```Question 316866
if the solution has to be in the first quadrant then the value of y and the value of x have to be in the first quadrant.

An equation like y > sqrt(x) should satisfy this.

A graph of this equation would look like this:

{{{graph(600,600,-5,5,-5,5,sqrt(x))}}}

When x is negative, x is undefined, so the domain of this function has to be all values of x >= 0.

Since all values of x are >= 0, then y = sqrt(x) must be positive as well.

If y has to be greater than sqrt(x), then the equation y > sqrt(x) will be all values of y that are above the graph of the equation y = sqrt(x).

This has isolated both x and y to be greater than or equal to 0 which conforms to the requirement that the solution is in quadrant 1.

I think that's what they mean.

You have to select a domain that can only be positive and then create an equation where the value of y will also always be positive.

An equation where this would not work would be y > absolute value of x.

The graph of that equation looks like this:

{{{graph(600,600,-5,5,-5,5,abs(x))}}}

You can see from this graph, that while the value of y is always positive, the value of x can also be negative which places the solution in both quadrants 1 and 2.

In quadrant 1 both x and y are positive
In quadrant 2, x is negative and y is positive
In quadrant 3, both x and y are negative
In quadrant 4, x is positive and y is negative
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