Answer :

chetankachana

Answer:

(0,3)

(-1,1)

Step-by-step explanation:

Hello!

Since both equations are equal to y, we can set the right-hand side of the equations equal to each other.

  • [tex]y = 2x + 3\\y = x^2 + 3x + 3[/tex]
  • [tex]y = y[/tex]
  • [tex]2x + 3 = x^2 + 3x + 3[/tex]

Solve for x

  • [tex]2x + 3 = x^2 + 3x + 3[/tex]
  • [tex]0 = x^2 + x[/tex]
  • [tex]0 = x(x + 1)[/tex]
  • [tex]x = 0, x =-1[/tex]

X can either be 0 or -1. Remember, a solution to a system is not complete without a y-value. Plug in 0 and -1 for x in the first equation, and find the corresponding y-values.

[tex]y = 2(0) + 3\\y = 3[/tex]

And

[tex]y = 2(-1) + 3\\y = -2 + 3\\y = 1[/tex]

So the solutions to the system are (0,3) and (-1,1)

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