Respuesta :

Hi there!  

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I believe your answer is:  

(-3, -1) and (1, 3)

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Here’s why:  

  • I have graphed the two equations given on a graphing program.
  • When graphed, they pass at points (-3, -1) and (1,3). Therefore, they are the solutions to the system.

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See the graph attached.

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Hope this helps you. I apologize if it’s incorrect.  

Ver imagen Аноним

Answer:

[tex]{ \tt{y = x + 2 - - - (a)}} \\ { \tt{ {x}^{2} + {y}^{2} = 10 - - - (b) }} \\ substitute \: for \: y \: in \: (b) : \\ { \tt{ {x}^{2} + {(x + 2)}^{2} = 10 }} \\ { \tt{ {x}^{2} + {x}^{2} + 4x + 4 = 10 }} \\ { \tt{2 {x}^{2} + 4x - 6 = 0}} \\ { \tt{ {x}^{2} + 2x - 3 = 0 }} \\ { \boxed{ \bf{x = 1 \: and \: - 3}}} \\ in \: (a) : \\ y = x + 2 \\ { \boxed{ \bf{y = 3 \: and \: - 1}}}[/tex]