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

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

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

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