TITLE: Linear equation with square root
QUESTION [7 upvotes]: Linear equation in two variables must have the form $ax+by=c$. But can we see $\sqrt{x+y}=2$ as a linear equation because if we square both sides we get $x+y=4$ and both equations have the same solution set?

REPLY [6 votes]: Given any function $f$, we can always take the linear equation
$$ax+by=c,$$
and apply $f$ to both sides:
$$f(ax+by)=f(c).$$
The resulting equation will generally not be linear unless $f$ is a linear function. Recall that a linear function is defined as satisfying the property
$$f(ax+by)=a f(x) + b f(y),$$
so in that case we get
$$a f(x) + b f(y)=f(c),$$
which is a linear function in the variables $f(x)$ and $f(y)$.
In your case $f(x) = \sqrt{x}$, which is not a linear function, since it does not satisfy the property given above. Thus the equation obtained is not linear.
Addendum: Here I have used the common definition where a linear equation is an algebraic equation of degree 1 (and a quadratic equation is an algebraic equation of degree 2, and so on). Thus, even though the set of solutions to the equation $f(ax+by)=f(c)$ is a straight line, the equation is nonetheless nonlinear unless $f$ is a linear function. (See comments.)