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harina [27]
3 years ago
11

Jen Butler has been pricing​ Speed-Pass train fares for a group trip to New York. Three adults and four children must pay $ 79.

Two adults and three children must pay $ 56. Find the price of the​ adult's ticket and the price of a​ child's ticket.
Mathematics
1 answer:
xxMikexx [17]3 years ago
3 0

<u>Answer:</u>

Price of the​ adult's ticket = $13

Price of a​ child's ticket = $ 10

<u>Step-by-step explanation:</u>

Given:

Cost of train fares for Three adults and four children = $79

Cost of train fares Two adults and three children  = $ 56

To Find:

Price of the​ adult's ticket and the price of a​ child's ticket =?

Solution:

Let the cost of one adult's ticket be x  and

Let the cost of one child's ticket be y

Then

Three adults and four children must pay $ 79 be

3x + 4y = 79-----------------------------------------------(1)

Two adults and three children must pay $56

2x + 3y = 56----------------------------------------------(2)

multiply  eq(1) by 2, we get

6x + 8y = 158----------------------(3)

multiply  eq(2) by 3, we get

6x + 9y = 168----------------------(4)

Subracting (3) from (4)

 6x + 9y = 168

 6x + 8y = 158

(-)      (-)       (-)

------------------------------

0x +  y = 10

------------------------------

y=10

Now substitute the value of y in eq (1) to get the value of x

3x + 4(10)= 79

3x +  40 = 79

3x = 79 - 40

3x = 39

x = \frac{39}{3}

x= 13

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Answer:

a).

= (67×10)÷2

= 670 ÷ 2

= 335

Explanation:

a).

= (67×10)÷2

= 670 ÷ 2

= 335

So, 5×67 = (67×10)÷2

=> 335 = 335

b).

= (67x2)+(67×2)

= 134 × 134

= 268

So, 5×67 ≠ (67x2)+(67×2)

=> 268 ≠ 335

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= (6 ×7)×5

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= 210

So, 5×67 ≠ (6×7)×5

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= (5×6)+(5×7)

= 30 + 35

= 65

So, 5×67 ≠ (5×6)+(5×7)

=> 65 ≠ 335

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