Question

Solve the system of equations.
y=
–
15x+47
y=x2–15x+11
Write the coordinates in exact form. Simplify all fractions and radicals.
,
,

Answer 1

To solve the system of equations, we have the two equations:

1. \( y = -15x + 47 \)
2. \( y = x^2 - 15x + 11 \)

We can set the two expressions for \( y \) equal to each other:

\[
-15x + 47 = x^2 - 15x + 11
\]

Now, simplify this equation. First, we notice that the \( -15x \) terms on both sides can be eliminated:

\[
47 = x^2 + 11
\]

Next, we rearrange the equation:

\[
x^2 + 11 - 47 = 0
\]
\[
x^2 - 36 = 0
\]

Now, factor the equation:

\[
(x - 6)(x + 6) = 0
\]

This gives us:

\[
x - 6 = 0 \quad \text{or} \quad x + 6 = 0
\]
\[
x = 6 \quad \text{or} \quad x = -6
\]

Next, we substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. We'll use the first equation, \( y = -15x + 47 \).

1. For \( x = 6 \):
\[
y = -15(6) + 47 = -90 + 47 = -43
\]
So one solution is \( (6, -43) \).

2. For \( x = -6 \):
\[
y = -15(-6) + 47 = 90 + 47 = 137
\]
So another solution is \( (-6, 137) \).

Thus, the solutions to the system of equations are:

\[
(6, -43) \quad \text{and} \quad (-6, 137)
\]

So the coordinates in exact form are \( \boxed{(6, -43)} \) and \( \boxed{(-6, 137)} \).