find any four linear solutions to 4 x + 3 Y is equals to 12
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( 0 , 4 ) , ( 3 , 0 ) , ( 1 , 8/3 ) & ( 9/4 , 1 )
First Writing the Equation
we have
4x+3y=12
put ,
x = 0 ,
then 4*0 + 3y = 12
3y =12
y= 12/3 = 4
so (x,y) = (0,4)
put ,
y=0
then 4x+3*0=12
4x=12
x= 12/4 =3
so (x,y) = (3,0)
put
x=1
4*1+3y=12
3y=8
y=8/3
(x,y)=(1,8/3)
last one try yourself ☺️
put
y=1
4x+3*1=12
4x=9
x=9/4
(x,y)=(9/4,1)
First Writing the Equation
we have
4x+3y=12
put ,
x = 0 ,
then 4*0 + 3y = 12
3y =12
y= 12/3 = 4
so (x,y) = (0,4)
put ,
y=0
then 4x+3*0=12
4x=12
x= 12/4 =3
so (x,y) = (3,0)
put
x=1
4*1+3y=12
3y=8
y=8/3
(x,y)=(1,8/3)
last one try yourself ☺️
put
y=1
4x+3*1=12
4x=9
x=9/4
(x,y)=(9/4,1)
samirashan:
plz explain answer my plz
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