help with this algebra question?

Question

Jo has 37 coins (all nickels, dimes, and quarters) worth $5.50. She has 4 more quarters than nickels. How many DIMES does Jo have? Use only one variable

Answer 1

1 Nickel = 5 c
1 Dime = 10 c
1 Quarter = 25 c
5.50$ = 550 c

Assume number of nickels = X

Then number of quarters = X + 4

Then number of dimes = total number of coins – (number of nickels + number of quarters) = 37 – (X + X + 4) = 33 – 2X

number of dimes * dime value + number of nickels * nickel value + number of quarters * quarter value = Total value

(33 – 2X) * 10 + X * 5 + (X + 4) * 25 = 550
330 – 20X + 5X + 25X + 100 = 550
10X + 430 = 550
10X = 550 – 430
10X = 120
X = 120 / 10
X = 12

Number of nickels = X = 12
Number of dimes = 33 – 2X = 33 – 2 * 12 = 9
Number of quarters = X + 4 = 16

12 * 5c + 9 * 10c + 16 * 25c = 60c + 90c + 400c = 550 c = 5.50 $

12 + 9 + 16 = 37 coins

Answer: Jo has 9 dimes :-)

Answer 2

hi using only one variable trial and error method is best
since we need number of dimes lets assume dimes = x
from given data for example if nickles=1 then quarters=5 then dimes will be 31 the value of these will be $4.40 hence this not correct
assuming nickles = 2 , dimes=x quarters=6 value will be $4.50
which is also not correct, now if we observe carefully when we increased number of nickles by 1 the value of all the coins is increasing by 10 cents, now we are short of our target value of the coins $5.50 by a $1.0 which means 10 x 10cents
so lets add 10 to the last value of our assumed number of nickles
that is 10+2=12 nickles so quarters will be 16 ; hence dimes will be 9 and the total value now will be $5.50
so total number of dimes is 9

Answer 3

For increasing cubic binomials the overall formula is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^a million + 3*a^a million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^a million + 3*x^a million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D