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Question 1144508: I am trying to help my son with this but am lost there are no videos to help me that I can find his problem is...
3a-b-5c=22
4c=-12-a-2b
2a-19=6b+3c
Found 3 solutions by ikleyn, MathTherapy, greenestamps:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
From the first equation, express b = 3a - 5c - 22 and substitute it into the second and the third equations. Then you will get
4c = -12 - a - 2(3a - 5c - 22) (2') instead of the second equation, and
2a - 19 = 6(3a - 5c -22) + 3c (3') instead of the third equation.
In this way, you just have the system of 2 equations (2') and (3') in 2 unknowns "a" and "c".
Write this system in the standard form
7a - 6c = 32 (4)
16a - 27c = 113 (5)
At this point, you have the system of 2 equations in 2 unknowns in the standard form.
I hope, that from this point your son will be able to complete the solution on his own.
If not, let me know.
ANSWER. a= 2, b= -1, c= -3.
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There is a bunch of lessons in this site on solving systems of linear equations in three unknowns by the Substitution and Elimination methods
- Solving systems of linear equations in 3 unknowns by the Substitution method,
- BRIEFLY on solving systems of linear equations in 3 unknowns by the Substitution method,
- Solving systems of linear equations in 3 unknowns by the Elimination method
- BRIEFLY on solving systems of linear equations in 3 unknowns by the Elimination method
- OVERVIEW of LESSONS on solving systems of linear equations in three unknowns by the Substitution and/or Elimination methods
and by using determinants (Cramer's rule)
- Determinant of a 3x3 matrix
- Co-factoring the determinant of a 3x3 matrix
- HOW TO solve system of linear equations in three unknowns using determinant (Cramer's rule)
- Solving systems of linear equations in three unknowns using determinant (Cramer's rule)
- Solving word problems by reducing to systems of linear equations in three unknowns
Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic
"3x3-Matrices, determinants, Cramer's rule for systems in three unknowns"
Save the link to this textbook together with its description
Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson
into your archive and use when it is needed.
Answer by MathTherapy(10552)  (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
You have two good algebraic solutions from other tutors using substitution; one tutor chose to solve the first equation for b and substitute in equations 2 and 3; another chose to solve for a in the second equation and substitute in equations 1 and 2.
It's good to see two different solutions using the same basic method to show you that there are many ways you could start on the problem. You could even start by solving for c in equation 3 and substituting in equations 1 and 2; but that would introduce ugly fractions in the calculations. The two choices shown by the two tutors are the best choices, if you are going to solve the system using elimination.
The other basic method for solving systems like this is elimination. You should try both methods and see which works better for you. In my work with students, I find that most students find one method or the other much easier for them.
(But my recommendation is to understand -- and be able to use -- both methods.)
To start a solution using elimination, put the three equations in Ax+By+Cz=D form.
Now look to see which variable will be easiest to eliminate first. With the coefficients of b in the three equations being -1, +2, and -6, that looks easiest. So
multiply the first equation by 2 and add to the second to eliminate b:
6a - 2b - 10c = 44
a + 2b + 4c = -12
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7a - 6c = 32
and multiply the second equation by 3 and add to the third to also eliminate b:
3a + 6b + 12c = -36
2a - 6b - 3c = 19
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5a + 9c = -17
Now, looking at the coefficients of c in the two new equations, multiply the first equation by 3 and the second by 2 and add to eliminate c:
21a - 18c = 96
10a + 18c = -34
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31a = 62
So a = 62/31 = 2.
Substituting a=2 in the second of the preceding equations gives 20+18c = -34, which leads to c = -3.
And substituting a=2 and c=-3 in the second original equation gives 2+2b-12 = -12, which leads to b = -1
ANSWER: a=2; b=-1; c=-3.
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