Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

In exercises, 1 through 10, findall solutions of the linear systems using elimination.T hen check your solutions.
10. $\begin{array}{rcl} | \begin{matrix} x + 2 y + 3 z = 1 \\ 2 x + 4 y + 7 z = 2 \\ 3 x + 7 y + 11 z = 8 \end{matrix} | \end{array}$

The solution of system of equation is $x = - 9, y = 5, z = 0$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:Transforming the system

To get the solution, we will transform the value of x,y,z.

$|x + 2 y + 3 z = 1 . . . . . . (1) 2 x + 4 y + 7 z = 2 . . . . . . (2) 3 x + 7 y + 11 z = 8 . . . . . . (3)|$ Into the form $|x = . . . . y = . . . . z = . . .|$

Step 2: Eliminating the variables

In the given system of equations, we can eliminate the variables by adding or subtracting the equations.

In this system, we can eliminate the variable x from equation 2 and 3. First of all subtract the 2 times of equation 1 from equation 2 and then subtract the 3 times of first equation from equation 3.

$\begin{array}{rcl} |x + 2 y + 3 z = 1 2 x - 2 x + 4 y - 4 y + 7 z - 6 z = 0 3 x - 3 x + 7 y - 6 y + 11 z - 9 z = 5| & = & |x + 2 y + 3 z = 1 z = 0 y + 2 z = 5| \end{array}$

Now put the value of z which is in second equation to the first and second equation.

$\begin{array}{rcl} |x + 2 (y) + 3 (0) = 1 z = 0 y + 2 (0) = 5| & = & |x = 1 - 2 y z = 0 y = 5| = |x = - 9 z = 0 y = 5| \end{array}$

Hence, we get value of $x = - 9, y = 5, z = 0$

Step 3: Checking the solution

Now check the solution by putting the value of x, y and z in the given system of equation.

$- 9 + 2 (5) + 3 (0) = 1 2 (- 9) + 4 (5) + 7 (0) = 2 3 (- 9) + 7 (5) + 11 (0) = 8$

Which is true.

Hence, the solution of system of equation is $x = - 9, y = 5, z = 0$

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Linear Algebra With Applications

Answers without the blur.

Short Answer

In exercises, 1 through 10, findall solutions of the linear systems using elimination.T hen check your solutions.
10. $\begin{array}{rcl} | \begin{matrix} x + 2 y + 3 z = 1 \\ 2 x + 4 y + 7 z = 2 \\ 3 x + 7 y + 11 z = 8 \end{matrix} | \end{array}$

Step by Step Solution

Step 1:Transforming the system

Step 2: Eliminating the variables

Step 3: Checking the solution

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Linear Algebra With Applications

Answers without the blur.

Short Answer

In exercises, 1 through 10, findall solutions of the linear systems using elimination.T hen check your solutions.10.|x+2y+3z=12x+4y+7z=23x+7y+11z=8|

Step by Step Solution

Step 1:Transforming the system

Step 2: Eliminating the variables

Step 3: Checking the solution

Most popular questions for Math Textbooks

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Pure Maths

In exercises, 1 through 10, findall solutions of the linear systems using elimination.T hen check your solutions.
10. $\begin{array}{rcl} | \begin{matrix} x + 2 y + 3 z = 1 \\ 2 x + 4 y + 7 z = 2 \\ 3 x + 7 y + 11 z = 8 \end{matrix} | \end{array}$