Solve the system of equations.

y=

15x+47
y=x2–15x+11
Write the coordinates in exact form. Simplify all fractions and radicals.
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1 answer

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 \( \boxed{(-6, 137)} \).