Suppose there is a pile of quarters, dimes, and pennies with a total value of $1.08 A) How much of each coin can be present without being able to make change for a dollar?B) Explain why $1.19 is the greatest amount of money it is possible to have without being able to make change for a dollar.

a)

if 1 quarter = $ 0.25

  1 dime = $ 0.10

  1 penny = $ 0.01

so to make the total of $1.08 and its is also required to calculate the number of each coins present without being able to make change for a dollar

therefore we say;

1 Quarter + 8 dimes + 3 penny

: ( 1 × 0.25 ) + ( 8 × 0.10 ) + ( 3 × 0.01 )

: 0.25 + 0.80 + 0.03 = $ 1.08

b)

Now if you have a 4 Quarters, you have change for $1.

If you have 5 dimes, you have change for 2 Quarters.

If you have nickel; one of those can combine with 2 dimes to have a change for a Quarter.

If you have 5 pennies, you have enough change for 1 nickel

Therefore

(4-1)×0.25 + (5-1)×0.1 + (0×0.05) + (5-1)×0.01 = x

(3 × 0.25) + ( 4 × 0.1) + 0 + ( 4 × 0.01) = x

x = 0.75 + 0.4 + 0.04

x = $ 1.19

PROVED

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.