Use the substitution method to determine whether the linear-quadratic system y=x2−9x+39 , y=9x−42 has 0, 1, or 2 solutions.(1 point)

To determine the number of solutions for the system of equations given by:

1. \( y = x^2 - 9x + 39 \)
2. \( y = 9x - 42 \)

we can use the substitution method. Since both equations equal \( y \), we can set them equal to each other:

\[
x^2 - 9x + 39 = 9x - 42
\]

Now, we will rearrange this equation to one side to form a quadratic equation:

\[
x^2 - 9x - 9x + 39 + 42 = 0
\]
\[
x^2 - 18x + 81 = 0
\]

Next, we can simplify this quadratic equation:

\[
x^2 - 18x + 81 = 0
\]

We can attempt to factor this quadratic. We notice that:

\[
x^2 - 18x + 81 = (x - 9)^2
\]

Setting the factored form equal to zero gives us:

\[
(x - 9)^2 = 0
\]

This implies:

\[
x - 9 = 0 \quad \Rightarrow \quad x = 9
\]

Since this yields a single value for \( x \), we have one solution in \( x \). To find the corresponding value of \( y \), we can substitute \( x = 9 \) back into either equation. Let's use the second equation \( y = 9x - 42 \):

\[
y = 9(9) - 42 = 81 - 42 = 39
\]

Thus, the solution to the system is the point \( (9, 39) \).

In conclusion, the system of equations has **1 solution**.