Use the substitution method to determine whether the linear-quadratic system y=x2+3x+4 , y=2x+10 has 0, 1, or 2 solutions.

To determine the number of solutions to the system of equations using the substitution method, we have the following equations:

1. \( y = x^2 + 3x + 4 \)
2. \( y = 2x + 10 \)

Since both equations are equal to \( y \), we can set them equal to each other:

\[
x^2 + 3x + 4 = 2x + 10
\]

Next, we will move all terms to one side of the equation:

\[
x^2 + 3x + 4 - 2x - 10 = 0
\]

Simplifying this, we get:

\[
x^2 + (3x - 2x) + (4 - 10) = 0
\]
\[
x^2 + x - 6 = 0
\]

Now we can factor the quadratic equation:

\[
x^2 + x - 6 = (x + 3)(x - 2) = 0
\]

Setting each factor to zero gives us the solutions for \( x \):

1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x - 2 = 0 \) → \( x = 2 \)

So, we have two values for \( x \): \( x = -3 \) and \( x = 2 \).

Next, we will find the corresponding \( y \) values for each \( x \) value using the second equation, \( y = 2x + 10 \):

1. For \( x = -3 \):
\[
y = 2(-3) + 10 = -6 + 10 = 4
\]
So, one solution is \( (-3, 4) \).

2. For \( x = 2 \):
\[
y = 2(2) + 10 = 4 + 10 = 14
\]
So, the other solution is \( (2, 14) \).

In conclusion, the linear-quadratic system has **two solutions**: \( (-3, 4) \) and \( (2, 14) \).