Chapter 11: Q. 78 (page 757)

In Problems 75– 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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In Problems 75– 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

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In Problems 75– 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

Applied Mathematics

Logic and Functions

Statistics

Pure Maths

Decision Maths

Geometry

Study anywhere. Anytime. Across all devices.

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