Question

find four different solutions of the equation x+3y=6​

Answer 1

Answer:

Four solutions of the equation are :-

• (0, 2)
• (6, 0)
• (-3, 3)
• (1, 3)

Explanation:

Given equation,

$\rm \implies \: x + 3y = 6$

We need to find four solutions.

Since, we know that there are infinite solutions of a specific equation.

So,

Substituting, x = 0

$\rm \implies \: 0 + 3y = 6$

$\rm \implies \: 3y = 6$

$\rm \implies \: y = \dfrac{6}{3}$

$\rm \implies \: y = 2$

We have,

• x = 0 and y = 2

∴ Solution 1: (0, 2)

Substituting, y = 0

$\rm \implies \: x + 3(0) = 6$

$\rm \implies \: x + 0 = 6$

$\rm \implies \: x = 6$

∴ Solution 2: (6, 0)

Substituting, x = -3

$\rm \implies \: ( - 3) + 3y = 6$

$\rm \implies \: - 3 + 3y = 6$

$\rm \implies \: 3y = 6 + 3$

$\rm \implies \: 3y = 9$

$\rm \implies \: y = \dfrac{9}{3}$

$\rm \implies \: y = 3$

∴ Solution 3: (-3, 3)

Substituting, y = 1

$\rm \implies \: x + 3(1) = 6$

$\rm \implies \: x + 3 = 6$

$\rm \implies \: x= 6 - 3$

$\rm \implies \: x= 3$

∴ Solution 4: (1, 3)

Answer 2

Question :-

find four different solutions of the equation x+3y=6.

Solution :-

Given equation :-

➦ x + 3y = 6

Here, we put any number randomly on the x or y, then we get infinite solutions. But here we want only four solution they are as follows :

Substituting x = 0

⇒ 0 + 3y = 6

⇒ 3y = 6

⇒ y = 6/3

⇒ y = 2

First Solution :- (x, y) = (0, 2)

Substituting x = 1

⇒ 1 + 3y = 6

⇒ 1 + y = 6/3

⇒ 1 + y = 2

⇒ y = 2 - 1

⇒ y = 1

Second Solution :- (x, y) = (1, 1)

Substituting x = 2

⇒ 2 + 3y = 6

⇒ 2 + y = 6/3

⇒ 2 + y = 2

⇒ y = 2 - 2

⇒ y = 0

Third Solution :- (x, y) = (2,0)

Substituting x = 3

⇒ 3 + 3y = 6

⇒ 3 + y = 6/2

⇒ y = 2 - 3

⇒ y = -1

Fourth Solution :- (x, y) = (3, -1)

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9. Find four different solutions for each of the following

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(¡¡¡) 4x + y = 9

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(v) x = 2y

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