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Nookie1986 [14]
2 years ago
9

Marcie bought a total of 20 used books and cds during a yard sale for a total of 54.50$. of books cost 1.50$ each and cds 5$ eac

h, how many of each did she buy?
1 define your variables ,- what are you solving for
2 set up equations - using the information given
3 solve the system using your method of choice
Mathematics
1 answer:
melomori [17]2 years ago
5 0
Let numbers of books be 'b' and numbers of CDs be 'c'

We can set up two equations:
Equation [1] ⇒ b+c=20
Equation [2] ⇒ 1.50b+5c=54.50

We are solving for the number of books and the number of CDs bought

When we have two equations in terms of two different variables; b and c, that we need to solve, then this becomes a simultaneous equation problem. 

First, rearrange Equation [1] to make either b or c the subject:
b+c=20
b=20-c

Then we substitute b=20-c into Equation [2]
1.50b+5c=54.50
1.50(20-c)+5c=54.50
30-1.50c+5c=54.50
5c-1.5c=54.50-30
3.5c=24.50
c=7

Now we know the value of c which is c=7, substitute this value into b=20-c we have b=20-7=13

Answer:
Numbers of books = 13
Numbers of CDs = 7
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The correct option is option B.

Step-by-step explanation:

The given function is

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We can rearrange the function as

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⇒ 3(x - 4)² - 2 = 18y - 36

⇒ 3(x - 4)² = 18y - 34

⇒ (x - 4)^{2} = 6y - \frac{34}{3}

⇒ x - 4 = \pm \sqrt{6y - \frac{34}{3}}

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