SOLUTION: a person has 63 coins in pennies,dimes, and quarters with a combined value of $7.65 there are twice as many dimes as pennies. I think one of the equation is x+y=63 but I can

Algebra ->  Matrices-and-determiminant -> SOLUTION: a person has 63 coins in pennies,dimes, and quarters with a combined value of $7.65 there are twice as many dimes as pennies. I think one of the equation is x+y=63 but I can      Log On


   



Question 830966: a person has 63 coins in pennies,dimes, and quarters with a combined value of $7.65 there are twice as many dimes as pennies.
I think one of the equation is x+y=63 but I cant think of the other equations

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
let x be the number of pennies, then 2x is the number of dimes and y is the number of quarters. We now have two equations
x +2x +y = 63
(.01)x +(.10)2x +(.25)y = 7.65
solve first equation for y and substitute in second equations
y = 63 -3x
.01x +.20x +.25(63-3x) = 7.65
.21x +15.75 -.75x = 7.65
-.54x = -8.1
x = 15
therefore we have 15 pennies, 30 dimes, now we can find out the number of quarters
y = 63 - (15 +30)
y = 18
15 pennies, 30 dimes and 18 quarters
.15 + 3 + 4.5 = 7.65
7.65 = 7.65
answer checks :-)