   Chapter 7.6, Problem 5SWU ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# In Exercises 1-6, solve the system of linear equations. { 2 x − y + z = 3 2 x + 2 y + z = 4 − x + 2 y + 3 z = − 1

To determine

To calculate: The values of x,y,z for the system of linear equation 2xy+z=3, 2x+2y+z=4 and x+2y+3z=1.

Explanation

Given Information:

The system of linear equation 2xy+z=3, 2x+2y+z=4 and x+2y+3z=1.

Formula used:

Step 1: Reduce any of the given equation in one variable.

Step 2: Substitute the value into other two equations.

Step 3: Solve the system of equations for two variables.

Calculation:

Consider the given equation

2xy+z=3z=32x+y

Substitute this value in equation 2x+2y+z=4,

2x+2y+(32x+y)=42x+2y+32x+y=43y+3=43y=1

And,

y=13

Now, substitute the values in equation x+2y+3z=1

x+2

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