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?

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?

Answer

Explanation:

Step1: Create equation for Job A

Job A has an initial salary of $75000$ and a fixed - amount raise of $5000$ per year. This is an arithmetic - sequence problem. The general formula for an arithmetic sequence is $y=a+(x - 1)d$, where $a$ is the first - term, $x$ is the number of terms, and $d$ is the common difference. Here, when $x = 1$ (first year), $y_1=75000$, and $d = 5000$. So the equation is $y_1=75000+5000(x - 1)=70000 + 5000x$.

Step2: Create equation for Job B

Job B has an initial salary of $55000$ and a percentage raise of $8%$ per year. This is a geometric - sequence problem. The general formula for a geometric sequence is $y=a(1 + r)^{x-1}$, where $a$ is the first - term, $r$ is the rate of growth, and $x$ is the number of terms. Here, $a = 55000$, $r=0.08$, so the equation is $y_2=55000(1 + 0.08)^{x - 1}$.

Step3: Complete the table

For $x = 0$:

  • $y_1=70000+5000\times0 = 70000$
  • $y_2=55000(1 + 0.08)^{-1}=\frac{55000}{1.08}\approx50925.93$ For $x = 1$:
  • $y_1=70000+5000\times1 = 75000$
  • $y_2=55000(1 + 0.08)^{0}=55000$ For $x = 2$:
  • $y_1=70000+5000\times2 = 80000$
  • $y_2=55000(1 + 0.08)^{1}=55000\times1.08 = 59400$ For $x = 3$:
  • $y_1=70000+5000\times3 = 85000$
  • $y_2=55000(1 + 0.08)^{2}=55000\times1.08^{2}=55000\times1.1664 = 64152$ For $x = 4$:
  • $y_1=70000+5000\times4 = 90000$
  • $y_2=55000(1 + 0.08)^{3}=55000\times1.08^{3}=55000\times1.259712 = 69284.16$ For $x = 5$:
  • $y_1=70000+5000\times5 = 95000$
  • $y_2=55000(1 + 0.08)^{4}=55000\times1.08^{4}=55000\times1.36048896 = 74826.8928$ For $x = 6$:
  • $y_1=70000+5000\times6 = 100000$
  • $y_2=55000(1 + 0.08)^{5}=55000\times1.08^{5}=55000\times1.4693280768 = 80813.044224$ For $x = 7$:
  • $y_1=70000+5000\times7 = 105000$
  • $y_2=55000(1 + 0.08)^{6}=55000\times1.08^{6}=55000\times1.586874322944 = 87278.08776192$ For $x = 8$:
  • $y_1=70000+5000\times8 = 110000$
  • $y_2=55000(1 + 0.08)^{7}=55000\times1.08^{7}=55000\times1.71382426877952 = 94260.3347828736$ For $x = 9$:
  • $y_1=70000+5000\times9 = 115000$
  • $y_2=55000(1 + 0.08)^{8}=55000\times1.08^{8}=55000\times1.8509302002818816 = 101701.161015503488$ For $x = 10$:
  • $y_1=70000+5000\times10 = 120000$
  • $y_2=55000(1 + 0.08)^{9}=55000\times1.08^{9}=55000\times1.999004616304432128 = 109945.25389674376704$ For $x = 11$:
  • $y_1=70000+5000\times11 = 125000$
  • $y_2=55000(1 + 0.08)^{10}=55000\times1.08^{10}=55000\times2.15892499760878679808 = 118740.8748684832738944$ For $x = 12$:
  • $y_1=70000+5000\times12 = 130000$
  • $y_2=55000(1 + 0.08)^{11}=55000\times1.08^{11}=55000\times2.3316389974174907419264 = 128230.144857962990805952$

Step4: Find when Job B earns more than Job A

We need to solve the inequality $y_2>y_1$, i.e., $55000(1 + 0.08)^{x - 1}>70000+5000x$. We can solve this by trial - and - error or by using a graphing utility. By trial - and - error, when $x = 9$, $y_1 = 115000$ and $y_2\approx101701.16$, when $x = 10$, $y_1 = 120000$ and $y_2\approx109945.25$, when $x = 11$, $y_1 = 125000$ and $y_2\approx118740.87$, when $x = 12$, $y_1 = 130000$ and $y_2\approx128230.14$. So Job B earns more than Job A in the $12$th year.

Step5: Choose a job

If you are looking for short - term gain (first 11 years), Job A is a better choice as it pays more. However, if you are looking for long - term gain (12th year and beyond), Job B is a better choice because its salary will exceed that of Job A due to the compounding effect of the percentage raise.

Answer:

  1. Equation for Job A: $y_1=70000 + 5000x$; Equation for Job B: $y_2=55000(1 + 0.08)^{x - 1}$
Year, $x$ Job A Yearly Income, $y_1$ Job B Yearly Income, $y_2$
0 70000 $\approx50925.93$
1 75000 55000
2 80000 59400
3 85000 64152
4 90000 69284.16
5 95000 74826.8928
6 100000 80813.044224
7 105000 87278.08776192
8 110000 94260.3347828736
9 115000 101701.161015503488
10 120000 109945.25389674376704
11 125000 118740.8748684832738944
12 130000 128230.144857962990805952
  1. 12th year
  2. Short - term (first 11 years): Job A; Long - term (12th year and beyond): Job B.