Another math problem help please?

Question

A parking meter takes only dimes and quarters. at the end of the day, the number of dimes in the meter is eight more than twenty times the number of quarters. the value of the coins in the meter is $7.55. find the number of quarters and dimes in the meter. please show a step by step solving method using a variable expression representing a quantity, thanks

Answer 1

let x be the dime and y the quarters

x=20y+8 acc to the equation

so x-20y=8---------------------------------...

if there is a total of 47.55 that means the value of dimes and quarters is $7.55

so 10x+25y=7.55 or 755 (in cents)----------------(2)

set up both equations and solve simultaneously

10(x-20y)=10(8)

10x+25y=755

so 10x-200y=80

     10x+25y=755     minus both

______________________

       -225y= -675

            y=3

and then find x

Answer 2

0.1a + 0.25b = 7.55

a=20b+8

Substitute:

0.1(20b+8) +0.25b = 7.55

2b+0.8+0.25b=7.55

2.25b = 6.75

b=3

Replace b with 3

a=20(3)+8

a=68

Therefore 68 dimes and 3 quarters are in the meter.

Answer 3

First you know that:

$7.55=.1d+.25q (d= # of dimes, q= # of quarters)

20q+8=d

From there, I guessed and checked.

20(1)+8=d

28dimes, 1 quarter X

20(2)+8=d

48 dimes 2 quarters X

20(3)+8=d

68 dimes 3 quarters= $7.55 (=