x-2y=0 and 3x-y^2=0 solve the system by the method of substitution??

x = 2y
6 y - y ² = 0
y (6 - y) = 0
y = 0 , y = 6
x = 0 , x = 12

Solution is (0 , 0) , (12 , 6)

x-2y=0
3x-y^2=0
I would prefer to use the first equation to do the substituting because it's simpler, but you could use the other one if you prefer. First, x-2y=0, add the 2y to both sides which is
x=2y. Take x=2y and substitue it in the equation 3x-y^2=0.
3(2y)-y^2=0
6y-y^2=0
Factor the equation
y(6-y)=0
y=0 or 6-y=0
y=0 or y=6
Now, substitue that in the equation 3x-y^2=0
3x-0^2=0
3x=0
x=0
3x-6^2=0
3x-36=0
3x=36
x=12
So your answer would be x={0, 12} and y={0, 6}
Hope this helps

Add 2y to both sides of the first equation to get

x - 2y + 2y = 2y

x = 2y

Replace x by 2y in the second equation

3(2y) - y^2 = 0

6y - 2^y = 0

Factor out u to get

y(6 - y) = 0

so y = 0 and y = 6 are the two solutions. Putting these two values back into x - 2y = 0 we get

x = 0 and x = 12

first make first equation in terms of x
so: x-2y=0
+2y
x=2y
substitute in x into 2nd equation so:
3(2y)-y^2=0
6y-y^2=0
then factorise
y(6-y)=0
therefore y=0 or 6
put both values of y into first equation to find x
x-2*0=0 and x-2*6=0
therefore x= 0 if y=0 and x=12 if y=6

x-2y=0
3x-y^2=0

for the method of substitution you have to sub one equation for another so rearange the first equation x-2y=0 for x=2y
and sub x=2y into the second equation so the second equation will be

3(2y) - Y^2=0
6y-y^2=0 take y out
y(6 - y) = 0
therefore y = 0 and y = 6

x-2y=0 therefore,
x=2y by shifting 2y to the right forming equation1

considering 3x-y^2=0 as equation 2,substitute equation 1 into it by substituting the value of x which is 2y giving ,

3(2y)-y^2=0
6y-y^2=0
y^2-6y=0
y(y-6)=0
therefore y= 6 or 0,substituting back into the equation of x=2y, when y=6,x=12
when y=0,x=0

x-2y=0 and 3x-y^2=0 Set x = 2y and substitute into second to get 2y - y^2 = 0 => y(2-y) = 0 => y = 0 or y = 2 then x= 0 or 4

x - 2y = 0
x = 2y
3x = 6y


3x - y^2 = 0
6y - y^2 = 0
y(6 - y) = 0

y = 0

6 - y = 0
y = 6


x - 2y = 0
x - 2(0) = 0
x - 0 = 0
x = 0

x - 2y = 0
x - 2(6) = 0
x - 12 = 0
x = 12