find four solutions of 4x+3y=12
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Answered by
254
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)
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)
Answered by
93
given that
4x+3y = 12
if y=0, 4x+3(0)=12
4x=12
x=12/4
x=3 x=3,y=0
if y=1,4x+3(1)=12
4x+3=12
x=9/4 x=9/4,y=1
if y=2,4x+3(2)=12
4x=6
x=3/2 x=3/2,y=2
if y=3, 4x+3(3)=12
4x=12-9
x = 3/4 x=3/4,y=3
4x+3y = 12
if y=0, 4x+3(0)=12
4x=12
x=12/4
x=3 x=3,y=0
if y=1,4x+3(1)=12
4x+3=12
x=9/4 x=9/4,y=1
if y=2,4x+3(2)=12
4x=6
x=3/2 x=3/2,y=2
if y=3, 4x+3(3)=12
4x=12-9
x = 3/4 x=3/4,y=3
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