Question

in the system shown below, what are the coordinates of the solution that lies in quadrant IV? x2+y2=25 x-y2=-5

Answer 1

Answer:

$x=4,y=-3$, or as an ordered pair: (4, -3)

Step-by-step explanation:

We have the system of equations

$x^2+y^2=25$ (1)

$x-y^2=-5$ (2)

Let's solve our system step-by-step:

Step 1. Solve for $y^2$ in equation (1)

$x^2+y^2=25$

$y^2=25-x^2$ (3)

Step 2. Replace equation (3) in equation (2) and solve for $x$

$x-y^2=-5$

$x-(25-x^2)=-5$

$x-25+x^2=-5$

$x^2+x-20=0$

$(x+5)(x-4)=0$

$x+5=0,x-4=0$

$x=-5,x=4$

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 $x=4$ (4)

Step 3. Replace equation (4) (the positive value of x) in equation (1)

$x^2+y^2=25$

$4^2+y^2=25$

$16+y^2=25$

$y^2=25-16$

$y^2=9$

$y=\sqrt{9} ,y=-\sqrt{9}$

$y=3,y=-3$

Since y is negative in quadrant IV, we take $y=-3$

We can conclude that the solution of our system of equations that lies in the quadrant IV is $x=4,y=-3$, or as an ordered pair: (4, -3)