Help me!

you have two equations in four unknowns, but you have a constraint all numbers of quarters,numbers of Dimes, Number of Nickles, and Numbers of pennys must either be zero or a positives integer.
If a,b,c, &d represent the numbers of quarters,numbers of Dimes, Number of Nickles, and Numbers of pennys respectively the two equations become
a+b+c+d=48
and
25a+10b+5c+d=100

Since a has the least variability start by setting a=0
so
b+c+d=48
and
10b+5c+d=100
eliminating d these equations become
9b+4c=52
or
c=13-9b/4
Moreover since c is positive b<7
and b has to be evenly divisible by 4
for b ranging from 0 to 6 the only b's divisible by 4 are b=0 and b=4

Try the first of a=0 & b=0 then c=13 and d=35
check 0+0+13+35=48
25*0+10*0+5*13+35=100
so this is a solution

Now try b=4, then c=4 & d=40
check 0+4+4+40=48
25*0+10*4+5*4+40=100

Now try a=1
b+c+d=47
10b+5c+d=75
9b+4c=28
There just aren't enough coins in a dollar to have the 48 coins. So there is no solution with any quarters and a has to be equal to zero.

So there are two solutions

no quarters, no dimes, 13 nickels, and 35 cents
and
no quarters, four dimes, four nickels, and 40 cents.

Rap

Been thinking of more solutions.

if the number of quarters (a) equal 1 then

b+c+d=47
10b+5c+d=75
and
9b+4c=28 or c=7-9b/4
so b<3 and divisible by zero--so b=0 & c=7
d=40
check
1+0+7+40=48
25*1+10*0+7*5+40=100

now try a=2
b+c+d=46
b+c+d=50
9b+4c=4 or c=1-9b/4 or b<1 and divisible by 4
so a=2, b=0, c=1 and d=45
check
2+0+1+45=58
25*2+0*10+1*5+45=100

now try a=3
b+c+d=45
10b+5c+d=25
9b+4c=-20
now there is no solution

So
1 quarter, seven nickles, and 40 cents is also a solution
&
2 quarters, 1 nickle and 45 cents is a solution

so there are four possible solution.

Rap

In matrix form the solution can be expressed as

|0_0_13_35||1|
|0_4__4_40||1|
|1_0__7_40||1|=48
|2_0__1_45||1|

|0_0_13_35||25|
|0_4__4_40||10|
|1_0__7_40||05|=100
|2_0__1_45||01|

Then there's one solution, but it's a matrix.

Rap