Algebra 2
Algebra 2
1st Edition
ISBN: 9780078884825
Author: McGraw-Hill/Glencoe
Publisher: Glencoe/McGraw-Hill School Pub Co
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Chapter 3.5, Problem 11PPS
To determine

To calculate: The solution set of system of equations,

  4r+6s−t=−183r+2s−4t=−24−5r+3s+2t=15

Expert Solution & Answer
Check Mark

Answer to Problem 11PPS

The solution set of the system of equations is (−2,−1,4).

Explanation of Solution

Given information:

The system of equations is provided below,

  4r+6s−t=−183r+2s−4t=−24−5r+3s+2t=15

Formula used:

A system of equations with three variables can be solved by elimination method and substitution method. Select any two pairs of equations from the system and eliminate the same variable by adding or subtracting and then simplify it.

Calculation:

Consider the provided system of equations as,

  4r+6s−t=−183r+2s−4t=−24−5r+3s+2t=15

Recall that a system of equations with three variables can be solved by elimination method and substitution method. Select any two pairs of equations from the system and eliminate the same variable by adding or subtracting and then simplify it.

So, multiply the third equation by 2 and then add resulting and the second equation to eliminate t .

  2(−5r+3s+2t)=2(15)−10r+6s+4t=30

Now add the resulting equation and the second equation as,

  −10r+6s+4t=30(+)3r+2s−4t=−24−7r+8s=6

Now, multiply first equation by 2 and add resulting equation and third equation to eliminate x ,

  2(4r+6s−t)=2(−18)8r+12s−2t=−36

Add resulting equation and the third equation as,

  8r+12s−2t=−36(+)−5r+3s+2t=153r+15s=−21

Multiply 3r+15s=−21 by 7 and −7r+8s=6 by 3 and add resulting equations to eliminate r as,

  7(3r+15s)=7(−21)21r+105s=−147

  3(−7r+8s)=3(6)−21r+24s=18

Add both the equations and solve to find the value of s ,

  21r+105s=−147(+)−21r+24s=18129s=−129

Simplify 129s=−129 as,

  129s=−129s=−129129s=−1

Substitute s=−1 in −7r+8s=6 to get the value of r ,

  −7r+8(−1)=6−7r−8=6−7r=14r=−2

Now, substitute r=−2 and s=−1 in equation second to get the value of t as,

  3(−2)+2(−1)−4t=−24−6−2−4t=−24−4t=−16t=4

To check whether the solution set is correct, substitute the values of r , s and t in first equation as,

  4r+6s−t=−184(−2)+6(−1)−4=−18−8−6−4=−18−18=−18

Since, left-hand side is equal to right-hand side, the solution set is true for all the equations.

Thus, the solution set of the system of equations is (−2,−1,4).

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Chapter 3.5, Problem 10PPS
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Chapter 3.5, Problem 12PPS

Chapter 3 Solutions

Algebra 2

Ch. 3.1 - Prob. 1ACYPCh. 3.1 - Prob. 1BCYPCh. 3.1 - Prob. 2ACYPCh. 3.1 - Prob. 2BCYPCh. 3.1 - Prob. 3CYPCh. 3.1 - Prob. 4ACYPCh. 3.1 - Prob. 4BCYPCh. 3.1 - Prob. 4CCYPCh. 3.1 - Prob. 4DCYPCh. 3.1 - Prob. 1CYU
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Prob. 30HPCh. 3.5 - Prob. 31HPCh. 3.5 - Prob. 32HPCh. 3.5 - Prob. 33HPCh. 3.5 - Prob. 34STPCh. 3.5 - Prob. 35STPCh. 3.5 - Prob. 36STPCh. 3.5 - Prob. 37STPCh. 3.5 - Prob. 38STPCh. 3.5 - Prob. 39SCh. 3.5 - Prob. 40SCh. 3.5 - Prob. 41SCh. 3.5 - Prob. 42SCh. 3 - Prob. 1QCCh. 3 - Prob. 2QCCh. 3 - Prob. 3QCCh. 3 - Prob. 4QCCh. 3 - Prob. 5QCCh. 3 - Prob. 6QCCh. 3 - Prob. 7QCCh. 3 - Prob. 8QCCh. 3 - Prob. 9QCCh. 3 - Prob. 10QCCh. 3 - Prob. 11QCCh. 3 - Prob. 12QCCh. 3 - Prob. 13QCCh. 3 - Prob. 14QCCh. 3 - Prob. 1MCQCh. 3 - Prob. 2MCQCh. 3 - Prob. 3MCQCh. 3 - Prob. 4MCQCh. 3 - Prob. 5MCQCh. 3 - Prob. 6MCQCh. 3 - Prob. 7MCQCh. 3 - Prob. 8MCQCh. 3 - Prob. 9MCQCh. 3 - Prob. 10MCQCh. 3 - Prob. 11MCQCh. 3 - Prob. 12MCQCh. 3 - Prob. 13MCQCh. 3 - Prob. 14MCQCh. 3 - Prob. 15MCQCh. 3 - Prob. 16MCQCh. 3 - Prob. 17MCQCh. 3 - Prob. 18MCQCh. 3 - Prob. 19MCQCh. 3 - Prob. 20MCQCh. 3 - Prob. 21MCQCh. 3 - Prob. 22MCQCh. 3 - Prob. 1SGRCh. 3 - Prob. 2SGRCh. 3 - Prob. 3SGRCh. 3 - Prob. 4SGRCh. 3 - Prob. 5SGRCh. 3 - 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