SOLUTION: This word problem has to do with coins, but I need to solve it by setting up and solving a system of two linear equations in two variables.
Cindy has 43 coins consisting of nick
Algebra.Com
Question 60293This question is from textbook Elementary and Intermediate Algebra
: This word problem has to do with coins, but I need to solve it by setting up and solving a system of two linear equations in two variables.
Cindy has 43 coins consisting of nickels and dimes. The total value of the coins is $3.40. How many coins of each kind does she have?
I have gotten this far:
n + d = 3.40
is that right?
This question is from textbook Elementary and Intermediate Algebra
Found 3 solutions by checkley71, funmath, Earlsdon:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
N+D=43 & 5N+10D=340
N=43-D
5(43-D)+10D=340
215-5D+10D=340
5D=340-215
5D=125
D=125/5
D=25 NUMBER OF DIMES
N+25=43
N=43-25
N=18 NUMBER OF NICKELS
PROOF
5*18+10*25=340
90+250=340
340=340
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
This word problem has to do with coins, but I need to solve it by setting up and solving a system of two linear equations in two variables.
Cindy has 43 coins consisting of nickels and dimes. The total value of the coins is $3.40. How many coins of each kind does she have?
:
I like to do these in terms of cents to avoid decimals.
Let number of nickles be:n
Let the number of dimes be:d
When we add the coins together we have 43 coins: n+d=43
Nickels are worth 5 cents, so the amount of money in nickels we have is: 5n
Dimes are worth 10 cents: 10d
Total cents when added together: 5n+10d=340
:
E1: n+d=43
E2: 5n+10d=340
Solve E1, for one of the variables like d and substitute into E2 and solve for n.
:
E1: n+d=43 --> d=43-n
5n+10(43-n)=340
5n+430-10n=340
-5n+430=340
-5n+430-430=340-430
-5n=-90
-5n/-5=-90/-5
n=18
Substitute that into E1 to see how many dimes there were:
d=43-18
d=25
There were 18 nickels and 25 dimes.
Sanity Check: does 18 nickels and 25 dimes = $3.40? You decide if we're right.
Happy Calculating!!!
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Er...not quite!
Let n = the number of nickels ($0.05) and d = the number of dimes ($0.10).
The sum of the nickels and dimes (coins) can be written:
1) n + d = 43
The total value of the coins can be written:
2) n($0.05) + d($0.10) = $3.40
This is the required system of two equations in two unknowns (n and d).
You can solve this system by substitution.
Solve equation 1) for n: (n = 43 - d) and substitute this into equation 2) then solve for d.
2a) (43-d)($0.05) + d($0.10) = $3.40 Simplify and solve for d.
2.15 - 0.05d + 0.10d = 3.40
2.15 + 0.05d = 3.40 Subtract 2.15 from both sides.
0.05d = 1.25 Divide both sides by 0.05
d = 25 There are 25 dimes.
n = 43 - d
n = 43 - 25
n = 18 There are 18 nickels.
Check:
n + d = 18 + 25 = 43
18($0.05) + 25($0.10) = $0.90 + $2.50 = $3.40
RELATED QUESTIONS
I need some help with this word problem, setting it up to solve. Thanks.
Joe has a... (answered by vleith,solver91311)
Find two numbers whose sum is -2 and whose difference is 35 by setting up a system of... (answered by stanbon)
I have to come up with my own system of equations with two variables (x,y). I am starting (answered by stanbon)
Solve the problem by using two equations with two variables:
A twin-engined aircraft... (answered by scott8148)
This is a system of two equations word problem. please set up the equations: At the Boy... (answered by mxgirl22)
We are solving word problems using system of equations in my Algebra II class right now.
(answered by josmiceli)
Help! I'm stuck on this problem that looks pretty simple but I can't can't figure out the (answered by lkwtcnj)
"The sum o two numbers is 94 and their difference is 20. Find the two numbers by setting... (answered by edjones)
A river has a current of 5mph. It takes Al half an hour longer to paddle upstream 1.2... (answered by ankor@dixie-net.com)