how do you solve that pair of simultaneous equation^2=4-y x=y+3?
To solve the pair of simultaneous equations:
1) x^2 = 4 - y
2) x = y + 3
We'll substitute equation 2) into equation 1) and solve for y.
Substituting equation 2) into equation 1), we get:
(y + 3)^2 = 4 - y
Expanding the square on the left hand side:
y^2 + 6y + 9 = 4 - y
Rearranging the equation:
y^2 + 6y + y + 9 - 4 = 0
Combining like terms:
y^2 + 7y + 5 = 0
Now, we have a quadratic equation in terms of y. We can solve it using factorization, completing the square, or the quadratic formula.
Factoring the quadratic equation:
(y + 5)(y + 1) = 0
Setting each factor equal to zero:
y + 5 = 0 or y + 1 = 0
Solving for y, we get:
y = -5 or y = -1
Now that we have the values of y, we can substitute them back into equation 2) to find the corresponding values of x.
For y = -5:
x = y + 3
x = -5 + 3
x = -2
For y = -1:
x = y + 3
x = -1 + 3
x = 2
Therefore, the solutions to the simultaneous equations are:
x = -2, y = -5
x = 2, y = -1