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salantis [7]
3 years ago
12

Find two integers whose product is 138 such that one of the integers is seven less than five times the other integer.

Mathematics
1 answer:
Oksana_A [137]3 years ago
4 0

Answer: the integers are 6 and 23

Step-by-step explanation:

Let x represent one of the integers.

Let y represent the other integer.

The product of the integers is 138. This means that

xy = 138 - - - - - - - - - -1

One of the integers is seven less than five times the other integer. This means that

x = 5y - 7

Substituting x = 5y - 7 into equation 1, it becomes

y(5y - 7) = 138

5y² - 7y = 138

5y² - 7y - 138 = 0

5y² + 23y - 30y - 138 = 0

y(5y + 23) - 6(5y + 23)

y - 6 = 0 or 5y + 23 = 0

y = 6 or y = - 23/5

The only possible value of y is 6

x = 138/y = 138/6

x = 23

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olganol [36]
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margin of error = critical value x standard error

To compute for the critical value:
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Now plug in the following in the formula:

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3 years ago
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