SOLUTION: I am supposed to set these up as an equation and solve I thought I had it but my answers look incorrect could you do it so that I can see what I did wrong? Thank you so much.
Jo
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Quadratic Equations and Parabolas
-> SOLUTION: I am supposed to set these up as an equation and solve I thought I had it but my answers look incorrect could you do it so that I can see what I did wrong? Thank you so much.
Jo
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Question 149203: I am supposed to set these up as an equation and solve I thought I had it but my answers look incorrect could you do it so that I can see what I did wrong? Thank you so much.
Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45. How many dimes does he have?
and
Walt made an extra $9000 last year from a part-time job. He invested part of the money at 9% and the rest at 8%. He made a total of $770 in interest. How much was invested at 8%? Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45. How many dimes does he have?
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Let n = number of nickels
and d = number of dimes
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Because we have two unknowns we need two equations:
From: "Joe has a collection of nickels and dimes that is worth $5.65." we get
.05n + .10d = 5.65 (equation 1)
.
From: "If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45." we get:
.05(n+8) + .10(2d) = 10.45
.05n + .40 + .20d = 10.45
.05n + .20d = 10.05 (equation 2)
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n = 25 (nickels)
d = 44 (dimes)
.
Below shows the details:
Solve: We'll use substitution. After moving 0.1*d to the right, we get: , or . Substitute that
into another equation: and simplify: So, we know that d=44. Since , n=25.
Answer: .
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Walt made an extra $9000 last year from a part-time job. He invested part of the money at 9% and the rest at 8%. He made a total of $770 in interest. How much was invested at 8%?
.
Let x = amount invested at 9%
9000-x = amount invested at 8%
.
.09x + .08(9000-x) = 770
.09x + 720 - .08x = 770
.09x - .08x = 50
.01x = 50
x = $5000 (invested at 9%)
9000-5000 = $4000 (invested at 8%)