How do you find the exact solutions to the system x^2/36-y^2/4=1 and x=y?

2 Answers
Shwetank Mauria
Feb 4, 2017

No solution.

Explanation:

To solve these systems of equations let us put x=y in x^2/36-y^2/4=1

and we get x^2/36-x^2/4=1

i.e. x^2/36-(9x^2)/36=1

or (-8x^2)/36=1

or -2x^2=9 or 2x^2+9=0

But for no real number we have 2x^2+9=0

Hence, we do not have any solution.

Observe that while x^2/36-y^2/4=1 represents hyperbola x=y is a line passing through origin with slope of 1 and the two do not intersect.
graph{(x-y)(x^2-9y^2-36)=0 [-20, 20, -10, 10]}

George C.
Feb 5, 2017

x = y = +-(3sqrt(2))/2i

Explanation:

This system only has Complex solutions...

Given:

{ (x^2/36-y^2/4 = 1), (x=y) :}

Substitute y=x in the first equation to get:

x^2/36-x^2/4 = 1

Multiply both sides by 36 to get:

x^2-9x^2 = 36

That is:

-8x^2 = 36

Divide both sides by -8 to get:

x^2 = -36/8 = -18/4 = ((3sqrt(2))/2i)^2

Hence:

x = y = +-(3sqrt(2))/2i

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