Can someone help me solve this simultaneous equation please?

Question

x^2 = 4-y
x = y + 2

Listen me I need a clear comprehendable answer please, and need I a step by step explanation in order to give this ten points.

Answer 1

Just plug the value for x into the first equation and solve:

(y + 2)² = 4 - y
y² + 4y + 4 = 4 - y
y² + 5y = 0
y(y + 5) = 0

So y can either be 0 or -5.

If y is 0, x = 0 + 2 = 2, so this solution is the point (2, 0).

If y is -5, x = -5 + 2 = -3, so this is (-3, -5).

Check the solutions to verify:

(2, 0):
2² = 4 - 0, check
2 = 0 + 2, check

(-3, -5):
(-3)² = 4 - (-5), check
-3 = -5 + 2, check

Answer 2

transfer the y's to the left side and they change the signs so it would be x^2+y=4 and x-y=2 and add the y's and they became cancelled and it would be also the same thing for the others x^2+x=4+2 so its x^2+x=6 and think of a number that is squared and added to itself so its 2 and x=2 substitute x into x-y=2 so 2-y=2 so -y=2-2 because when we transfer the numbers or variables we change the signs so -y=0 then y=0 answer is x=2 , y=0

Answer 3

x² = 4 - y- - - - - -Equation 1
x = y + 2- - - - - -Equation 2
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Insert the x value equation 2 into equation 1

x² = 4 - y

(y + 2)² = 4 - y

y² + 4Y + 4 = 4 - y

y² + 4Y + 4 - 4 = 4 - y - 4

y² + 4y = - y

y² + 4y + y = - y + y

y² + 5y = 0

y(y + 5) = 0

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Roots

y = 0

Insert the y value into equation 1

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y + 5 = 0

y + 5 - 5 = 0 - 5

y = - 5. .<=..not used

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x² = 4 - y

x² = 4 - 0

x² = 4

x² = √4

x = 2

The answer is x = 2

Insert the x value into equation 1

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Check for equation 1

x² = 4 - y

(2)² = 4 - 0

4 = 4

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Check for equation 2

x = y + 2

2 = 0 = 2

2 = 2

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The solutioin set is { 2, 0 }

- - - - - - - - - - - -s-

Answer 4

x=y+2 so replace x in the 1st equation with y+2 like so: (y+2)^2=4-y,
now (y+2)^2 can be rewritten like this: (y+2)(y+2) which when simplified into a trinomial looks like this:y^2+4y+4, so now you can stick that back into the equation and solve for y. Once you solve for y you can go back to the original equations and solve for x.

Answer 5

Ok Listen boy here comes the answer

x = y + 2 THUS x^2 = (y+2)^2

is that clear ?

if yes i wil;l continue otherwise it is useless to continue

Answer 6

First off substitute x in the first eqn with the y+2 in the second.

Now expand the polynomial and move over the 4-y giving you:

y^2 + 5y = 0

Factor the quadratic and then solve for y. Now with these two values of y substitute them back into the second eqn, giving you the two values of x.