Answer:
[tex]x=4,y=-3[/tex], or as an ordered pair: (4, -3)
Step-by-step explanation:
We have the system of equations
[tex]x^2+y^2=25[/tex] (1)
[tex]x-y^2=-5[/tex] (2)
Let's solve our system step-by-step:
Step 1. Solve for [tex]y^2[/tex] in equation (1)
[tex]x^2+y^2=25[/tex]
[tex]y^2=25-x^2[/tex] (3)
Step 2. Replace equation (3) in equation (2) and solve for [tex]x[/tex]
[tex]x-y^2=-5[/tex]
[tex]x-(25-x^2)=-5[/tex]
[tex]x-25+x^2=-5[/tex]
[tex]x^2+x-20=0[/tex]
[tex](x+5)(x-4)=0[/tex]
[tex]x+5=0,x-4=0[/tex]
[tex]x=-5,x=4[/tex]
Remember that in the fourth quadrant x is positive and y is negative, so we are taking the positive value of x; in other words [tex]x=4[/tex] (4)
Step 3. Replace equation (4) (the positive value of x) in equation (1)
[tex]x^2+y^2=25[/tex]
[tex]4^2+y^2=25[/tex]
[tex]16+y^2=25[/tex]
[tex]y^2=25-16[/tex]
[tex]y^2=9[/tex]
[tex]y=\sqrt{9} ,y=-\sqrt{9}[/tex]
[tex]y=3,y=-3[/tex]
Since y is negative in quadrant IV, we take [tex]y=-3[/tex]
We can conclude that the solution of our system of equations that lies in the quadrant IV is [tex]x=4,y=-3[/tex], or as an ordered pair: (4, -3)
