Question

What are the solutions of the following system?

Answer 1

Answer:

Option C is correct

the solution for the given system of equation is, (0, -5) and (-4, 3)

Step-by-step explanation:

Using identity:

$(a+b)^2=a^2+b^2+2ab$

Given the equation:

$x^2+y^2=25$        .....[1]

$2x+y=-5$               .....[2]

We can write [2] as:

$y = -5-2x$

Substitute this in [1] we have;

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

Using identity rule;

$x^2+25+4x^2+20x = 25$

Combine like terms;

$5x^2+20x+25 = 25$

Subtract 25 from both sides we have;

$5x^2+20x=0$

⇒$5x(x+4)=0$

By zero product property we have;

x = 0 and x+4 = 0

x = 0 and x = -4

Substitute these in [2] we have;

for x = 0

2(0)+y = -5

⇒$y = -5$

For x = -4 ,

$2(-4)+y = -5$

-8+y = -5

add 8 to both sides we have;

y = 3

Therefore, the solution for the given system of equation is, (0, -5) and (-4, 3)

Answer 2

So what you do is solve for y in 2nd equation
minus 2x boh sides

y=-2x-5
sub that for y in other equation

x^2+(-2x-5)^2=25
x^2+4x^2+20x+25=25
5x^2+20x+25=25
mminus 25 both sides
5x^2+20x=0
factor
(x)(5x+20)=0
set to zero
x=0
5x+20=0
5x=-20
x=-4

sub back

y=-2x-5
y=-2(0)-5
y=-5
a point is (0,-5)
y=-2x-5
y=-2(-4)-5
y=-8-5
y=-13
another is (-4,-13)


(0,-5) and (-4,13)

I think it is C, but you forgot to put that 1 infront of the 3