One positive number is 5 less than twice a second number, and their product is 117. find the two numbers.

x,y - the numbers

one positive number is 5 less than twice a second number

(1)    x - 5 = 2y

their product is 117

(2)    xy = 117

(1)     x - 5 = 2y     add 5 to both sides

x = 2y + 5   substitute it to (2)

(2y + 5)y = 117     use distributive property

(2y)(y) + (5)(y) = 117

2y² + 5y = 117      subtract 117 from both sides

2y² + 5y - 117 = 0

2y² + 18y - 13y - 117 = 0

2y(y + 9) - 13(y + 9) = 0

(y + 9)(2y - 13) = 0 ↔ y + 9 = 0 ∨ 2y - 13 = 0

y + 9 = 0    subtract 9 from both sides

y = -9

2y - 13 = 0     add 13 to both sides

2y = 13     divide both sides by 2

y = 6.5

substitute the values of y to (1)

x = 2(-9) + 5 = -18 + 5 = -13 < 0

x = 2(6.5) + 5 = 13 + 5 = 18

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

the answer for this question is a= 13,622.84

step-by-step explanation: