Answer: \((x,y)=\left(\frac52,\frac72\right)\) or \((3,2)\).

The key point is to reduce the equation to a standard trigonometric equation, then use the given interval to list all valid solutions.

From

\(y+3x=11,\)

we have

\(y=11-3x.\)

Substitute this into \(xy+x^2=15\):

\(x(11-3x)+x^2=15.\)

Hence

\(11x-2x^2=15,\)

so

\(2x^2-11x+15=0.\)

Factorise:

\((2x-5)(x-3)=0.\)

Therefore

\(x=\frac52\quad\text{or}\quad x=3.\)

If \(x=\frac52\), then

\(y=11-3\left(\frac52\right)=\frac72.\)

If \(x=3\), then

\(y=11-9=2.\)

So the solutions are