Answer:

(i) 2x+y=7

y=7–2x

When x=0, then

y=7–2×0

y=7–0

y=7

When x=1 then

y=7–2×1

y=7–2

y=5

When x=2 then

y=7–2×2

y=7−4

y=3

When x=3 then

y=7−2×3

y=7−6

y=1

Hence, four solutions for equation 2x+y=7 are (0,7),(1,5),(2,3),(3,1).

(ii) πx+y=9

y=9−πx

When x=−1 then

y=9−π×−1

y=9+π

When x=0 then

y=9−π×0

y=9

When x=1 then

y=9−π×1

y=9−π

When x=2 then

y=9−π×2

y=9−2π

Hence, four solution for equation πx+y=9 are (−1,9+π),(0,9),(1,9−π),(2,9−2π).

(iii) x=4y

When y=0, then

x=4×0

x=0

When y=1 then

x=4×1

x=4

When y=−1 then

x=4×−1

x=−4

When y=2 then

x=4×2

x=8

Hence, four solutions for equation x=4y are (0,0),(4,1),(−4,−1),(8,2).

Verified answer

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Given polynomial p(x) :

• 2x + y = 10______(1)
• x + y = 7________(2)

Given Options :

1. (3,4)
2. (4,3)
3. (2,4)
4. (4,2)

• On Solving :

1. x = 3, y = 4

In equation (1)

[tex] \large \red \implies \sf\: 2(3) + 4 = 10 \\ \large \red \implies \sf \: 6 + 4 = 10 \\ \large \green \implies \bold{10 = 10}[/tex]

L.H.S = R.H.S

In equation (2)

[tex] \large \red \implies \sf\: 3 + 4 = 7 \\ \large \green \implies \bold{7 = 7}[/tex]

L.H.S = R.H.S

∴ right option is 1.

For x = 4 , y = 3

In equation (1)

⇒ 2(4) + 3 = 10

⇒ 8 + 3 = 10

∴ 11 ≠ 10

• Now satisfied

In equation (2)

4 + 3 = 7

7 = 7

L.H.S = R.H.S

• Satisfied

Option 2. satisfy only 2nd equation