Find the best tutors and institutes for Class 9 Tuition
Search in
Which one of the following options is true, and why? has (i) a unique solution,(ii) only two solutions,(iii) infinitely many solutions
The equation y = 3x + 5 is a linear equation in two variables and has infinite possible solutions.
Since for every value of x, there will be a value of y satisfying the above equation and vice-versa.
Therefore, the correct answer is (iii).
Write four solutions for each of the following equations:(i) (ii)
(iii)
(i) 2x + y = 7
For x = 0,
2(0) + y = 7
⇒ y = 7
Therefore, (0, 7) is a solution of this equation.
For x = 1,
2(1) + y = 7
⇒ y = 5
Therefore, (1, 5) is a solution of this equation.
For x = −1,
2(−1) + y = 7
⇒ y = 9
Therefore, (−1, 9) is a solution of this equation.
For x = 2,
2(2) + y = 7
⇒y = 3
Therefore, (2, 3) is a solution of this equation.
(ii) πx + y = 9
For x = 0,
π(0) + y = 9
⇒ y = 9
Therefore, (0, 9) is a solution of this equation.
For x = 1,
π(1) + y = 9
⇒y = 9 − π
Therefore, (1, 9 − π) is a solution of this equation.
For x = 2,
π(2) + y = 9
⇒ y = 9 − 2π
Therefore, (2, 9 −2π) is a solution of this equation.
For x = −1,
π(−1) + y = 9
⇒ y = 9 + π
⇒ (−1, 9 + π) is a solution of this equation.
(iii) x = 4y
For x = 0,
0 = 4y
⇒ y = 0
Therefore, (0, 0) is a solution of this equation.
For y = 1,
x = 4(1) = 4
Therefore, (4, 1) is a solution of this equation.
For y = −1,
x = 4(−1)
⇒ x = −4
Therefore, (−4, −1) is a solution of this equation.
For x = 2,
2 = 4y
⇒
Therefore, 
is a solution of this equation.
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) (v) (1, 1)
(i) (0, 2)
Putting x = 0 and y = 2 in the L.H.S of the given equation,
x − 2y = 0 − 2×2 = − 4 ≠ 4
L.H.S ≠ R.H.S
Therefore, (0, 2) is not a solution of this equation.
(ii) (2, 0)
Putting x = 2 and y = 0 in the L.H.S of the given equation,
x − 2y = 2 − 2 × 0 = 2 ≠ 4
L.H.S ≠ R.H.S
Therefore, (2, 0) is not a solution of this equation.
(iii) (4, 0)
Putting x = 4 and y = 0 in the L.H.S of the given equation,
x − 2y = 4 − 2(0)
= 4 = R.H.S
Therefore, (4, 0) is a solution of this equation.
(iv) 
Putting 
and 
in the L.H.S of the given equation,
L.H.S ≠ R.H.S
Therefore, is not a solution of this equation.
(v) (1, 1)
Putting x = 1 and y = 1 in the L.H.S of the given equation,
x − 2y = 1 − 2(1) = 1 − 2 = − 1 ≠ 4
L.H.S ≠ R.H.S
Therefore, (1, 1) is not a solution of this equation.
Find the value of k, if x = 2, y = 1 is a solution of the equation
Putting x = 2 and y = 1 in the given equation,
2x + 3y = k
⇒ 2(2) + 3(1) = k
⇒ 4 + 3 = k
⇒ k = 7
Hence, the value of k is 7.
How helpful was it?
How can we Improve it?
Please tell us how it changed your life *
Please enter your feedback
UrbanPro.com helps you to connect with the best Class 9 Tuition in India. Post Your Requirement today and get connected.
Find best tutors for Class 9 Tuition Classes by posting a requirement.
Get started now, by booking a Free Demo Class
