How do you find the exact solutions to the system #x^2/36-y^2/4=1# and #x=y#?
2 Answers
No solution.
Explanation:
To solve these systems of equations let us put
and we get
i.e.
or
or
But for no real number we have
Hence, we do not have any solution.
Observe that while
graph{(x-y)(x^2-9y^2-36)=0 [-20, 20, -10, 10]}
Explanation:
This system only has Complex solutions...
Given:
#{ (x^2/36-y^2/4 = 1), (x=y) :}#
Substitute
#x^2/36-x^2/4 = 1#
Multiply both sides by
#x^2-9x^2 = 36#
That is:
#-8x^2 = 36#
Divide both sides by
#x^2 = -36/8 = -18/4 = ((3sqrt(2))/2i)^2#
Hence:
#x = y = +-(3sqrt(2))/2i#