Dr. B. answered • 03/27/13

Math Problems?PhD, Gentle tutor for K12-College(Also AP Calc,SAT,ACT!)

Let's see how this starts, in thousands of dollars:

Yr | A | B

1 |20 | 25

2 |22.5 | 27

3| 25 | 29

etc

Notice every two years A goes up by 5 (in thousands) and B goes up by 4 so later we'll have:

5 | 30 | 33

7 | 35 | 37

9| 40 | 41

11|45|45

One thing to notice is that "the year" and "x" are not the same: when you're in year #1, x=0 since x is the number of years that have passed. So when x=1, "after x years" means after 1 year, so you're in year 2, and so forth. The table above shows that in the 11th year, when x=10 years have passed, the salaries are equal, but let's look at the equations:

A: y = 20 + (2.5)x is the salary in thousands after x years

B: y = 25 + (2)x is the salary in thousands after x years

In single dollars this would be 20,000 + 2500x and 25,000+ 2000x but let's use thousands so we can use smaller numbers. You can see that (8, 35) is not a solution if you plug in x=8.

Let's solve these two equations:

20 + (5/2)x = 25 + 2x

Subtract 20 from each side to get:

(5/2)x = 5 + 2x

Now notice that (5/2)x is equal to (2.5)x so subtract 2x from each side to get:

(2.5)x - 2x = 5 which is the same as (0.5x)=5 or x=10.

When x=10, 10 years have passed so you're starting the 11th year (again, think of "x=1 so "1 year have passed" so you're in the 2nd year, and earning not 20,000 and 25,000 but 22,500 and 27,000 respectively)

This matches our "table of values" solution we got earlier we we saw that the salaries are equal in the 11th year, and that before then, the "B" job had a higher salary, so if staying for only a few years, and if salary is the only factor, B is the better choice.