Math 1010

Finance

1.6 Patterns and Linear Equations

1. Julia feels like she needs to spring clean her house, but she also works full time so the time to complete her cleaning is limited. She has 25 items on her very detailed list of tasks and plans to accomplish one item on the list each day for the next few weeks.
   1. After 3 days, how many tasks does Julia have remaining on her list?

      22 tasks

   2. Julia has been working on her list for 16 days. How many tasks does she still have on her list?

      9 tasks

   3. After one week, how many tasks are still on Julia’s list?

      18 tasks

   4. Choose a variable to represent the amount of time that Julia has been working on completing the items on her list and use this variable to write a rule for the number of tasks remaining on her list.

      $25-d$

   5. If Julia has only 6 items left on her list, how many days has she been working on her spring cleaning?

      $25-d =6$

      $25-d-25=6-25$

      $-d=-19$

      $\frac{-d}{-1}=\frac{-19}{-1}$

      19 days

2. Rachel is planting a new dogwood tree in her yard. The tree is 12 feet tall now and expected to grow another foot every year.
   1. How tall will the tree be one year from now?

   13 feet

   2. How tall will the tree be five years from now?

   17 feet

   3. How tall will the tree be 15 years from now?

   27 feet

   4. Use a variable to represent the number of years from now, and write an expression to represent the height of the tree.

   $12+y$

3. A certain credit card company is offering its customers 5% cash back on all purchases made using their card during a calendar year.

   $0.05x$

   1. If the amount of purchases you made using the card for the year is $24,000, how much will the company give back to you?

   $1200

   2. If you purchase fifteen thousand dollars of items using the card one year, what will the company give back to you at the end of the year?

   $750

   3. If you have received $60.00 back from the credit card company, what was the total amount of purchases you made using the card for the year?

   $0.05x=60$

   $1200

   4. If the company sent you a check for $1826.20, what is the total amount of purchases you made for the year on their card?

   $0.05x=1826.20$

   $36,524

Solve for the variable. Show all your steps.

