Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN: 9780134463216
Author: Robert F. Blitzer
Publisher: PEARSON
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**Question 3**

Consider the system of equations below where \((-2, -1)\) represents one solution on the coordinate plane.

\[
y = -x^2 + 3
\]
\[
y = x + 1
\]

What is the \((x, y)\) coordinate of the other solution?

**Explanation:**

This problem involves solving a system of equations to find the points of intersection. One of the solutions is given as \((-2, -1)\).

1. **Equation 1**: \(y = -x^2 + 3\) represents a downward opening parabola.
2. **Equation 2**: \(y = x + 1\) is a linear equation representing a straight line.

To find the intersection points, set the equations equal to each other:

\[
-x^2 + 3 = x + 1
\]

Rearrange into a standard quadratic equation:

\[
-x^2 - x + 3 - 1 = 0 \quad \Rightarrow \quad -x^2 - x + 2 = 0
\]

Simplify:

\[
-x^2 - x + 2 = 0 \quad \Rightarrow \quad x^2 + x - 2 = 0
\]

Factor the quadratic equation:

\[
(x + 2)(x - 1) = 0
\]

Solutions for \(x\) are:

\[
x = -2 \quad \text{or} \quad x = 1
\]

For \(x = 1\), substitute back into \(y = x + 1\):

\[
y = 1 + 1 = 2
\]

Thus, the other solution is \((1, 2)\).
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Transcribed Image Text:**Question 3** Consider the system of equations below where \((-2, -1)\) represents one solution on the coordinate plane. \[ y = -x^2 + 3 \] \[ y = x + 1 \] What is the \((x, y)\) coordinate of the other solution? **Explanation:** This problem involves solving a system of equations to find the points of intersection. One of the solutions is given as \((-2, -1)\). 1. **Equation 1**: \(y = -x^2 + 3\) represents a downward opening parabola. 2. **Equation 2**: \(y = x + 1\) is a linear equation representing a straight line. To find the intersection points, set the equations equal to each other: \[ -x^2 + 3 = x + 1 \] Rearrange into a standard quadratic equation: \[ -x^2 - x + 3 - 1 = 0 \quad \Rightarrow \quad -x^2 - x + 2 = 0 \] Simplify: \[ -x^2 - x + 2 = 0 \quad \Rightarrow \quad x^2 + x - 2 = 0 \] Factor the quadratic equation: \[ (x + 2)(x - 1) = 0 \] Solutions for \(x\) are: \[ x = -2 \quad \text{or} \quad x = 1 \] For \(x = 1\), substitute back into \(y = x + 1\): \[ y = 1 + 1 = 2 \] Thus, the other solution is \((1, 2)\).
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