parmenides51 wrote: Find all values of the real parameter $a$, for which the system $(|x| + |y| - 2)^2 = 1$ $y = ax + 5$ has exactly three solutions This is just basic boring algebra casework except if you look at graphs : You just have to count intersections between line $y=ax+5$ and two squares C1 : $|x|+|y|=1$ and C3 : $|x|+|y|=3$ If the line encounters the upper part of a square (entering it), it encounters also the lower part (exiting it) and so the number of intersections is even except when the line encounters the square on x-axis which occurs only in four cases : $a=\pm\frac 53$ : one intersection on outer square $\boxed{a=\pm 5}$ : one intersection on inner square and two intersections on outer square.