$x+2=-9$

$x+2-2=-9-2$

$x=-11$

$\frac{w}{-3.4}=1.5$

$\frac w{-3.4}\left(-3.4\right)=1.5\left(-3.4\right)$

$w=-5.1$

$15 x=-225$

$\frac{15x}{15}=\frac{-225}{15}$

$w=-15$

$a-8=-7$

$a-8+8=-7+8$

$a=1$

$-4x=-18$

$\frac{-4x}{-4}=\frac{-18}{-4}$

$x=4.5$

$\frac{w}{2}=-4.4$

$\frac w{2}\left(2\right)=-4.4\left(2\right)$

$a=-8.8$

$13+y=-3$

$13+y-13=-3-13$

$y=-16$

$x-20=-1$

$x-20+20=-1+20$

$x=19$

$-x=-16$

$\frac{-x}{-1}=\frac{-16}{-1}$

$x=16$

$\frac{a}{-5}=15$

$\frac a{-5}\left(-5\right)=15\left(-5\right)$

$a=-75$

$-63=3x$

$\frac{-63}{3}=\frac{3x}{3}$

$-21=x$

$-x+16=-10$

$-x+16-16=-10-16$

$-x=-26$

$\frac{-x}{-1}=\frac{-26}{-1}$

$x=26$

$15-x=1$

$15-x-15=1-15$

$-x=-14$

$\frac{-x}{-1}=\frac{-14}{-1}$

$x=14$

Two-step Equations: Whole Class Problem

4. Daniel and his wife need to purchase a new water heater for their home. They have chosen to purchase a 65-gallon heater because of the size of their family. The water heater will cost $700. The cost for them to have the old water heater removed, the area prepared for the new one and the new water heater installed will cost them $90 per hour.
   1. What quantities are important in this problem? Cost of new water heater, number of hours plumber works, plumber charge per hour, total cost of job
   2. Which of these quantities are variable quantities? Number of hours plumber works, total cost of job
   3. What is the independent quantity? Number of hours plumber works (time)
   4. What is the dependent quantity? Total cost
   5. Why is this quantity called dependent? The total cost depends on the number of hours the plumber works.
   6. The units refer to how you will measure each of the quantities. What units will you use to measure the independent quantity? hours
   7. What units will you use to measure the dependent quantity? Dollars
   8. Define a variable that represents the independent quantity and use this variable to write a rule for the dependent quantity. $C=90x+700$
   9. If the plumber works for 2 hours, what did Daniel pay for the water heater and installation? $880
   10. If the plumber works for five hours, what is Daniel’s cost? $1150
   11. If Daniel is invoiced for $1285.00, how long did the plumber work?

       $90x+700=1285$

       $90x+700-700=1285-700$

       $90x=585$

       $\frac{90x}{90}=\frac{\displaystyle585}{90}$

       $x=6.5$

       The plumber worked 6.5 hours.

   12. If Daniel does not have to pay anything at all for buying and installing his new water heater, how many hours has the plumber worked? Does this answer make sense? Please defend your answer.

       $90x+700=0$

       $90x+700-700=0-700$

       $90x=-700$

       $\frac{90x}{90}=\frac{\displaystyle-700}{90}$

       $x=-7.8$

       This does not make sense because the plumber cannot work a negative number of hours.

5. Phyllis has been saving money for a long time to take a vacation with her daughter. After many months, she has saved $1875. She wants to spend her money wisely to make it stretch throughout the entire vacation, so she plans to spend $150 per day.
   1. Define a variable for the number of days Phyllis and her daughter will spend on vacation and use this variable to write a rule for the amount of vacation money they have left to spend. $1875-150d$
   2. If Phyllis and her daughter are away on vacation for one week, how much money will they have left? $1875-150(7)= \$825$
   3. If the two of them decide to spend 10 days on vacation, how much money do they still have to spend?

      $1875-150(10)=\$375$

   4. If they spend $1500, how long is their vacation?

   $1875-150d=375$

   $1875-150d-1875=375-1875$

   $-150d=-1500$

   $\frac{-150D}{-150}=\frac{\displaystyle-1500}{-150}$

   $d=10$

   Their vacation was 10 days long.

   5. If they have no money left when they return and they stick to their $150 per day budget, how many days is their vacation?

      $1875-150d=0$

      $1875-150d-1875=0-1875$

      $-150d=-1875$

      $\frac{-150D}{-150}=\frac{\displaystyle-1875}{-150}$

      $d=12.5$

      Their vacation was 12.5 days long.

   6. Would it make sense for the independent quantity to be a negative number? Please defend your answer.

      No, because you can't spend a negative number of days on vacation.

   7. Is it possible for the dependent quantity to be a negative number? Explain your answer.

      Yes, you can overspend your budgeted amount of money.

6. Jim is concerned that the grass he cuts is growing too slowly and he will not be mowing enough yards to earn school money. He finds that the grass measures 2 inches when he cuts it and grows at a rate of 1/8 inch per day.
   1. What are the quantity names and units for this problem?

      Time in days and Length of grass in inches

   2. Define a variable for the time, and use this variable to write a rule for the length of the grass. $\frac18d+2$
   3. What will be the height of the grass 4 days after it is cut?

      $\frac18(4)+2$

      $\frac48+2$

      $\frac12+2$

      $2\frac12$

      The grass will be 2.5 inches tall.

   4. What will be the height of the grass 2 weeks after it is cut?

      $\frac18(14)+2$

      $\frac{14}8+2$

      $\frac74+2$

      $3.75$

      The grass will be 3.75 inches tall.

   5. In how many days after it is cut will the grass be 3 inches tall?

      $\frac18d+2=3$

      $\frac18d+2-2=3-2$

      $\frac18d=1$

      $\frac81\cdot\frac18d=1\cdot\frac81$

      $d=8$

      The grass will be 3 inches tall 8 days after it is cut.

