$$ if \mathrm A = \{ (x,y): x^2 + y^2 = 25\}$$ $$ and \; \mathrm B = \{ (x,y): x^2 + 9y^2 = 144 \}$$ $$ \text Find \; n( \text A \; \cap \; \text B)$$
Answer Given in Textbook = 4.
Given, x and y are 2 variables, which are real numbers.
Problem - We have only studied simple algebra in one and two variables (equations using 2 variables using substitution, cancellation, cross-multiplication method), Hyperbola and Graph related stuff are not introduced.
In Sets - Basic Operations i.e. Union, Intersection, Belongs to, subsets, subtraction and complimentary.
So, is there any way to solve this problem without Using Complex Graphs and Hyperbola and Parabola?
So, there are two ways I've tried.
First - Manually using Hit and Trial Method, which gets (3, 4), (4, 3), (-3, 4) and (4, -3) which satisfies the first equation in set A, but none of those satisfies the value for the Second equation for the set B, so it's clear that there are no integers which are part of the four answers.
Second - $$ x^2 + y^2 = 25 \implies x^2 = 25 - y^2 $$ Now, using this value of $ x^2 $ in the second equation to get value of $ y = 4$ but that doesn't seems to work.
I'm not sure if there's any other way. Please keep the solution simple.
Hint
$$x^2=25-y^2$$
then
$$x^2+9y^2=144\to 25-y^2+9y^2=144\to y^2=\frac{119}{8}\to y=\pm\sqrt{\frac{119}{8}}$$
can you solve for $x$ and find $4$ solutions?