algebra help!?

This is a system of equations. Your two equations are:

number of quarters + number of dimes = 20

q + d = 20

and

.25 per quarter plus .10 per dime equals 3.05

.25q + .10d = 3.05

so now, you solve for one variable, and substitute that in the other equation, solve for the variable

q + d = 20

d = 20 - q

.25q + .10(20 - q)= 3.05

.25q + 2 + -.1q = 3.05

2 + .15q = 3.05

.15q = 1.05

q = 7

There are seven quarters.

Now substitute that to find the number of dimes.

q + d = 20

(7) + d = 20

d = 20-7

d = 13

There are 13 dimes.

Now check:

.25(7) + .10(13) = 3.05

1.75 + 1.30 = 3.05

3.05 = 3.05

[1]

Let the no of quarters be x and therefore the no of dimes are 20-x

By the problem,

0.25x+0.10(20-x)=3.05

0.25x+2-0.10x=3.05

0.15x=3.05-2=1.05

x=1.05/0.15=7

Therefore,no of quarters is 7 and the no of dimes is 20-7=13

let q=quarter d=dime

q + d = 1/4 + 1/10 = 0.35 =====> q = 0.35 - d

Let a be number of quarter and b be number of dimes

a+b = 20 ==========> a = 20 - b

and

aq + bd =3.05

Solve it

x=number of quarters, y=number of dimes

.25x+.1y=3.05

x+y=20

solve these two equations

x=20-y

.25(20-y)+.1y=3.05

5-.25y+.1y=3.05

-.15y=-1.95

**y=13

x=20-y

x=20-(13)

**x=7

so 7 quarters, and 13 dimes

13 dimes and 7 quarters. please let this be the best answer!