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 9PPS
To determine

To calculate: The solution set of system of equations,

  4a+5b−6c=2−3a−2b+7c=−15−a+4b+2c=−13

Expert Solution & Answer
Check Mark

Answer to Problem 9PPS

The solution set of the system of equations is (−3,−2,−4).

Explanation of Solution

Given information:

The system of equations is provided below,

  4a+5b−6c=2−3a−2b+7c=−15−a+4b+2c=−13

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,

  4a+5b−6c=2−3a−2b+7c=−15−a+4b+2c=−13

  4a+5b−6c=2−3a−2b+7c=−15−a+4b+2c=−13

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 4 and then add resulting equation and the first equation to eliminate a .

  4(−a+4b+2c)=4(−13)−4a+16b+8c=−52

Now add this equation and the first equation as,

  −4a+16b+8c=−52(+)4a+5b−6c=221b+2c=−50

Now, multiply third equation by −3 and add resulting equation and second equation to eliminate a .

  −3(−a+4b+2c)=−3(−13)3a−12b−6c=39

Add second equation and the resulting equation as,

  3a−12b−6c=39(+)−3a−2b+7c=−15−14b+c=24

Multiply −14b+c=24 by −2 and then add resulting equation and 21b+2c=−50 to eliminate c as,

  −2(−14b+c)=−2(24)28b−2c=−48

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

  21b+2c=−50(+)28b−2c=−4849b=−98

Simplify 49b=−98 as,

  49b=−98b=−9849b=−2

Substitute b=−2 in −14b+c=24 to get the value of c ,

  −14(−2)+c=2428+c=24c=24−28c=−4

Now, substitute b=−2 and c=−4 in equation second to get the value of a as,

  −3a−2(−2)+7(−4)=−15−3a+4−28=−15−3a=9a=−3

To check whether the solution set is correct, substitute the values of a , b and c in first equation as,

  4a+5b−6c=24(−3)+5(−2)−6(−4)=2−12−10+24=22=2

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 (−3,−2,−4).

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Chapter 3.5, Problem 8PPS
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Chapter 3.5, Problem 10PPS

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
Ch. 3.1 - Prob. 2CYUCh. 3.1 - Prob. 3CYUCh. 3.1 - Prob. 4CYUCh. 3.1 - Prob. 5CYUCh. 3.1 - Prob. 6CYUCh. 3.1 - Prob. 7CYUCh. 3.1 - Prob. 8CYUCh. 3.1 - Prob. 9CYUCh. 3.1 - Prob. 10CYUCh. 3.1 - Prob. 11CYUCh. 3.1 - Prob. 12CYUCh. 3.1 - Prob. 13PPSCh. 3.1 - Prob. 14PPSCh. 3.1 - Prob. 15PPSCh. 3.1 - Prob. 16PPSCh. 3.1 - Prob. 17PPSCh. 3.1 - Prob. 18PPSCh. 3.1 - Prob. 19PPSCh. 3.1 - Prob. 20PPSCh. 3.1 - Prob. 21PPSCh. 3.1 - Prob. 22PPSCh. 3.1 - Prob. 23PPSCh. 3.1 - Prob. 24PPSCh. 3.1 - Prob. 25PPSCh. 3.1 - Prob. 26PPSCh. 3.1 - Prob. 27PPSCh. 3.1 - Prob. 28PPSCh. 3.1 - Prob. 29PPSCh. 3.1 - Prob. 30PPSCh. 3.1 - Prob. 31PPSCh. 3.1 - Prob. 32PPSCh. 3.1 - Prob. 33PPSCh. 3.1 - Prob. 34PPSCh. 3.1 - Prob. 35PPSCh. 3.1 - Prob. 36PPSCh. 3.1 - Prob. 37PPSCh. 3.1 - Prob. 38PPSCh. 3.1 - Prob. 39PPSCh. 3.1 - Prob. 40PPSCh. 3.1 - Prob. 41PPSCh. 3.1 - Prob. 42HPCh. 3.1 - Prob. 43HPCh. 3.1 - Prob. 44HPCh. 3.1 - Prob. 45HPCh. 3.1 - Prob. 46HPCh. 3.1 - Prob. 47HPCh. 3.1 - Prob. 48HPCh. 3.1 - Prob. 49HPCh. 3.1 - Prob. 50HPCh. 3.1 - Prob. 51STPCh. 3.1 - Prob. 52STPCh. 3.1 - Prob. 53STPCh. 3.1 - Prob. 54STPCh. 3.1 - Prob. 55STPCh. 3.1 - Prob. 56STPCh. 3.1 - Prob. 57STPCh. 3.1 - Prob. 58STPCh. 3.1 - Prob. 59STPCh. 3.1 - Prob. 60STPCh. 3.1 - Prob. 61SCh. 3.1 - Prob. 62SCh. 3.1 - Prob. 63SCh. 3.1 - Prob. 64SCh. 3.1 - Prob. 65SCh. 3.1 - Prob. 66SCh. 3.1 - Prob. 1ECh. 3.1 - Prob. 2ECh. 3.1 - Prob. 3ECh. 3.1 - Prob. 4ECh. 3.1 - Prob. 5ECh. 3.1 - Prob. 6ECh. 3.1 - Prob. 7ECh. 3.1 - Prob. 8ECh. 3.1 - Prob. 9ECh. 3.1 - Prob. 10ECh. 3.2 - Prob. 1ACYPCh. 3.2 - Prob. 1BCYPCh. 3.2 - Prob. 1CCYPCh. 3.2 - Prob. 2ACYPCh. 3.2 - Prob. 2BCYPCh. 3.2 - Prob. 2CCYPCh. 3.2 - Prob. 3CYPCh. 3.2 - Prob. 4ACYPCh. 3.2 - Prob. 4BCYPCh. 3.2 - Prob. 1CYUCh. 3.2 - Prob. 2CYUCh. 3.2 - Prob. 3CYUCh. 3.2 - Prob. 4CYUCh. 3.2 - Prob. 5CYUCh. 3.2 - Prob. 6CYUCh. 3.2 - Prob. 7CYUCh. 3.2 - Prob. 8CYUCh. 3.2 - Prob. 9CYUCh. 3.2 - Prob. 10CYUCh. 3.2 - Prob. 11CYUCh. 3.2 - Prob. 12CYUCh. 3.2 - Prob. 13CYUCh. 3.2 - Prob. 14CYUCh. 3.2 - Prob. 15PPSCh. 3.2 - 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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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