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Mnenie [13.5K]
3 years ago
15

HEY PLEASE HELP. The student's savings box had accumulated 35 twenty and fifty cent coins in one month, in the total amount of 1

0.3 euros. Several 20 and several 50 cent coins are in the box.
Mathematics
1 answer:
stira [4]3 years ago
6 0

Answer: 11 and 24

Step-by-step explanation:

Given

There are 35- 50 cents and 20 cents coins

The total of them is 10.3 euros

Suppose no of 50 cents coins is x and 20 cents is y

so we can write

\Rightarrow 0.5x+0.2y=10.3\quad \ldots(i)

Also,

\Rightarrow x+y=35\quad \ldots(ii)

Solving (i) and (ii) we get

x=11; y=24

There are 11 fifty cent and 24 twenty cents coins

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5x-4y+-13 and 3x-4y+-11
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Solving
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Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '4y' to each side of the equation.
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Simplifying
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Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '11' to each side of the equation.
-11 + 3x + 11 + -4y = 0 + 11

Reorder the terms:
-11 + 11 + 3x + -4y = 0 + 11

Combine like terms: -11 + 11 = 0
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Add '4y' to each side of the equation.
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Combine like terms: -4y + 4y = 0
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Simplifying
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