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 16PPS
To determine

To calculate: The solution set of system of equations,

  −9a+3b−2c=618a+7b+5c=−1385a−5b+8c=−45

Expert Solution & Answer
Check Mark

Answer to Problem 16PPS

The solution set of the system of equations is (−8,−7,−5).

Explanation of Solution

Given information:

The system of equations is provided below,

  −9a+3b−2c=618a+7b+5c=−1385a−5b+8c=−45

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,

  −9a+3b−2c=618a+7b+5c=−1385a−5b+8c=−45

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 7 and second equation by 5 and then add both the resulting equations to eliminate b .

  7(5a−5b+8c)=7(−45)35a−35b+56c=−315

  5(8a+7b+5c)=5(−138)40a+35b+25c=−690

Now add both the resulting equations as,

  40a+35b+25c=−690(+)35a−35b+56c=−31575a+81c=−1005

Now, multiply first equation by 5 and third equation by 3 and add resulting equations to eliminate b ,

  5(−9a+3b−2c)=5(61)−45a+15b−10c=305

  3(5a−5b+8c)=3(−45)15a−15b+24c=−135

Add both the resulting equations as,

  −45a+15b−10c=305(+)15a−15b+24c=−135−30a+14c=170

Multiply 75a+81c=−1005 by 2 and −30a+14c=170 by 5 and add resulting equations to eliminate a as,

  2(75a+81c)=2(−1005)150a+162c=−2010

  5(−30a+14c)=5(170)−150a+70c=850

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

  150a+162c=−2010(+)−150a+70c=850232c=−1160

Simplify 232c=−1160 as,

  232c=−1160c=−1160232c=−5

Substitute c=−5 in −30a+14c=170 to get the value of a ,

  −30a+14(−5)=170−30a−70=170−30a=240a=−8

Now, substitute a=−8 and c=−5 in equation second to get the value of b as,

  8(−8)+7b+5(−5)=−138−64+7b−25=−1387b=−49b=−7

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

  −9a+3b−2c=61−9(−8)+3(−7)−2(−5)=6172−21+10=6161=61

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 (−8,−7,−5).

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Chapter 3.5, Problem 15PPS
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Chapter 3.5, Problem 17PPS

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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