Solve the following system of equations in terms of x: (1) x-3y+z= -2 (2) -4x+9y=7 * Student can enter max 2000 characters Use the pape

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
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Chapter9: Quadratic Functions And Equations
Section9.7: Solving Systems Of Linear And Quadratic Equations
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### 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
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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