Find the points of intersection of the graphs of the equations. x2 + y = 17 -2x + y = 14 (x, y) (smaller x-value) (x, y) )- ) (larger x-value)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Finding Points of Intersection of Quadratic and Linear Equations**

To determine the points of intersection for the following equations:

\[ x^2 + y = 17 \]
\[ -2x + y = 14 \]

we solve these equations simultaneously. 

First, we express \( y \) from the second equation:

\[ y = 2x + 14 \]

Next, we substitute \( y \) in the first equation with \( 2x + 14 \):

\[ x^2 + (2x + 14) = 17 \]
\[ x^2 + 2x + 14 = 17 \]
\[ x^2 + 2x - 3 = 0 \]

We then solve for \( x \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = 2 \), and \( c = -3 \):

\[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-3)}}{2(1)} \]
\[ x = \frac{-2 \pm \sqrt{4 + 12}}{2} \]
\[ x = \frac{-2 \pm 4}{2} \]

Thus, the possible values of \( x \) are:

\[ x = \frac{2}{2} = 1 \]
\[ x = \frac{-6}{2} = -3 \]

Correspondingly, the values of \( y \) are found using \( y = 2x + 14 \):

For \( x = 1 \):

\[ y = 2(1) + 14 = 16 \]
Resulting in the point of intersection \( (1, 16) \).

For \( x = -3 \):

\[ y = 2(-3) + 14 = 8 \]
Resulting in the point of intersection \( (-3, 8) \).

The problem statement requests these points sorted by the x-values, smaller to larger:

\[ (x, y) = (-3, 8) \]
\[ (x, y) = (1, 16) \]

The provided image has fields for entering these two points with additional labels indicating the requirement for points with smaller and larger x-values:

1. **
Transcribed Image Text:**Finding Points of Intersection of Quadratic and Linear Equations** To determine the points of intersection for the following equations: \[ x^2 + y = 17 \] \[ -2x + y = 14 \] we solve these equations simultaneously. First, we express \( y \) from the second equation: \[ y = 2x + 14 \] Next, we substitute \( y \) in the first equation with \( 2x + 14 \): \[ x^2 + (2x + 14) = 17 \] \[ x^2 + 2x + 14 = 17 \] \[ x^2 + 2x - 3 = 0 \] We then solve for \( x \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 2 \), and \( c = -3 \): \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-3)}}{2(1)} \] \[ x = \frac{-2 \pm \sqrt{4 + 12}}{2} \] \[ x = \frac{-2 \pm 4}{2} \] Thus, the possible values of \( x \) are: \[ x = \frac{2}{2} = 1 \] \[ x = \frac{-6}{2} = -3 \] Correspondingly, the values of \( y \) are found using \( y = 2x + 14 \): For \( x = 1 \): \[ y = 2(1) + 14 = 16 \] Resulting in the point of intersection \( (1, 16) \). For \( x = -3 \): \[ y = 2(-3) + 14 = 8 \] Resulting in the point of intersection \( (-3, 8) \). The problem statement requests these points sorted by the x-values, smaller to larger: \[ (x, y) = (-3, 8) \] \[ (x, y) = (1, 16) \] The provided image has fields for entering these two points with additional labels indicating the requirement for points with smaller and larger x-values: 1. **
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