create an equation for each situation, then use the equation to answer the questions that follow. suppose…

create an equation for each situation, then use the equation to answer the questions that follow. suppose you have to decide which of two jobs you will work: job a: $75,000 for the first year with a $5,000 raise every year thereafter. job b: $55,000 for the first year with an 8% raise every year thereafter. 1. create an equation for each situation. 2. use each equation to complete the table to the right. 3. when will job b earn more than job a? 4. which job would you choose and why? year, x job a yearly income, y1 job b yearly income, y2 0 1 2 3 4 5 6 7 8 9 10 11 12

create an equation for each situation, then use the equation to answer the questions that follow. suppose you have to decide which of two jobs you will work: job a: $75,000 for the first year with a $5,000 raise every year thereafter. job b: $55,000 for the first year with an 8% raise every year thereafter. 1. create an equation for each situation. 2. use each equation to complete the table to the right. 3. when will job b earn more than job a? 4. which job would you choose and why? year, x job a yearly income, y1 job b yearly income, y2 0 1 2 3 4 5 6 7 8 9 10 11 12

Answer

Explanation:

Step1: Create equation for Job A

Job A has an initial salary of $a_1 = 75000$ and an arithmetic - sequence increase of $d = 5000$ per year. The formula for an arithmetic sequence is $y_1=a_1+(x - 1)d$. So, $y_1=75000+5000(x - 1)=75000 + 5000x-5000=70000 + 5000x$, where $x$ is the number of years after the first - year and $y_1$ is the yearly income.

Step2: Create equation for Job B

Job B has an initial salary of $b_1 = 55000$ and a geometric - sequence increase of $r=1 + 0.08=1.08$ per year. The formula for a geometric sequence is $y_2=b_1r^{x - 1}$. So, $y_2 = 55000\times(1.08)^{x - 1}$ (when $x\geq1$), and for $x = 0$, we can consider the initial state, and the formula still holds.

Step3: Complete the table

For $x = 0$:

  • $y_1=70000+5000\times0 = 75000$
  • $y_2=55000\times(1.08)^{0}=55000$ For $x = 1$:
  • $y_1=70000+5000\times1 = 75000$
  • $y_2=55000\times1.08 = 59400$ For $x = 2$:
  • $y_1=70000+5000\times2=80000$
  • $y_2=55000\times(1.08)^{2}=55000\times1.1664 = 64152$ For $x = 3$:
  • $y_1=70000+5000\times3 = 85000$
  • $y_2=55000\times(1.08)^{3}=55000\times1.259712 = 69284.16$ For $x = 4$:
  • $y_1=70000+5000\times4 = 90000$
  • $y_2=55000\times(1.08)^{4}=55000\times1.36048896 = 74826.8928$ For $x = 5$:
  • $y_1=70000+5000\times5 = 95000$
  • $y_2=55000\times(1.08)^{5}=55000\times1.4693282768 = 80813.055224$ For $x = 6$:
  • $y_1=70000+5000\times6 = 100000$
  • $y_2=55000\times(1.08)^{6}=55000\times1.586874338944 = 87278.08864192$ For $x = 7$:
  • $y_1=70000+5000\times7 = 105000$
  • $y_2=55000\times(1.08)^{7}=55000\times1.71382428605952 = 94260.3357332736$ For $x = 8$:
  • $y_1=70000+5000\times8 = 110000$
  • $y_2=55000\times(1.08)^{8}=55000\times1.8509302289442816 = 101701.162591935488$ For $x = 9$:
  • $y_1=70000+5000\times9 = 115000$
  • $y_2=55000\times(1.08)^{9}=55000\times1.999004647259824128 = 109945.25559929032704$ For $x = 10$:
  • $y_1=70000+5000\times10 = 120000$
  • $y_2=55000\times(1.08)^{10}=55000\times2.15892407104060005888 = 118740.8239072330032384$ For $x = 11$:
  • $y_1=70000+5000\times11 = 125000$
  • $y_2=55000\times(1.08)^{11}=55000\times2.3316380067238480635904 = 128239.090369811643497472$ For $x = 12$:
  • $y_1=70000+5000\times12 = 130000$
  • $y_2=55000\times(1.08)^{12}=55000\times2.518179047261755908677632 = 138499.84759939657497726976$

Step4: Find when Job B earns more than Job A

We need to solve the inequality $y_2>y_1$, i.e., $55000\times(1.08)^{x - 1}>70000 + 5000x$. By checking the values in the table, we find that when $x = 8$, $y_1 = 110000$ and $y_2=101701.16$; when $x = 9$, $y_1 = 115000$ and $y_2=109945.26$; when $x = 10$, $y_1 = 120000$ and $y_2=118740.82$; when $x = 11$, $y_1 = 125000$ and $y_2=128239.09$. So, Job B earns more than Job A in the 11th year.

Step5: Choose a job

If you plan to work for a short - term (less than 11 years), Job A is a better choice because it pays more. If you plan to work for a long - term (11 years or more), Job B is a better choice because of the compound - growth nature of its salary increase, which will result in higher earnings over time.

Answer:

  1. Equation for Job A: $y_1=70000 + 5000x$; Equation for Job B: $y_2 = 55000\times(1.08)^{x - 1}$
Year, $x$ Job A Yearly Income, $y_1$ Job B Yearly Income, $y_2$
0 75000 55000
1 75000 59400
2 80000 64152
3 85000 69284.16
4 90000 74826.8928
5 95000 80813.055224
6 100000 87278.08864192
7 105000 94260.3357332736
8 110000 101701.162591935488
9 115000 109945.25559929032704
10 120000 118740.8239072330032384
11 125000 128239.090369811643497472
12 130000 138499.84759939657497726976
  1. In the 11th year.
  2. If working for less than 11 years, choose Job A for higher earnings. If working for 11 years or more, choose Job B for higher long - term earnings.