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

5

step-by-step explanation:

first, you divide the triangle in two, tô form a retangle triangle. after, using pythagoream theorem:

18^2 = 14^2 + (x/2)^2

(x^2/4) + 196 = 324

x^2/4 = 324 - 196

x^2/4 = 128

x^2 = 512

x = √512

x = √2^9

x = 16√2 cm

i hope i you.