Question 76853
solve the sytem algebraically

y^2 = x^2 - 32
x^2 + y^2 = 40
Substitute x^2-32 in for y^2 in the second equation.
{{{x^2+(x^2-32)=40}}}
{{{2x^2-32=40}}}
{{{2x^2-32+32=40+32}}}
{{{2x^2=72}}}
{{{2x^2/2=72/2}}}
{{{x^2=36}}}
{{{sqrt(x^2)=+-sqrt(36)}}}
x=-6 or x=6
Let x=-6 or x=6 (They will both have the same outcome because x is squared.) in the first equation and solve for y.
{{{y^2=(-6)^2-32}}}
{{{y^2=36-32}}}
{{{y^2=4}}}
{{{sqrt(y^2)=+-sqrt(4)}}}
y=-2 or y=2
Therefore two solutions are (-6,-2) and (-6,2)
{{{y^2=(6)^2-32}}}
{{{y^2=36-32}}}
{{{y^2=4}}}
{{{sqrt(y^2)=+-sqrt(4)}}}
y=-2 or y=2
Therefore two solutions are (6,-2) and (6,2)
There are four solutions: (-6,-2), (-6,2), (6,-2), and (6,2)
It helps to look at the graph.
Graphically it looks like this:
{{{graph(300,200,-10,10,-10,10,sqrt(x^2-32),-1*sqrt(x^2-32),sqrt(40-x^2),-1*sqrt(40-x^2))}}}
Happy Calculating!!!