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5. Use the elimination method to solve the linear-quadratic system y=x^2+x-42 & y=-x+6

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5 answers

To solve the linear-quadratic system of equations using the elimination method, we'll start with the two equations you provided:

  1. \( y = x^2 + x - 42 \) (Quadratic equation)
  2. \( y = -x + 6 \) (Linear equation)

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

\[ x^2 + x - 42 = -x + 6 \]

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

  1. Add \( x \) and subtract \( 6 \) from both sides:

\[ x^2 + x + x - 42 - 6 = 0 \]

This simplifies to:

\[ x^2 + 2x - 48 = 0 \]

Next, we can factor the quadratic equation \( x^2 + 2x - 48 \):

To factor, we need two numbers that multiply to \(-48\) and add up to \(2\). The numbers \(8\) and \(-6\) work because:

\[ 8 \cdot -6 = -48 \quad \text{and} \quad 8 + (-6) = 2 \]

Thus, we can write the factored form:

\[ (x + 8)(x - 6) = 0 \]

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

  1. \( x + 8 = 0 \) → \( x = -8 \)
  2. \( x - 6 = 0 \) → \( x = 6 \)

Now, we have two values for \( x \). Next, we will substitute each of these values back into the linear equation \( y = -x + 6 \) to find the corresponding \( y \) values.

For \( x = -8 \): \[ y = -(-8) + 6 = 8 + 6 = 14 \] So, one solution is \( (-8, 14) \).

For \( x = 6 \): \[ y = -(6) + 6 = -6 + 6 = 0 \] So, another solution is \( (6, 0) \).

Final Solutions

The solutions to the linear-quadratic system are:

\[ (-8, 14) \quad \text{and} \quad (6, 0) \]

Shorter please

Certainly! Here's a shorter version:

We have the system:

  1. \( y = x^2 + x - 42 \)
  2. \( y = -x + 6 \)

Set them equal:

\[ x^2 + x - 42 = -x + 6 \]

Rearranging gives:

\[ x^2 + 2x - 48 = 0 \]

Factoring:

\[ (x + 8)(x - 6) = 0 \]

Thus, the solutions for \( x \) are:

  • \( x = -8 \)
  • \( x = 6 \)

Now, substitute back to find \( y \):

  1. For \( x = -8 \): \[ y = -(-8) + 6 = 14 \] → Solution: \( (-8, 14) \)

  2. For \( x = 6 \): \[ y = -(6) + 6 = 0 \] → Solution: \( (6, 0) \)

Final Solutions:

\[ (-8, 14) \quad \text{and} \quad (6, 0) \]

Shorter

Sure! Here’s an even shorter summary:

  1. Set equations equal: \[ x^2 + x - 42 = -x + 6 \] becomes: \[ x^2 + 2x - 48 = 0 \]

  2. Factor: \[ (x + 8)(x - 6) = 0 \] Solutions: \( x = -8 \) and \( x = 6 \)

  3. Find \( y \):

    • For \( x = -8 \): \( y = 14 \) → \( (-8, 14) \)
    • For \( x = 6 \): \( y = 0 \) → \( (6, 0) \)

Solutions:

\[ (-8, 14) \quad \text{and} \quad (6, 0) \]