money math problem?

Question

A collection 17 coins of consisting of nickels and dimes is worth $1.15. How many dimes are there?

Answer 1

let
x = # of nickles
y = # of dimes

...collection 17 coins ...
x + y = 17
x = 17 - y (eq 1)

...is worth $1.15...
0.05x + 0.10y = 1.15
5x + 10y = 115
x + 2y = 23 (eq 2)

plug eq 1 into eq 2
x + 2y = 23
(17 - y) + 2y = 23
17 + y = 23
y = 23 - 17
y = 6

plug this into eq 1
x + 2y = 23
x + 2(6) = 23
x + 12 = 23
x = 23 - 12
x = 11

Answer 2

x = # of dimes y = # nickels

x + y = 17 where x = number of dimes and y = number of nickels. Then

x*$0.10 + y*$.05 = $1.15 or 10x +5y = 115 (multiply through by 100)

with x = (17 - y) from the first equation, the second equation becomes

(17-y)*10 + y*5 = 115 ==> 170 - 10y + 5y = 115 or

5y = 55 ==> y = 11 and then x = 6

A bit clumsy. There may be a more direct way

Answer 3

nickels plus dimes = 17 coins
n + d = 17
5n + 10d = 115 (this is in cents; each nickel is worth five cents, so you have 5 times as much cents as nickels)

n+d=17 is the same as n = 17-d
substitute that into the second equation

5(17-d) +10d = 115
85 - 5d + 10d = 115
5d = 30
d = 6 (six dimes)
17-6 = 11 nickels

Check
6*10 + 11*5 = 60 + 55 = 115
checks!

6 dimes and 11 nickels

Answer 4

6

Answer 5

There are 6 dimes, 11 nickles.

Answer 6

.05 x + .10 (17-x) = 1.15
.05 x + 1.70 - 0.10 x = 1.15
-0.05 x + 1.70 = 1.15
0.55 = 0.05 x
x = 11 .05 x 11 = 0.55 and 1.15 - 0.55 = 0.60
y = 17-11=6

11 nickels and 6 dimes

Answer 7

let n = number of nickels
let d = number of dimes

n + d = 17
.05n + .10d = 1.15 (OR 5n + 10d = 115)
NOW solve the equations using elimination method
-10(n+d=17) becomes
-10n-10d=-170
5n+10d=115
THEREFORE
-5n=-55
n=11
ANSWER 11 nickels, 6 dimes

Answer 8

The 6 - 5- a million answer is obviously marvelous and that i exchange into able to remedy it via "bull artwork" -- yet ought to a pair mathematician obtainable instruct us the elegant formulation? How do you remedy for 3 unknowns with in basic terms 2 equations?