Question

Help please! How many solutions are there in the system below?
2x^2+y^2=33
x^2+y^2+2y=19
a)4
b)2
c)1
d)3

Answer 1

Eliminate:$$2x^2+y^2=33\\x^2+y^2+2y=19\\2x^2+2y^2+4y=38\\2x^2+2y^2+4y-2x^2-y^2=38-33\\y^2+4y=5\\y^2+4y-5=0\\(y-1)(y+5)=0\\y=1,-5$$... now substitute:$$2x^2+y^2=33\\2x^2+1=33\\2x^2=32\\x^2=16\\x=\pm4$$and$$2x^2+y^2=33\\2x^2+25=33\\2x^2=8\\x^2=4\\x=\pm2$$hence we get solutions \((-2,-5),(2,-5),(-4,1),(4,1))\)