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**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 solutionsSolution:** Option (iii) is true because a linear equation has an infinitely many solutions.

Solution:

When x = 0, 2(0) + y = 7

â‡’ 0 + y = 7

â‡’ y= 7

âˆ´ Solution is (0, 7).

When x = 1, 2(1) + y = 7

â‡’ y = 7 â€“ 2

â‡’ y= 5

âˆ´ Solution is (1, 5).

When x = 2, 2(2) + y = 7

â‡’ y = 7 â€“ 4

â‡’ y = 3

âˆ´ Solution is (2, 3).

When x = 3 2(3) + y = 7 â‡’ y = 7 â€“ 6

â‡’ y = 1

âˆ´ Solution is (3, 1).

(ii) Ï€x + y = 9

When x = 0 Ï€(0) + y = 9

â‡’ y = 9 â€“ 0

â‡’ y= 9

âˆ´ Solution is (0, 9).

When x = 1, Ï€(1) + y = 9

â‡’ y = 9 â€“ Ï€

âˆ´ Solution is {1, (9 â€“ Ï€)}.

When x = 2 Ï€(2) + y = 9

â‡’ y = 9 â€“ 2Ï€

âˆ´ Solution is {2, (9 â€“ 2Ï€)}.

When x = â€“1, p(â€“1) + y = 9

â‡’ â€“Ï€ + y = 9

â‡’ y = 9 + Ï€

âˆ´ Solution is {â€“1, (9 + Ï€)}.

(iii) x = 4y When x = 0, 4y = 0

â‡’ y = 0

âˆ´ Solution is (0, 0).

When x = 1, 4y = 1

â‡’ y = (1/4)

âˆ´ Solution is (1, (1/4))

When x = 4, 4y = 4

â‡’ y= (4/4) = 1

âˆ´ Solution is (4, 1).

When x = â€“4, 4y = â€“4

â‡’ y = (-4/4) = â€“1

âˆ´ Solution is (â€“4, â€“1).

**Question 3: Check which of the following are solutions of the equation x â€“ 2y = 4 and which are not: (i) (0, 2) (ii) (2, 0) (iii) (4, 0) (iv) (2 , 4âˆš2****) ( v ) (1, 1)Solution:**

Putting x = 0 and y = 2 in x â€“ 2y = 4,

we have L.H.S. = 0 â€“ 2(2) = â€“4

But R.H.S. = 4

âˆ´ L.H.S. â‰ R.H.S.

âˆ´ x = 0, y = 0 is not a solution.

**(ii)** (2, 0) means x = 2 and y = 0

âˆ´ Putting x = 2 and y = 0 in x â€“ 2y = 4,

we get L.H.S. = 2 â€“ 2(0) = 2 â€“ 0 = 2

But R.H.S. = 4

âˆ´ L.H.S. â‰ R.H.S.

âˆ´ (2, 0) is not a solution.

**(iii)** (4, 0) means x = 4 and y = 0

Putting x = 4 and y = 0 in x â€“ 2y = 4,

we get L.H.S. = 4 â€“ 2(0) = 4 â€“ 0 = 4

But R.H.S. = 4

âˆ´ L.H.S. = R.H.S.

âˆ´ (4, 0) is a solution.

**(iv)** (2 , 4âˆš2) means x = 2 and y = 4âˆš2

Putting x = 2 and y = 4âˆš2 in x â€“ 2y = 4,

we get L.H.S. =âˆš2 â€“ 2 ( 4âˆš2) = âˆš2 â€“ 8âˆš2

= âˆš2(1 â€“ 8) = â€“7âˆš2

But R.H.S. = 4 âˆ´ L.H.S. â‰ R.H.S.

âˆ´ (âˆš2, 4âˆš2) is not a solution.

**(v)** (1, 1) means x = 1 and y = 1

Putting x = 1 and y = 1 in x â€“ 2y = 4,

we get L.H.S. = 1 â€“ 2(1) = 1 â€“ 2 = â€“1

But R.H.S. = 4

â‡’ L.H.S. â‰ R.H.S

âˆ´ (1, 1) is not a solution.**Question 4: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.****Solution:** We have 2x + 3y = k

Putting x = 2 and y = 1 in 2x + 3y = k,

we get 2(2) + 3(1) = k

â‡’ 4 + 3 = k

â‡’ 7 = k

Thus, the required value of k = 7.

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