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Answer 1

Answer:

1.(iii) is true because for every value of x, we get a corresponding value of y and vice-versa in the given equation.

4.We have 2x + 3y = k

putting x = 2 and y = 1 in 2x+3y = k,we get

2(2) + 3(1) ⇒ k = 4 + 3 – k ⇒ 7 = k

Thus, the required value of k is 7

Step-by-step explanation:

2.

i.)When x = 0, 2(0) + y = 7 ⇒ y = 7

∴ Solution is (0, 7)

When x =1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5

∴ Solution is (1, 5)

When x = 2, 2(2) + y =7y = 7 – 4 ⇒ y = 3

∴ Solution is (2, 3)

When x = 3, 2(3) + y = 7y = 7 – 6 ⇒ y = 1

∴ Solution is (3, 1).

ii.)When x = 0, π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9

∴ Solution is (0, 9)

When x = 1, π(1) + y = 9 ⇒ y = 9 – π

∴ Solution is (1, (9 – π))

When x = 2, π(2) + y = 9 ⇒ y = 9 – 2π

∴ Solution is (2, (9 – 2π))

When x = -1,π(-1) + y = 9 ⇒ y = 9 + π

∴ Solution is (-1, (9 + π))

iii.)When x = 0, 4y = 1 ⇒ y = 0

∴ Solution is (0, 0)

When x = 1, 4y = 1 ⇒ y = 14

∴ Solution is (1,14 )

When x = 4, 4y = 4 ⇒ y = 1

∴ Solution is (4, 1)

When x = 4, 4y = 4 ⇒ y = -1

∴ Solution is (-4, -1)

3.

i.)(0,2) means x = 0 and y = 2

Puffing x = 0 and y = 2 in x – 2y = 4, we get

L.H.S. = 0 – 2(2) = -4.

But R.H.S. = 4

∴ L.H.S. ≠ R.H.S.

∴ x =0, y =2 is not a solution

sorry it will take time for other sums

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