Q. 78 In Problems 75鈥� 82, algebraica... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 11: Q. 78 (page 757)

In Problems 75鈥� 82, algebraically solve each system of equations using any method you wish.
$\{\begin{cases} 2 x + 3 y - z = - 2 \\ 4 x + 3 z = 6 \\ 6 y - 2 z = 2 \end{cases}$

Short Answer

Expert verified

The values of unknown variables are $x = - 1.5, y = 1.67 a n d z = 4$

Step by step solution

Step 1. Given information

The given equations $\{\begin{cases} 2 x + 3 y - z = - 2 \\ 4 x + 3 z = 6 \\ 6 y - 2 z = 2 \end{cases}$

Step 2. Finding the values of x,y

Consider the equations $2 x + 3 y - z = - 2, 4 x + 3 z = 6$ and $6 y - 2 z = 2$ .To solve these equations,we first convert them into matrix form and then transform it into reduced echelon form

$[\begin{matrix} 2 & 3 & - 1 \\ 4 & 0 & 3 \\ 0 & 6 & - 2 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} - 2 \\ 6 \\ 2 \end{matrix}] \{R_{1} = \frac{R_{1}}{2}\} [\begin{matrix} 1 & \frac{3}{2} & - \frac{1}{2} \\ 4 & 0 & 3 \\ 0 & 6 & - 2 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} - 1 \\ 6 \\ 2 \end{matrix}] \{R_{2} = R_{2} - 4 R_{1}\} [\begin{matrix} 1 & \frac{3}{2} & - \frac{1}{2} \\ 0 & - 6 & 5 \\ 0 & 6 & - 2 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} - 1 \\ 10 \\ 2 \end{matrix}] \{R_{2} = - \frac{R_{2}}{6}\} [\begin{matrix} 1 & \frac{3}{2} & - \frac{1}{2} \\ 0 & 1 & - \frac{5}{6} \\ 0 & 6 & - 2 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} - 1 \\ - \frac{5}{3} \\ 2 \end{matrix}] \{R_{1} = R_{1} - \frac{3}{2} R_{2}; R_{3} = R_{3} - 6 R_{2}\} [\begin{matrix} 1 & 0 & \frac{3}{4} \\ 0 & 1 & - \frac{5}{6} \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} \frac{3}{2} \\ - \frac{5}{3} \\ 4 \end{matrix}] \{R_{1} = R_{1} - \frac{3}{4} R_{3}; R_{2} = R_{2} + \frac{5}{6} R_{3}\} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} - 1.5 \\ 1.67 \\ 4 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} - 1.5 \\ 1.67 \\ 4 \end{matrix}]$

Thus,we have solved the equations and the values of unknown variables are $x = - 1.5, y = 1.67$ and $z = 4$

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91影视

In Problems 75鈥� 82, algebraically solve each system of equations using any method you wish.
$\{\begin{cases} 2 x + 3 y - z = - 2 \\ 4 x + 3 z = 6 \\ 6 y - 2 z = 2 \end{cases}$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Finding the values of x,y

One App. One Place for Learning.

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91影视

In Problems 75鈥� 82, algebraically solve each system of equations using any method you wish.2x+3y-z=-24x+3z=66y-2z=2

Short Answer

Step by step solution

Step 1. Given information

Step 2. Finding the values of x,y

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Logic and Functions

Probability and Statistics

Applied Mathematics

Statistics

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

In Problems 75鈥� 82, algebraically solve each system of equations using any method you wish.
$\{\begin{cases} 2 x + 3 y - z = - 2 \\ 4 x + 3 z = 6 \\ 6 y - 2 z = 2 \end{cases}$