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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) ​

 Dec 18, 2018

Best Answer 

 #1
avatar+4296 
+3

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. 

 Dec 18, 2018
 #1
avatar+4296 
+3
Best Answer

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. 

tertre Dec 18, 2018

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