How do you solve each system of nonlinear equations?

Question

2x + y + 3 = 0
x^2 + y^2 = 5

Answer 1

2x+y+3=0 --- (i)
x^2+y^2=5 ---(ii)

from(i) 2x+y= -3
or, y = -2x-3

putting this value in (ii) we get

x^2+(-2x-3)^2-5=0
or, x^2+(-2x)^2+(-3)^2+2(-2x)(-3)-5=0
or, x^2+4x^2+9+12x-5=0
or, 5x^2+12x+4=0
or, 5x^2+10x+2x+4=0
or, 5x(x+2)+2(x+2)=0
or, (5x+2)(x+2)=0

or, x = -2 and -2/5

now putting this value in equation (ii) we get

4+y^2=5
or y^2=1
or y = 1

and

4/25+y^2=5
or, 4+25y^2=125
or, 25y^2 = 121
or, y = Sqrt (121)/5

so we have two sets of (x,y) value and they are

(-2, 1) and (-2/5, sqrt121/5)

cool :)

Answer 2

It's always best to use substitution from the linear equation, and then transfer the variable over to the other one.

2x + y + 3 = 0
x^2 + y^2 = 5

Solving for y from the first equation, we get

y = -2x - 3

Now, we plug in y into the second equation, to obtain

x^2 + [-2x - 3]^2 = 5. Expanding the binomial, we get

x^2 + [4x^2 + 12x + 9] = 5
5x^2 + 12x + 9 = 5
5x^2 + 12x + 4 = 0

We now factor this.
(5x + 2) (x + 2) = 0, yielding
x = -2/5 or x = -2

Now, we plug each of these values in individually for
y = -2x - 3

Plug in x = -2/5: y = -2(-2/5) - 3 = 4/5 - 3 = -11/5
Plug in x = -2: y = -2(-2) - 3 = 4 - 3 = 1

Therefore, your solutions are:
{x = -2/5, y = -11/5}
{x = -2, y = 1}

Answer 3

first make first equation in terms of x like that
y = -2x-3

then substitute into y

x^2 + (-2x-3)^2 = 5

x^2 +(4x^2+12x+9)=5

5x^2+12x+4=0

(5x + 2 )(x+2) = 0
x = -2/5
x = -2

then plug x into any of the equation

-4 + y + 3 = 0
y = 1
-4/5 + y + 3 = 0
y = -11/5

Answer:
x = -2/5
x = -2
y = 1
y = -11/5

Answer 4

I won't do it for you, but I can give you some strategies to solve this system of equations. First, you would likely need to solve the first equation for y. Then, substitute it into the bottom equation. This will give you two distinct answers for x, but you are going to have two ordered pairs that will satisfy this system of equations.

Answer 5

sorry can't help