Value of coins.....................

Pleas someone help me to solve this problem!!!! :'(

Larry has 25 coins consist of only Nicole's,n,and the quarter,q.the total value of these coins is$ 3.45 how many of each coin does he have??

Heheh at "Nicole's". I assume that's "nickels".

Q + N = 25
25Q + 5N = 345 (I'm using cents instead of dollars because it's easier to interpret)

First one can be rewritten as Q = -N + 25. Substitute that into the second equation, since we know they're equal.

25Q + 5N = 345
25(-N + 25) + 5N = 345
-25N + 625 + 5N = 345
-20N + 625 = 345
-20N = -280
N = 14

That means Q = 11.

Check: 11 quarters is $2.75. 14 nickels is $0.70. Add those together and we get 25 coins worth $3.45, which is right.

So .25q+.05n=3.45. This means that 25 cents times the number of quarters plus 5 cents times the number of nickels is equal to $3.45.
You also have the equation q+n=25 because the total number of coins is 25.
Now, if you find q in terms of n you get n=25-q.
Next, plug this back into the first equation for n and solve for q:
0.25q+0.05(25-q)=3.45
0.25q+1.25-0.05q=3.45
0.20q=2.20
q=11
So you have 11 quarters. Plus, you know that you have 25 coins total so you must have 14 nickels!
Quarters=11
Nickels=14

assuming you mean nickles, not Nicole's:

Equation 1: n + q = 25 ----> re-arrange to n = 25-q
Equation 2: 0.05n + 0.25q = 3.45

Sub eq 1 into eq 2:

0.05 (25-q) + 0.25q = 3.45
1.25 - 0.05q + 0.25q = 3.45
0.2q = 2.2 ----> multiply both sides by 5
q = 11
Sub q = 11 into eq 1:
n = 25- (11)
n = 14
Therefore 11 quarters, 14 nickels

n +q = 25 ==> n = 25-q
5n + 25q = 345
==> 5 (25-q) +25q =345
125 +20q = 345
20q = 220
q = 11
n = 25-11 = 14

n+q= 25 .05n + .25q = $3.45