robert7248
contestada

In the system shown below, what are the coordinates of the solution that lies in quadrant I?

Write your answer in the form (a,b) without using spaces.

[tex]x^{2} +y^2=25[/tex]
[tex]x-y^2=-5[/tex]

Respuesta :

absor201

Answer:

(4,3)

Step-by-step explanation:

Given

x^2+y^2=25

x-y^2=-5

In order to solve the equations, from equation 2 we get

-y^2= -5-x

y^2=5+x

Putting the value of y^2 in equation 1

x^2+5+x=25

x^2+5-25+x=0

x^2+x-20=0

x^2+5x-4x-20= 0

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

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

So

x+5=0  x-4=0

x=-5   x=4

Now for x=-5

x^2+y^2=25

(-5)^2+y^2=25

25+y^2=25

y^2=25-25

y^2=0

so Y=0  

And for x = 4

x^2+y^2=25

(4)^2+y^2=25

16+y^2=25

y^2=25-16

y^2=9

y= ±3

So the solution to the system of equations is

(-5,0) , (4,3), (4,-3)

The only solution that belongs to first quadrant is (4,3)

lucic

Answer:

(4,3)

Step-by-step explanation:

solve the simultaneous equations

x²+y²=25.........................(i)

x-y²= -5.............................(ii) ⇒make y² the subject of the equation;

y²=x+5.................................(iii)

Substitute equation (iii) in (i)

x² + x+5 =25

x²+x=25-5

x²+x-20=0..........solve for the quadratic equation

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

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

x-4=0

x=  4 or

x+5=0

x=  -5

finding value of y

if y²=x+5 then;

when x=4, y²=4+5=9⇒y=√9 = ±3...................y= ±3

when x= -5  , y²= -5+ 5=0............... y=0

Coordinates = (4,3) and (-5,0)

Coordinates that lie in the 1st quadrant is (4,3)

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