Answer:
$0.18
Step-by-step explanation:
we need to set up two simultaneous equations
using variables, pencils = p and erasers = e
3 pencils and 2 erasers, the cost is $0.76
3p + 2e = $0.76
2 pencils and 4 erasers, the cost is $1.04.
2p + 4e = $1.04
we now have
3p + 2e = $0.76
2p + 4e = $1.04
to make the number of erasers the same, multiply the first equation by 2 to give 4e
2(3p + 2e = $0.76)
6p + 4e = $1.52
now we have the same number of erasers for both equations
6p + 4e = $1.52
2p + 4e = $1.04
subtract across: 6p - 2p = 4p, 4e - 4e = 0, $1.52 - $1.04 = $0.48
we are left with 4p = $0.48
divide both sides by 4
p = $0.12
1 pencil = $0.12
go back to the start of both equations and use one of them to find 1 eraser. I'll use 3p + 2e = $0.76
input $0.12 in p
3($0.12) + 2e = $0.76
$0.36 + 2e = $0.76
subtract $0.36 on both sides
2e = $0.76 - $0.36
2e = $0.40
divide 2 on both sides
e = $0.20
1 eraser = $0.20
How much more does 1 eraser cost than 1 pencil?
we now know 1 pencil = $0.12 and 1 eraser = $0.20
find the difference between them
$0.20 - $0.12 = $0.18
final answer= $0.18
