Solve the system of equations by the addition method.?

Question

{x / 3+ y / 6 = 1

{x / 2 - y / 4 = 0

a. (3, 3 / 2)
b. (3 / 2, 3)
c. infinite number of solutions
d. no solution

Answer 1

Lets multiply first equation by -1/2 and 2nd by 1/3, giving us
-x/6-y/12=-1/2
x/6-y/12=0

Add the equations:
-2y/12=-1/2
-y/6=-1/2

So, y=(-1/2)(-6)=3.

And x/3+3/6=1
x/3=1-3/6=1/2
So, x=(1/2)*3=3/2

So, its b).

Answer 2

x/3 + y/6 = 1
Multiply through by 6 :
2x + y = 6 (Equation 1)

x/2 - y/4 = 0
Multiply through by 4 :
2x - y = 0 (Equation 2)

Now add Equation 1 to Equation 2 :
4x = 6, so, x = 3/2.

Substituting this into Equation 2 gives :
2(3/2) - y = 0, so, y = 3.

Therefore, answer b. is the correct one.

Answer 3

x / 3+ y / 6 = 1

x / 2 - y / 4 = 0

Multiply first equation by 6 and second by 4

2x + y = 6
2x - y = 0

Now add them, the y cancels

4x = 6
x= 1.5

Substitute x = 1.5 into 2x - y = 0

2(1.5) - y = 0

y = 3

Check values in first equation

x / 3+ y / 6 = 1

(1.5)/3 + 3/6 = 1

0.5 + 0.5 = 1

1 = 1

the answers are proved.
.

Answer 4

2x + y = 6
2x - y = 0---------ADD

4x = 6
x = 3/2

3 + y = 6
y = 3

x = 3/2 , y = 3