Is y = 6?

1. y^2 = 36
You have either y=6 or y=-6
So it is insufficient.

2. y^2- 7y + 6 = 0
We can write it like that
(y-6)(y-1)=0
So here the solutions are either y=6 or y=1
So it is insufficient.

1 and 2.
y=6 or y=-6 and y=6 or y=1
So it is 6 !
C

This is exactly what I had: y=6 or y=-6 and y=6 or y=1.

However, I chose E since y could be +/-6 or 1. This is an elementary question, but are we supposed to pick the common value 6 in such scenarios?

In this case, you cannot chose y=1 or y=-6
You can see
that for y=1, y^2 is different from 36
for y=-6, y^2- 7y + 6 is different from 0
But with y=6 we have y^2 = 36 and y^2- 7y + 6 = 0

What you look is a solution which suits both statements, your solution cannot contradict with one equation but be wrong with another equation. You should practice DS questions to be used to what they ask you.

Thanks pepeprepa. You cleared my doubt. Later!