Solve the system of equations.

y=

14x+13
y=x2–14x–23
Write the coordinates in exact form. Simplify all fractions and radicals.
,
,

1 answer

To solve the system of equations given by:

  1. \( y = -14x + 13 \)
  2. \( y = x^2 - 14x - 23 \)

we can set the two equations for \( y \) equal to each other:

\[ -14x + 13 = x^2 - 14x - 23 \]

Next, we can simplify this equation. First, we can eliminate \(-14x\) from both sides:

\[ 13 = x^2 - 23 \]

Now, add \( 23 \) to both sides:

\[ 13 + 23 = x^2 \] \[ 36 = x^2 \]

Next, we take the square root of both sides to solve for \( x \):

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

Now that we have the \( x \) values, we can substitute them back into either equation to find the corresponding \( y \) values. We'll use the first equation \( y = -14x + 13 \):

  1. For \( x = 6 \):

\[ y = -14(6) + 13 = -84 + 13 = -71 \] So we have the point \((6, -71)\).

  1. For \( x = -6 \):

\[ y = -14(-6) + 13 = 84 + 13 = 97 \] So we have the point \((-6, 97)\).

Thus, the solutions to the system of equations are:

\[ (6, -71) \quad \text{and} \quad (-6, 97) \]

In exact form, the coordinates are:

\[ \boxed{(6, -71)} \quad \text{and} \quad \boxed{(-6, 97)} \]

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