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Digiron [165]
2 years ago
8

During the first week of school Andrew purchased two pounds of Jellies and two pounds of Bonbons for $38. The second week of sch

ool he purchased two pounds of Jellies and four pounds of Chocolates for $74 and the third week he purchased two pounds of Jellies, two pounds of Bonbons and one pound of Chocolates for $53. What is the cost of one pound of Chocolates?
Mathematics
1 answer:
Ira Lisetskai [31]2 years ago
7 0

Answer:

j = $7

b = $12

c = $15

Step-by-step explanation:

Let

j = cost of one pound of jelly

b = cost of one pound of Bonbon

c = cost of one pound of chocolate

First week:

two pounds of Jellies and two pounds of Bonbons for $38.

2j + 2b = 38

Second week:

two pounds of Jellies and four pounds of Chocolates for $74

2j + 4c = 74

Third week:

two pounds of Jellies, two pounds of Bonbons and one pound of Chocolates for $53.

2j + 2b + c = 53

Recall,

2j + 2b = 38

So,

38 + c = 53

c = 53 - 38

= 15

c = $15

Substitute c = 15 into

2j + 4c = 74

2j + 4(15) = 74

2j + 60 = 74

2j = 74 - 60

2j = 14

j = 14/2

= 7

j = $7

Substitute j = 7 into

2j + 2b = 38

2(7) + 2b = 38

14 + 2b = 38

2b = 38 - 14

2b = 24

b = 24/2

= 12

b = $12

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A minor league baseball team plays 101 games in a season. If the team won 14 more than twice as many games as they lost, how man
vagabundo [1.1K]
A minor league plays 101 games,
They win 14 more than twice as many as they lost.
How many wins and loses?

Okay, simple.
First, write a simple addition equation. In this case wins = w and losses = l.
l + w = 101.

Now, we have to figure out a way to make one of the terms the same term as the other, in this case we can change the terms of w to l.
l = l
w = 2l +14 (14 more than twice the amount)

Okay. So plug in the new amount for w.
l + 2l + 14 = 101. Great! we now have a simple equation. lets solve.
put similar terms together.
3l + 14 = 101
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So, we have 29 loses, and
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To check,
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Hope this helps!

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