+108 Find the point of intersection.
\(x+y=5\)
and
\(\frac{x^2}{16}+\frac{y^2}{9}=1\)
+108 Hmm looks like I was originally correct, just made some calculation errors...
This is my working, I'm sure there is a more efficient way to do it however...
\(\frac{x^2}{16}+\frac{(5-x)^2}{9}=1\)
\(\frac{9x^2+16(5-x)^2}{144}=1\)
\(9x^2+16(5-x)^2=144\)
\(9x^2+16(25-10x+x^2)=144\)
\(9x^2+400-160x+16x^2=144\)
\(25x^2-160x=-256\)
\(25x^2-160x+256=0\)
And then just solve for x using the quadratic formula...
\(x=\frac{160+-\sqrt{25600-25600}}{50}\)
x = 3.2
y = 1.8
Thus point of intersection = (3.2,1.8)
The answer in the book however is (doesn't show any working):
(16/5, 9/5) which of course is the same answer, but there must be a more efficient way to get to this answer if the answer is a fraction.
