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)
Answer:
D
Step-by-step explanation:
Consider the equation:
Use distributive property in both sides:
Separate terms with x into the left part and terms without x into the right part:
Since you get true equality 0 = 0, the equation is true for all real values of x. So, the equation has infinitely many solutions.
The answer would be 48...
