Question 686341: how do you slove this: mary has $2.35 in nickels and dimes. if she has 33 cions in all, find the number of nickels and dimes?
Answer by RedemptiveMath(80)   (Show Source):
You can put this solution on YOUR website! Find the common theme(s) or idea(s) given in the information. Since this is basic algebra, it is more likely that you will run across system of equations only using two variables. Notwithstanding, you must figure out what the common pieces are here. We are dealing with coins. But what specifically about coins are we dealing with? Well, we are dealing with the amount of coins Mary has. The information tells us that she has 33 coins in all, so the first theme we could use is the amount of coins she has physically. This gives us our first equation:
n + d = 33.
We know that between nickels and dimes, she has 33 total coins, but we do not know exactly how many of each coin she has. So, that is what the above equation shows. Next, we deal with the amount of money Mary has from the coins. The information tells us that the total amount of money she has from these coins is $2.35. This gives us our second equation:
$0.05n + $0.10d = $2.35.
We know that between nickels and dimes she has $2.35, but we still do not know how many of each she has. We do know, however, that nickels are worth 5 cents and dimes are worth 10 cents. So, we could write the equation expressing that the sum of nickels (5 cents per nickel) and dimes (10 cents per dime) that Mary has is equal to $2.35. You could also do use a shortcut and write the equation like this:
5n + 10d = 235.
We can do so because we can manipulate the decimal equation by multiplying each term by 100 to remove the decimals completely. This will not affect your answer when you solve.
Using our two equations that we have derived from the information given, I will show to methods used to solve systems of linear equations. First, I will use substitution by manipulating the first equation:
n + d = 33
n = 33 - d (get n by itself on one side of the equation)
Now we can plug this in for n in the second equation:
5(33-d) + 10d = 235
165 - 5d + 10d = 235
5d = 70
d = 14 dimes
OR
$0.05(33-d) + $0.10d = $2.35
$1.65 - $0.05d + $0.10d = $2.35
$0.05d = $0.70
d = 14 dimes.
Now we can plug what we have found for d into either equation or its equivalents to get n:
n + 14 = 33
n = 19 nickels
OR
5n + 10(14) = 235
5n + 140 = 235
5n = 95
n = 19 nickels
OR
$0.05n + $0.10(14) = $2.35
$0.05n + $1.4 = $2.35
$0.05n = $0.95
n = 19 nickels.
We could have started out by solving for n by manipulating the first equation for d (d = 33 - n). We could have also manipulated the second equation first and then plugged it into the first equation. All of these constitute substitution, and if done correctly you should get the constant answers of d = 14 and n = 19.
Now I will show you elimination. First, we must manipulate one of the equations so that it has common coefficient connected to both of one of the variables in both equations. Or, manipulating the first equation by multiplying each term by 5:
n + d = 33
5n + 5d = 165
So we can subtract the 5n's:
5n + 5d = 165
5n + 10d = 235
-5d = -70
d = 14 dimes.
Elimination can be done using the original second equation (the one that I manipulated by removing the decimals), and it can be done for both of the variables. Be cautious of your signs though.
Whether you have used elimination or substitution, you should have come up with answers of 14 and 19 for dimes and nickels respectively.
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