Answer:
C) The solution for the given system of equations are A(0,-5) and B(-4,3)
Step-by-step explanation:
The given system of equation are : [tex]x^{2} + y^{2} = 25\\2x + y = -5[/tex]
from equation 2, we get y = -5 - 2x .
Put the above value of y in the equation (1).
We get: [tex]x^{2} + y^{2} = 25 \implies x^{2} + (-5-2x)^{2} = 25[/tex]
By ALGEBRAIC IDENTITY:
[tex](a+b)^{2} = a^{2} + b^{2} + 2ab\\ (-5-2x)^{2} = (-5)^{2} + (-2x)^{2} + 2(-5)(-2x)[/tex]
or, [tex]x^{2} + (-5-2x)^{2} = 25 \implies x^{2} + (25 + 4x^{2} + 20x) = 25[/tex]
or, [tex]5x^{2} + 20x = 0 \implies x(5x + 20) = 0[/tex]
⇒ x = 0 or, x = -20/5 = -4
So, the possible values for x are: x = 0 or x = -4
If x = 0, y = -5-2x = -5-2(0) = -5
and if x = -4, y = -5 -2(-4) = -5 + 8 = 3
Hence, the solution for the given system of equations are A(0,-5) and B(-4,3)
