Question 1036738
five baskets contain coconuts. the first and second 
baskets together have a total of 54 coconuts. the 
second and third baskets have 38 coconuts. the third 
and fourth baskets have 28 coconuts. the fourth and 
fifth baskets have 25 coconuts. and the first and 
fifth baskets have 49 coconuts. How many coconuts 
are in each basket?
<pre><b>
Notice that I have changed the number of coconuts in
the baskets.  I did that so I wouldn't be doing your
work for you.  You do your problem the exact same way.
Use this as a model:

Five baskets contain coconuts. 

A = number of coconuts in the 1st basket.
B = number of coconuts in the 2nd basket.
C = number of coconuts in the 3rd basket.
D = number of coconuts in the 4th basket.
E = number of coconuts in the 5th basket.

the first and second baskets together have a total of 54 coconuts. 
A + B = 54

the second and third baskets have 38 coconuts. 

    B + C = 38


the third and fourth baskets have 28 coconuts. 

        C + D = 28

the fourth and fifth baskets have 25 coconuts. 

            D + E = 25

and the first and fifth baskets have 49 coconuts. 

A             + E = 49

So the system of equations is:
 
(1)  A + B             = 54
(2)      B + C         = 38
(3)          C + D     = 28
(4)              D + E = 25
(5)  A             + E = 49

Solve (5) for A:  A = 49-E, substitute in (1)

49-E + B = 54.  Solve for B: B = 5+E.  Substitute in (2)

 5+E + C = 38.  Solve for C: C = 33-E. Substitute in (3)

33-E + D = 28.  Solve for D: D = E-5. Substitute in (4)

E-5 + E = 25.  Solve for E: 2E = 30, E = 15

Substitute E=15 in D = E-5 = 15-5 = 10
Substitute E=15 in C = 33-E = 33-15 = 18
Substitute E=15 in B = 5+E = 5+15 = 20
Substitute E=15 in A = 49-E = 49-15 = 34

Now use the above to solve your problem the exact same way.  

Edwin</pre>