The given equations are
y = x^2 + 2x + 7
y = x + 7
We would substitute y = x + 7 into the first equation. It becomes
x + 7 = x^2 + 2x + 7
Collecting like terms, it becomes
x^2 + 2x - x + 7 - 7 = 0
x^2 + x = 0
By factorising x, it becomes
x(x + 1) = 0
Thus,
x = 0 or x + 1 = 0
x = 0 or x = - 1
Substituting x = 0 into y = x + 7, it becomes
y = 0 + 7
y = 7
Thus, one solution set is (0, 7)
Substituting x = - 1 into y = x + 7, it becomes
y = - 1 + 7
y = 6
Thus, another solution set is (- 1, 6)
Therefore, the solution sets are
{(0, 7), (- 1, 6)}
Option A is correct
