Question

find four different values of 4x+3y=12

Answer 1

The given equation is 4x + 3y = 12

For x=0, 

4×0 + 3y = 12

⇒ 3y = 12 - 0

⇒ 3y = 12

⇒ y = 12/3 

⇒ y = 4

So (0,4) is a solution.

For x=1, 

4×1 + 3y = 12

⇒ 3y = 12 - 4

⇒ 3y = 8

⇒ y = 8/3 

So (1,8/3) is a solution.

For x=2, 

4×2 + 3y = 12

⇒ 3y = 12 - 8

⇒ 3y = 4

⇒ y = 4/3 

So (2,4/3) is a solution.

For x=3, 

4×3 + 3y = 12

⇒ 3y = 12 - 12

⇒ 3y = 0

⇒ y = 0/3 

⇒ y = 0

So (3,0) is a solution.

Four solutions are: (0,4); (1,8/3); (2,4/3); (3,0)

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