Agebra 2; system of equations using three variables.
in Algebra 2 Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

2 Answers

l;aslkcs;lkzcx;lkzxc;lck;lacs;lkc;lzcxmlksamklklc xnmlkasnmklcmaslkm .,zMxlk;ashckjdS ;HFSL;SAHLKAS;HKLCH;CL;JHCASHKLAS;HKLCSAHKCAS;KLSACLK;hncnsDLZKJHIUEWHIO;P;lnckjESB FIO;hijacshkljQWUEWQOKLCNWSKISJDJD SKASKSKSKSKS WIOKJaw;HJDKJLQWIUODHCNASXNDS9DSIOJADWKJSHOIYAS098U21IUO12IOU12091209092109209209120912091209210912HJSKLCSH0OH09ADS09AS09ADS09AS09SA0909S90AS09AS09S89DS76AS67AS6A4SASD87897ADS09SAA-DA98765SAD4879DD87SA68D

by
What is the answer to 8x - 6z = 38 ,
2x - 5y + 3z = 5 , x + 10y - 4z = 8?

1)  8x - 6z = 38
2)  2x - 5y + 3z = 5
3)  x + 10y - 4z = 8

Multiply equation two by 2 and add to equation one.
2 * (2x - 5y + 3z) = 5 * 2
  4x - 10y + 6z = 10
+(8x        - 6z = 38)
-----------------------------
  12x - 10y     = 48
4)  12x - 10y = 48

Multiply equation two by 4.
4 * (2x - 5y + 3z) = 5 * 4
5)  8x - 20y + 12z = 20

Multiply equation three by 3.
3 * (x + 10y - 4z) = 8 * 3
6)  3x + 30y - 12z = 24

Add equation six to equation five.
   8x -  20y + 12z = 20
+(3x + 30y -  12z = 24)
---------------------------------
  11x + 10y         = 44
7)  11x + 10y = 44

By adding equation seven to equation four, we
can eliminate y, and solve for x.
  12x -   10y = 48
+(11x + 10y = 44)
--------------------------
  23x          = 92
23x = 92
x = 4 <<<<<<<<<<<<<<<<<<<<<<<<<<<<<

We can plug that into both equation four and
equation seven, solving for y and verifying at
the same time.

Four:
12x - 10y = 48
12(4) - 10y = 48
48 - 10y = 48
-10y = 48 - 48
-10y = 0
y = 0 <<<<<<<<<<<<<<<<<<<<<<<<<<<<<

Seven:
11x + 10y = 44
11(4) + 10y = 44
44 + 10y = 44
10y = 44 - 44
10y = 0
y = 0    same answer for y

We can substitute both of those values into the
original three equations to solve for z. Using
all three equations gives us the chance to verify
the results.

One:
8x - 6z = 38
8(4) - 6z = 38
32 - 6z = 38
-6z = 38 - 32
-6z = 6
z = -1 <<<<<<<<<<<<<<<<<<<<<<<<<<<<<

Two:
2x - 5y + 3z = 5
2(4) - 5(0) + 3z = 5
8 - 0 + 3z = 5
3z = 5 - 8
3z = -3
z = -1     same answer for z

Three:
x + 10y - 4z = 8
(4) + 10(0) - 4z = 8
4 + 0 - 4z = 8
-4z = 8 - 4
-4z = 4
z = -1     and verified again

x = 4, y = 0, z = -1

Why did they have y equal to zero? To cloud your
thinking, to confuse the issue, to see if you would
stay with it or walk away. There are times when
y is zero; it is unavoidable. Accept it.
by Level 11 User (78.4k points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
87,516 questions
100,285 answers
2,420 comments
735,467 users