Janis was offered two different jobs when she graduated from college. she made the graph and table to show how much she would earn over time at each job. earnings over time for job 1 earnings over time for job 2. when will janis’s salary be the same for job 1 and job 2, and how much will she be earning at that point? the salaries will be the same in year 20, and she will be earning $80,000. the salaries will be the same in year 16, and she will be earning $70,000. the salaries will be the same in year 12, and she will be earning $60,000. the salaries will be the same in year 10, and she will be earning $55,000.

The salaries will be the same in year 20 and she will be earning $80,000.

We first create an equation to represent the data in the table.  To find the slope we use the formula
m = (y₂-y₁)/(x₂-x₁) = (60,000-55,000)/(12-10) = 5,000/2 = 2500

Writing this in point-slope form, we have
y-y₁=m(x-x₁)
y-55,000 = 2500(x-10)

Converting to slope-intercept form, we first use the distributive property:
y-55,000 = 2500*x - 2500*10
y-55,000 = 2500x - 25,000

Adding 55,000 to both sides,
y = 2500x+30,000

Now we set this equal to the equation from the graph:
2500x+30,000 = 2000x+40,000

Subtract 2000x from both sides:
2500x+30,000-2000x = 2000x+40,000-2000x
500x+30,000 = 40,000

Subtract 30,000 from both sides:
500x+30,000-30,000 = 40,000-30,000
500x = 10,000

Divide both sides by 500:
500x/500 = 10,000/500
x = 20

The salaries will be the same in year 20.
y=2000(20)+40,000
y=40,000+40,000 = 80,000

The salary in year 20 will be $80,000.