Lizzie has 30 coins that total $4.80. all of her coins are dimes, d, and quarters, q. algebraically determine how many of each type of coin did lizzie have?

Lizzie has 18 dimes and 12 quarters

Solution:

Let "d" be the number of dimes

Let "q" be the number of quarters

We know that,

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

Given that LIzzie has 30 coins

number of dimes + number of quarters = 30

d + q = 30 ---- eqn 1

Also given that the coins total $ 4.80

number of dimes x  value of 1 dime + number of quarters x  value of 1 quarter = 4.80

0.1d + 0.25q = 4.8 ------ eqn 2

Let us solve eqn 1 and eqn 2

From eqn 1,

d = 30 - q ---- eqn 3

Substitute eqn 3 in eqn 2

0.1(30 - q) + 0.25q = 4.8

3 - 0.1q + 0.25q = 4.8

0.15q = 1.8

From eqn 3,

d = 30 - 12

Thus she has 18 dimes and 12 quarters

$32,660

step-by-step explanation:

damm bro the struggle

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