Question

Need to find out how much each box of fruit weighs, and do not know how to start:
A+B=48
B+C=54
C+D=46
D+E=32
A+C+E=64

Answer 1

There are several ways of approaching a large system of equations. Do you know how to reduce matrices into reduced row echelon form?

Answer 2

You can also solve the system by successive addition and subtraction if you noticed that the variables are chained in a near diagonal form.
The answers turned out to be all even (by chance, I guess).

Answer 3

I really don't know how to even start thinking....
I don't know the rules. For instance...If I take one part of the above problem one step further and say B=48-A and B=C-54. That means that 48-A=C-54
Then I can say C-A=6. My question then is...Could I then draw the conclusion that B=C-A and B=6...? Or am I waaaay out of order?

Answer 4

I'm confused...
B=48-A
B=C-54
So, 48-A = C - 54
Then, it should be A+C = 48+54, ie, A+C = 102...

Answer 5

Reduction to the echelon form is the standard solution, perhaps a little more work for a 5x5 system.

I will start you off with successive elimination (that does not apply in all cases, only cases similar to this one).

A+B =48 ...(1)
B+c =54 ... (2)
Subtract
A -c = -6 (1A)
C+D=46 ...(3)
Add
A +D=40 ...(2a)
...
Continue this way until you get
A-E=8 ....(3A)
add to
A+C+E=64 ...(5)
to get
2A+C=72
Add to (1A) and solve for A.
Substitute A in (1a), (2a), (3a).... to get remaining variables.

Answer 6

use matrices! they're so much cleaner!

Answer 7

And... wait!
B+C = 54
B = 54 - C. not C-54

In that cause
B=48-A
B=54-C
So, 48 - A = 54-C
A-C = 48 - 54
A-C = -6
In other words, C-A = 6 , that make sense now :|

Answer 8

Thank you so much for your responses! Still a bit confused about the rules, but will work on this :0) Thanks again! And Mathmate, thank you so much for the detailed answer, will try to work through that!

Answer 9

You're welcome! :)

Answer 10

I'm thinking if this way works.
A+B=48 -(1)
B+C=54 -(2)
C+D=46 -(3)
D+E=32 -(4)
A+C+E=64 -(5)

(1)+(2)+(3)+(4)
A+E + 2(B+C+D) = 48+54+46+32
(64-C) + 2(B+C+D) = 180

From (2) , B = 54 - C
From (3), D = 46 - C

So,

(64-C) + 2 [(54-C) +C+ (46-C)] = 180
(64-C) + 2 (100-C) = 180
^ You can solve C from here. Then, solve the equations one by one to get the other values for other unknowns.

Answer 11

Thank you Callisto! Callisto and Mathmate, I will for sure come back to see you! Now I need to go and work on this! You're the best!