = 4 and y = 0.
On putting x = 4 and y = 0 in L.H.S., we get
L.H.S. = x – 2y = 4 – 2(0) =4-0 = 4
But R.H.S. = 4
∴ L.H.S. = R.H.S.
Hence, (4, 0) is a solution of x – 2y = 4.
(iv) Given equation is x – 2y = 4
(\( \sqrt { 2 }\), 4\( \sqrt { 2 }\)) means x = \( \sqrt { 2 }\) and y = 4\( \sqrt { 2 }\)
On putting x = \( \sqrt { 2 }\) and y = 4\( \sqrt { 2 }\) in L.H.S., we get
L.H.S. = x – 2y = \( \sqrt { 2 }\) – 2(4\( \sqrt { 2 }\)) = \( \sqrt { 2 }\)-8\( \sqrt { 2 }\)
= \( \sqrt { 2 }\)(1 – 8) = – 7\( \sqrt { 2 }\)
But R.H.S. = 4
∴ L.H.S ≠ R.H.S.
Hence (\( \sqrt { 2 }\), 4\( \sqrt { 2 }\)) is not a solution of x – 2y = 4
(v) Given equation is x – 2y = 4
(1, 1) means x = 1 and y = 1
On putting x = 1 and y = 1 in L.H.S. we get
L.H.S. = x – 2y = 1 – 2(1) = 1 – 2 = -1 But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
Hence, (1, 1) is not a solution of x – 2y = 4.
Ex 4.2 Class 9 Maths Question 4.
Find the value by k if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
If x = 2, y = 1 is a solution by the equation 2x + 3y = k, then these value will satisfy the equation.
∴ 2 x 2+3 x 1 = k ⇒ k=4+3=7
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