To solve the linear-quadratic system using the substitution method, we can start with the equations you've provided:
- \( y = x^2 + 3x + 5 \)
- \( y = 2x + 11 \)
Since both equations equal \( y \), we can set them equal to each other:
\[ x^2 + 3x + 5 = 2x + 11 \]
Next, we can rearrange this equation to form a standard quadratic equation:
\[ x^2 + 3x - 2x + 5 - 11 = 0 \]
This simplifies to:
\[ x^2 + x - 6 = 0 \]
Now we can factor the quadratic equation:
\[ (x + 3)(x - 2) = 0 \]
Setting each factor to zero gives us the solutions for \( x \):
- \( x + 3 = 0 \) → \( x = -3 \)
- \( x - 2 = 0 \) → \( x = 2 \)
Now we will find the corresponding \( y \) values for each \( x \).
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For \( x = -3 \): \[ y = 2(-3) + 11 = -6 + 11 = 5 \] This gives us the point \( (-3, 5) \).
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For \( x = 2 \): \[ y = 2(2) + 11 = 4 + 11 = 15 \] This gives us the point \( (2, 15) \).
Thus, the solutions to the system are \( (-3, 5) \) and \( (2, 15) \).
From the options given, the correct response is:
(2, 15) and (−3, 5).