Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter9: Quadratic Functions And Equations
Section9.7: Solving Systems Of Linear And Quadratic Equations
Problem 30HP
Related questions
Question
![### Solving Systems of Equations
This module will guide you through solving the following system of equations in terms of \( x \):
1. \( x - 3y + z = -2 \)
2. \(-4x + 9y = 7 \)
You need to find the values of \( x \), \( y \), and \( z \) that satisfy both equations simultaneously.
Below are the steps to solve a system of linear equations:
**Step 1: Express one variable in terms of the others (if possible).**
In equation (2), we can solve for \( y \):
\[ -4x + 9y = 7 \]
\[ 9y = 4x + 7 \]
\[ y = \frac{4x + 7}{9} \]
**Step 2: Substitute this expression into the other equation.**
Substitute \( y = \frac{4x + 7}{9} \) into equation (1):
\[ x - 3\left(\frac{4x + 7}{9}\right) + z = -2 \]
\[ x - \frac{12x + 21}{9} + z = -2 \]
\[ x - \frac{12x}{9} - \frac{21}{9} + z = -2 \]
\[ x - \frac{4x}{3} - \frac{7}{3} + z = -2 \]
**Step 3: Simplify and solve for \( x \) and \( z \).**
Combine like terms and solve for \( z \):
\[ \left(x - \frac{4x}{3}\right) + z = -2 + \frac{7}{3} \]
\[ \left(\frac{3x - 4x}{3}\right) + z = \frac{-6 + 7}{3} \]
\[ \left(\frac{-x}{3}\right) + z = \frac{1}{3} \]
\[ z = \frac{x}{3} + \frac{1}{3} \]
So now you have expressions for \( y \) and \( z \) in terms of \( x \):
\[ y = \frac{4x + 7}{9} \]
\[ z = \frac{x}{3} + \frac{1}{3} \]
**Step](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf6f4faa-bb31-4351-9b90-23a290e16aec%2F1d3060e1-f5f1-4a6e-960a-ebec89bc0572%2F2w77xe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving Systems of Equations
This module will guide you through solving the following system of equations in terms of \( x \):
1. \( x - 3y + z = -2 \)
2. \(-4x + 9y = 7 \)
You need to find the values of \( x \), \( y \), and \( z \) that satisfy both equations simultaneously.
Below are the steps to solve a system of linear equations:
**Step 1: Express one variable in terms of the others (if possible).**
In equation (2), we can solve for \( y \):
\[ -4x + 9y = 7 \]
\[ 9y = 4x + 7 \]
\[ y = \frac{4x + 7}{9} \]
**Step 2: Substitute this expression into the other equation.**
Substitute \( y = \frac{4x + 7}{9} \) into equation (1):
\[ x - 3\left(\frac{4x + 7}{9}\right) + z = -2 \]
\[ x - \frac{12x + 21}{9} + z = -2 \]
\[ x - \frac{12x}{9} - \frac{21}{9} + z = -2 \]
\[ x - \frac{4x}{3} - \frac{7}{3} + z = -2 \]
**Step 3: Simplify and solve for \( x \) and \( z \).**
Combine like terms and solve for \( z \):
\[ \left(x - \frac{4x}{3}\right) + z = -2 + \frac{7}{3} \]
\[ \left(\frac{3x - 4x}{3}\right) + z = \frac{-6 + 7}{3} \]
\[ \left(\frac{-x}{3}\right) + z = \frac{1}{3} \]
\[ z = \frac{x}{3} + \frac{1}{3} \]
So now you have expressions for \( y \) and \( z \) in terms of \( x \):
\[ y = \frac{4x + 7}{9} \]
\[ z = \frac{x}{3} + \frac{1}{3} \]
**Step
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