Asked by Anonymous
Jeremy had 34 nickels and quarters totaling $4.10. He had two less than twice as many nickels and quarters. How many of each did he have?
Answers
Answered by
Sue S.
Do you mean? -
twice as many nickels AS quarters.
twice as many nickels AS quarters.
Answered by
Anonymous
Sue S. Yes :/ Sorry
Answered by
DrBob222
No, you mean two less than twice as many nickels as quarters; however, you don't need that information to work the problem.
n = # nickels
q = # quarters
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n + q = 34
5n + 25q = 410 Solve these two simultaneously.
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5n + 25(34-n) = 410
Solve and n = 22 and q = 12
check:
22*0.05 = $1.10
12*0.25 = 3.00
Total is $4.10.
You will see that the number of nickels is 2 less than 24 and 24 is 2 x # q = 24,
n = # nickels
q = # quarters
----------------------------
n + q = 34
5n + 25q = 410 Solve these two simultaneously.
---------------------------
5n + 25(34-n) = 410
Solve and n = 22 and q = 12
check:
22*0.05 = $1.10
12*0.25 = 3.00
Total is $4.10.
You will see that the number of nickels is 2 less than 24 and 24 is 2 x # q = 24,
Answered by
Anonymous
Dr.Bob222 thank you. I’m used to solving the top one. However, my teacher wants me to use the second equation. I did come to the class later than everyone else. So I don’t know how you put those numbers into the equation. But for now I’ll use the top one
Answered by
DrBob222
I'm not sure what you mean by using the top equation or the second equation. You must use BOTH equations and solve them simultaneously.
#n + #q = 34 (eqn 1 tells you the total number of coins)
5*#n + 25*#q = 410 (eqn 2 tells you the value of the coins; i.e. 410 cents).
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Using eqn 1 solve for #q. That gives you #q = 34 - #n. Substitute that q into the q in eqn 2 as shown below.
5*#n + 25(34-#n) = 410
5*#n + 850 - 25^#n = 410
-20n = 410-850
-20n = -440
20n = 440
n = 440/20 = 22 = #nickels
Since n + q = 34 the q must be 34-22 = 12
#n + #q = 34 (eqn 1 tells you the total number of coins)
5*#n + 25*#q = 410 (eqn 2 tells you the value of the coins; i.e. 410 cents).
---------------------------
Using eqn 1 solve for #q. That gives you #q = 34 - #n. Substitute that q into the q in eqn 2 as shown below.
5*#n + 25(34-#n) = 410
5*#n + 850 - 25^#n = 410
-20n = 410-850
-20n = -440
20n = 440
n = 440/20 = 22 = #nickels
Since n + q = 34 the q must be 34-22 = 12
Answered by
Anonymous
i have the same question now in my algebra textbook. what grade you in me 7th