MEDAL AWARD. You have a total of 25 coins, all nickels and quarters. The total value is $3.85. How many of each type of coin do you have? Use a system of linear equations to solve. Use substitution or elimination. Show all work.
the first thing is 20 nickels = $1 4 quarters = $ 1 so... lets assume that u have "x" amount of nickels and "y" amount of quarters. since u have only 25 coins in your hand x + y = 25 and this is the first equation. now moving to the second data given.. the sum of money u have is $3.85 which means the total value of coins you got is $3.85 if you got 1 nickel u will have - >$ 1/20 = $ 0.05 if u got "x" amount of nickel u will have -> x times $ 0.05 = 0.05(x) in the same way if u have 1 quarter you will have - > $ 1/4 = $ 0.25 if u got "y" amount of quarters you will have -> y times $ 0.25 = 0.25(y) but the total is $3.85 which means 0.05(x) + 0.25(y) = 3.85 multiply the whole equation by 100 and u will get 5x + 25y = 385 so the system of equatio u have is 5x + 25y = 385 x + y = 25 now i think u can solve this for x and y .. and its all left to do :)
I usually find it easiest to write coin problems in terms of cents from the outset: \[n+q = 25\]\[5n+25q = 3.85*100\] I would probably solve this by multiplying the first equation by -5, then adding the two together: \[-5n-5q=-125\]\[~5n+25q = ~~~385\]----------------- \[-5n+5n-5q+25q=-125+385\]You should be able to do the rest...
Why do you multiply the first equation by -5?
that way when u add the first equation to the second one the term "5n" gets cancel out... the first equation have (-5n) since it was multiplied by (-5) the second equation already have (5n) add them together -5n + 5n = 0 so the only unknown variable remaining is "q" ... so u have a linear equation with 1 unknown term.. which is easy to solve :)
So 13 quarters?
And 12 nickels?
Yes, easily checked: 13*25 = 325 12*5 = 60 325+60 = 385 And thanks for the explanation, ISURU!
Okay thank you both!
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