the last post I made about this didn't work or something so yeah sorry for posting twice Melody

Solve the following system:
{n+54 = p
0.05 n+0.01 p = 2.7

In the first equation, look to solve for p:
{n+54 = p
0.05 n+0.01 p = 2.7

n+54 = p is equivalent to p = n+54:
{p = n+54
0.05 n+0.01 p = 2.7

Substitute p = n+54 into the second equation:
{p = n+54
0.05 n+0.01 (n+54) = 2.7

0.05 n+0.01 (n+54) = 0.05 n+(0.01 n+0.54) = 0.06 n+0.54:
{p = n+54
0.06 n+0.54 = 2.7

In the second equation, look to solve for n:
{p = n+54
0.06 n+0.54 = 2.7

0.06 n+0.54 = (3 n)/50+27/50 and 2.7 = 27/10:
(3 n)/50+27/50 = 27/10

Subtract 27/50 from both sides:
{p = n+54
(3 n)/50 = 54/25

Multiply both sides by 50/3:
{p = n+54
n = 36

Substitute n = 36 into the first equation:
{p = 90
n = 36

Collect results in alphabetical order:
Answer: |  n = 36                  and                p=90
 

Okay so Katie you understand how make an system out of this correct?

If so I'll show you my method of solving these types of problems.

Instead of thinking of it as money I think of the pennies and nickels as a value of .01 and .05, respectively. The total amount will take the value of 2.70.

So in order to solve a problem like this you need a system.

Let "p" be pennies and "n" be nickels.

If the total amount of pennies and nickels have a value of 2.70 and there are 54 more pennies than nickels you'll get a system like this:

n + 54 = p

n(0.05) + p(0.01) = 2.70

Now I think I should leave the rest to you. If you don't know systems let me know.

okay thank you I do know systems I just didn't know the formula for this problem thank you that helps

Very nice, HSC.......!!!!!