Question

ok so I need to solve this word problem algebraically I already have written out the problem but I'm not sure how to get the answer...

Answer 1

141=a+(47-z)+z+(36-z)

Answer 2

91=b+(36-z)+z+(47-z)

Answer 3

60=c+(36-z)+z+(36-z)

Answer 4

200=a+(47-z)+b+(36-z)+c+(36-z)+z+5

Answer 5

alrighty

Answer 6

first of all, simplify the four equations completely

Answer 7

You don't really have to, but I'd reccommend it to avoid confusion

Answer 8

any of yall got ps3

Answer 9

okay, so I have
141=a+83-z
91=b+83-z
60=c+72-z
200=a+124-2z+b+c

Answer 10

you should simplify the numbers. example: 141 and 83 in the first equation

put all of them in the form
A a + B b +C c + Z z = #
where A, B, C, Z are the coefficients

Answer 11

you end up with this system of equations
1a+ 0 b +0 c -1 z = 58
0 a+ 1 b+0 c - 1 z = 8
0 a +0 b +1c - 1 z = -12
1 a + 1 b+1 c - 2z = 76

if we just list the coefficients we get
1 0 0 -1 58
0 1 0 -1 8
0 0 1 -1 -12
1 1 1 -2 76

to solve, use elimination. The idea is to zero out all entries below the diagonal

the first step is multiply the first row by -1 and add that to the last row to get a new last row.

in other words: -1 * first row= -1 0 0 1 -58
add to the last row 1 1 1 -2 76
to get 0 1 1 -1 18 and you now have

1 0 0 -1 58
0 1 0 -1 8
0 0 1 -1 -12
0 1 1 -1 18

next you want to zero out the leading one in the bottom row.
to do that , multiply the second row by -1: 0 -1 0 1 -8
and add to the last two 0 1 1 -1 18
to get a new last row: 0 0 1 0 10
you now have

1 0 0 -1 58
0 1 0 -1 8
0 0 1 -1 -12
0 0 1 0 10

if a computer were doing this it would use the 3rd row to get rid of the leading one in the bottom row, but we don't have to. remember that the last row is short for

0a + 0b + 1c +0z =10
or simply c=10
now use c=10 in the 3rd row:
c - z = -12 but c=10, so 10 - z = -12
solve for z