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What are the solutions of the following system?

x^2+y^2=25

2x+y=-5

A(0, –5) and (–5, 5)

B.(0, –5) and (5, –15)

C.(0, –5) and (–4, 3)

D.(0, –5) and (4, –13)

Verified Answer :

apologiabiology

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

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)

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