Question

verify that which of the are solution of. equation x+2y=4: (0,2),(0,-2)),(2,1),(-2,0),(3,1\2)and(5, following 0).

Answer 1

x+2y=4----(1)
i)put (0,2) in (1)
0+2*2=4
4=4(true)
Therefore (0,2) is solution of (1)
ii)put (0, -2) in (1)
0+2*(-2)=4
-4=4 (false)
(0, -2) is not a solution of (1)
Do the same for all points

Answer 2

$ANSWER:$

$\\ x + 2y = 4$

Substitute x=0 and y=2

LHS =x+2y

=0+2(2)

=4

LHS=RHS

Therefore,

(0,2) is the solution of the equation.

Substituting x=0 and y=-2

LHS =x+2y

=0+2(-2)

=0-4

=-4

LHS≠RHS

Therefore,

(0,-2) is not the solution of the equation.

Substituting x=2 and y=1

LHS =x+2y

=2+2(1)

=2+2

=4

LHS=RHS

Therefore,

(2,1) is the solution of the equation.

Substituting x=-2 and y=0

LHS =x+2y

=-2+2(0)

=-2+0

=-2

LHS≠RHS

Therefore,

(-2,0)is not the solution of the equation.

Substituting x=3 and y=1/2

LHS =x+2y

=3+2(1/2)

=3+1

=4

LHS=RHS

Therefore,

(3,1/2)is the solution of the equation.

Substituting x=5 and y=0

LHS =x+2y

=5+2(0)

=5+0

=5

LHS≠RHS.

Therefore,

(5,0) is not the solution of the equation.

Therefore,

the solution of the equation are:

1.(0,2),

2.(2,1) &

3.(3,1/2)