Lenny is offered jobs at two different companies. at company a, he gets a base salary of $55,000 with manual raises of $2500. at company b, he gets a base salary of $62000 with annual raises of $2000.

a. how many years will it take for lenny make the same salary at both companies?

b. if lenny decides that he will work 20 years on his job, which company should he go to? explain

Before answering these questions, we need to write out both equations.
For Job 'A', his salary would be: 55,000 + 2,500n
This assumes 'n' equals the number of years Lenny spends at these companies. 
For Job 'B', his salary would be: 62,000 + 2,000n

A.
So basically we need to find a number were:
55,000 + 2,500n = 62,000 + 2,000n
is true. So basically solve ^that^ equasion.
55,000 + 2,500n = 62,000 + 2,000n
simplify
500n = 7,000
devide
n = 14
Fourteen years is your answer.

B.
     a. 55,000 + 2,500n
     b. 62,000 + 2,000n
So in order to answer this question, you basically just have to replace 'n' with the number '20', and see which one's bigger.
a. 
55,000 + 2,500(20)
55,000 + 50,000
105,000
b.
62,000 + 2,000(20)
62,000 + 40,000
102,000
105,000 > 102,000
So your answer is:
Lenny should go to company 'a' because 105,000 is greater than 102,000.
[[tip: If you want to wow, surprise, or confuse your teacher (depending on her actual intelligence) write something like, 'another reason Lenny should choose company 'a' is because it gives higher annual raises, which is better in long-term.' He/She may not get it though, lol, wrong class.]] 

Anyway, I hope this helped! :D

38+105= 143, 39+104=143, and you will find then adding the numbers on opposite sides, moving inwards yields the answer 138 every time. because of this, you’re going to multiply 138 times the number of pairs, which is 34. therefore, 34*138= 4692(i’m sorry if it’s the wrong numbers, but it’s the right concept)