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 1:
Given:
y = 3 x + 5 is a linear Equation.
For x = 0, y = 0 + 5 = 5
Therefore, (0, 5) is another solution.
For x = 1, y = 3 × 1 + 5 = 8
Therefore (1, 8) is another solution.
For y = 0 , 3x + 5 = 0 is the one solution
Clearly, for different values of x, we get another value of y. Thus, the chosen value of x together with this value of y constitutes another solution of the given equation. So, there is no end to different solutions of a linear equation in two variables. Therefore, a linear equation in two variables has infinitely many solutions
Hence (iii) is the correct answer
Question 2:
Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y
Solution 2:
(i)2x + y = 7
Given,
Linear Equation, 2x + y = 7
We can write the above equation as below by simplifying
y = 7- 2x ----------Equation (1)
Let us now take different values of x and substituting in the Equation (1), we get
For x = 0,
2(0) + y = 7
⇒y = 7
Hence, we get (x, y) = (0, 7)
For x = 1,
2(1) + y = 7
⇒ y = 5
Hence, we get (x, y) = (1, 5)
For x = 2,
2(2) + y = 7
⇒ y = 3
Hence, we get (x, y) = (2, 3)
For x = 3,
2(3) + y = 7
⇒ y = 1
Hence we get (x, y) = (3, 1)
Therefore the four solutions of the given equation are (0,7) , (1,5), (2, 3), (3,1)
(ii)Ï€x + y = 9
Given,
Linear Equation, πx + y = 9
We can write the above equation as below by Transposing
y = 9 - πx ----------Equation (1)
Let us now take different values of x and substituting in the Equation (1), we get
For x = 0,
y = 9 - π(0)
⇒y = 9
Hence we get (x, y) = (0, 9)
For x = 1,
y = 9- π (1)
= 9- π
Hence we get (x, y) = (1, 9- π )
For x = 2,
y = 9- π (2)
Hence we get (x, y) = (2, 9 −2Ï€)
For x = 3,
y = 9- π (3)
Hence we get (x, y) = (3, 9 - 3Ï€)
Therefore the four solutions of the given equation are (0, 9), (1, 9 - Ï€), (2, 9 −2Ï€ ), (3, 9 - 3Ï€)
(iii)x = 4y
Given,
Linear Equation, x = 4y
We can write the above equation as below by Transposing
y = x/4 ----------Equation (1)
Let us now take different values of x and substituting in the Equation (1), we get
For x = 0,
y = 0/4 = 0
Hence we get (x, y) = (0, 0)
For x = 1,
y = 1/4
Hence we get (x, y) = (1, 1/4)
For x = 2,
y = 2/4 = 1/2
Hence we get (x, y) = (2, 1/2)
For x = 3,
y = 3/4
Hence we get (x, y) = (3, 3/4)
Therefore the four solutions of the given equation are (0, 0), (1, 1/4), (2, 1/2), (3, 3/4)
Question 3:
Check which of the following solutions of the equation are x − 2y = 4 and which are not:
(i)(0, 2)
(ii)(2, 0)
(iii)(4, 0)
(iv) (√2,4√2)
(v)(1, 1)
Solution 3:
Given : x − 2y = 4 is a Linear Equation----------Equation(1)
(i)(0, 2)
B Substituting x = 0 and y = 2 in the L.H.S of the given Equation (1)
x − 2y
= (0) – (2)2
= − 4 ≠ 4 ≠ RHS
L.H.S ≠ R.H.S
Therefore, (0, 2) is not a solution of this equation.
(ii)(2, 0)
By Substituting, x = 2 and y = 0 in the L.H.S of the given Equation (1),
x − 2y = 2 – 2(0) = 2 ≠ 4 ≠ RHS
L.H.S ≠ R.H.S
Therefore, (2, 0) is not a solution of this equation.
(iii)(4, 0)
By Substituting, x = 4 and y = 0 in the L.H.S of the given Equation (1)
x − 2y
= 4 − 2(0)
= 4
L.H.S = R.H.S
Therefore, (4, 0) is a solution of this equation.
(iv)(√2,4√2)
By Substituting,
x=√2 and y=4√2 in the L.H.S of the given Equation (1)
x-2y
=√2-8√2
= -7√2 ≠ 4 ≠ R.H.S
L.H.S ≠ R.H.S
Therefore, is not a solution of this equation.
(v)(1, 1)
By Substituting, x = 1 and y = 1 in the L.H.S of the given Equation (1)
x − 2y
= 1 − 2(1)
= 1 − 2
= − 1 ≠ 4 ≠ RHS
L.H.S ≠ R.H.S
Therefore, (1, 1) is not a solution of this equation.
Question 4:
Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution 4:
Given : 2x + 3y = k is the Linear Equation ---------------- Equation (1)
x = 2
y = 1
k =?
By substituting the values of x and in the Equation (1),
2x + 3y = k
⇒ 2(2) + 3(1) = k
⇒ 4 + 3 = k
Therefore, the value of k is 7.
⇒ Hence, k = 7
Therefore, the value of k is 7.
