SOLUTION: a jar contains 53 coins. all of which are nickels and dimes. the total value of coins in the car is 4.25 how many of each coin do they have you solve the problem like so x+y=

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: a jar contains 53 coins. all of which are nickels and dimes. the total value of coins in the car is 4.25 how many of each coin do they have you solve the problem like so x+y=      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1055335: a jar contains 53 coins. all of which are nickels and dimes. the total value of coins in the car is 4.25 how many of each coin do they have
you solve the problem like so
x+y=53
.10x+.05y=4.25

How many of each coin would I have???
Found 4 solutions by Boreal, stanbon, Alan3354, MathTherapy:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x+y=53
.10x+.05y=4.25
multiply the second by 100 to remove the decimals
10x+5y=425; now multiply the first by -5 to eliminate the y when added.
-5x-5y=-265
5x=160
x=32 dimes or $3.20
y=21 nickels or $1.05
Add to $4.25

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
a jar contains 53 coins. all of which are nickels and dimes. the total value of coins in the car is 4.25 how many of each coin do they have
you solve the problem like so
x+y=53
.10x+.05y=4.25
------
10x + 10y = 530
10x + 5y = 425
------
Subtract and solve for "y"::
5y = 105
y = 21 (# of nickels)
x = 53-21 = 32 (# of dimes)
-------------
Cheers,
Stan H.
----------

How many of each coin would I have???

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
1 ? is sufficient.

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!

a jar contains 53 coins. all of which are nickels and dimes. the total value of coins in the car is 4.25 how many of each coin do they have
you solve the problem like so
x+y=53
.10x+.05y=4.25

How many of each coin would I have???
You could apply SUBSTITUTION, but!

x + y = 53 ------ eq (i)

.1x + .05y = 4.25 ------- eq (ii)

To apply ELIMINATION, and if you don't like working with decimals, simply multiply eq (ii) by 20 to get:

2x + y = 85 ------ eq (iii)

 x + y = 53 ------ eq (i)

Now SUBTRACT eq (i) from eq (iii) to ELIMINATE y and find value of x

Substitute value of x (number of dimes) into any of the 1st 2 equations, but note that eq (i) is the better of the two. 

This will give you the value of y, the number of nickels.