Determine two distinct real numbers a and b that satisfy this system of equations:
a
2-b = 73
b
2-a =73
Both equations have the same constant. In fact I have seen several versions of this problem differing only in the common constant term posed in various places. The common constant is usually of the form c^2+c+1, and for this problem c=8.
So likely the intended trick is to subtract the two equations, eliminating the constant and leaving behind
a^2-b^2+a-b = 0
Then factor into
(a-b)*(a+b+1) = 0
This forms two solution sets. a=b or a=-b-1. The first is to be discarded by the problem statement. So I will continue with the second and use the generalized constant.
Then b^2-(-b-1) = c^2+c+1
There is the easy solution of b=c, which makes a=-c-1. But going through the quadratic formula or factoring gets a second solution of b=-c-1, which makes a=-c-1.
These two solutions are the same pair just in the opposite order which makes sense from the symmetry in the original equation.
For this specific problem c=8. Then the two distinct real numbers a and b that satisfy the system of equations are a=8 and b=-8-1=-9, or the converse a=-9 and b=8.
Solution
