+476 What are the solutions to the system of equations?
y=−x2−5x−6
x+y=−3
Choices:
(−3, 0) and (−1,−2)
(−3, 0) and (−2, 0)
(−3, 0) and (−2,−1)
(−2, 0) and (−1,−2)
+4609 I think you mean: \(y=x^2-5x-6\) and \(x+y=-3.\) Taking \(x\) to the other side in our second equation, we have \(y=-3-x\) . This means \(-x^2-5x-6=-3-x\), so by using the quadratic formula, we get \(x=-3,\:x=-1.\) .
From here, first by plugging \(x=-3\) in the second equation to solve for \(y\), we get \(-3+y=-3, y=-3+3, y=0.\) Consequently, by plugging in \(x=-1\)\(\), we get \(-1+y=-3, y=-3+1, y=-2.\) Finally, our two ordered pairs are \(\boxed{(-3,0), (-1,-2)}\) , so (A) or the first option.
+4609 I think you mean: \(y=x^2-5x-6\) and \(x+y=-3.\) Taking \(x\) to the other side in our second equation, we have \(y=-3-x\) . This means \(-x^2-5x-6=-3-x\), so by using the quadratic formula, we get \(x=-3,\:x=-1.\) .
From here, first by plugging \(x=-3\) in the second equation to solve for \(y\), we get \(-3+y=-3, y=-3+3, y=0.\) Consequently, by plugging in \(x=-1\)\(\), we get \(-1+y=-3, y=-3+1, y=-2.\) Finally, our two ordered pairs are \(\boxed{(-3,0), (-1,-2)}\) , so (A) or the first option.