   6. In how many weeks after it is cut will the grass be 1 foot?

      $\frac18d+2=12$

      $\frac18d+2-2=12-2$

      $\frac18d=10$

      $\frac81\cdot\frac18d=10\cdot\frac81$

      $d=80$

      The grass will be 1 foot tall 80 days after it is cut.

7. A Fortune 500 company has assumed a 3.5 million dollar debt during the recent economic downfall and is continuing to assume more debt at a rate of $25,000 per month as they struggle to stay in business. Represent the amount of debt as a negative number.
   1. What are the quantity names and units for this problem?

      Time in months and Total debt in Dollars

   2. Define a variable for the time, and use this variable to write a rule for the amount of debt the company has assumed. $-3500000-25000m$
   3. How much debt will the company have after 6 months?

      $-350000-25000(6)=- \$3650000$

   4. If the company continues to assume more debt at this rate, what will their amount of debt reach after two years?

      $-3500000-25000(24)=- \$4100000$

   5. How many more months of debt has the company accumulated if their debt has reached $3875000?

      $-3500000-25000m=-3875000$

      $-3500000-25000m+3500000=-3875000+3500000$

      $-25000m=-375000$

      $\frac{-25000m}{-25000}=\frac{-375000}{-25000}$

      $m=15$

      The company's debt reached $3875000 after 15 months.

   6. If the company has been accumulating their debt at this rate for several months now, how long ago did they begin going into debt?

      $-3500000-25000m=0$

      $-3500000-25000m+3500000=0+3500000$

      $-25000m=3500000$

      $\frac{-25000m}{-25000}=\frac{3500000}{-25000}$

      $m=-140$

      The company started going into debt 14 months ago.

   Solve for the variable. Show all your steps.

$-9 x+1=-80$

$-9x+1-1=-80-1$

$-9x=-81$

$\frac{-9x}{-9}=\frac{-81}{-9}$

$m=9$

$8 n-7=25$

$8n-7+7=25+7$

$8n=32$

$\frac{8n}{8}=\frac{32}{8}$

$n=4$

$10-6 v=-104$

$10-6 v-10=-104-10$

$-6 v=-114$

$\frac{-6v}{-6}=\frac{-114}{-6}$

$v=19$

$-10=-10+7 m$

$-10+10=-10+7 m+10$

$0=7 m$

$\frac{0}{7}=\frac{7m}{7}$

$m=0$

$-9 x-13=-103$

$-9 x-13+13=-103+13$

$-9 x=-90$

$\frac{-9x}{-9}=\frac{-90}{-9}$

$x=10$

$\frac{x}{4}-6=-5$

$\frac{x}{4}-6+6=-5+6$

$\frac{x}{4}=1$

$4\cdot\frac x4=1\cdot4$

$x=4$

$6=\frac{w}{4}+2$

$6-2=\frac{w}{4}+2-2$

$4=\frac{w}{4}$

$4\cdot4=\frac{w}{4}\cdot4$

$16=w$

$\frac{a}{9}-1=-2$

$\frac{a}{9}-1+1=-2+1$

$\frac{a}{9}=-1$

$\frac{a}{9}\cdot9=-1\cdot9$

$a=-9$

$8+\frac{b}{-4}=5$

$8+\frac{b}{-4}-8=5-8$

$\frac{b}{-4}=-3$

$\frac{b}{-4}\cdot-4=-3\cdot-4$

$b=12$

$-2=2+\frac{v}{4}$

$-2-2=2+\frac{v}{4}-2$

$-4=\frac{v}{4}$

$-4\cdot4=\frac{v}{4}\cdot4$

$-16=v$

$\frac{m+9}{3}=8$

$\frac{m+9}{3}\cdot3=8\cdot3$

$m+9=24$

$m+9-9=24-9$

$m=15$

$-1=\frac{5+x}{6}$

$-1\cdot6=\frac{5+x}{6}\cdot6$

$-6=5+x$

$-6-5=5+x-5$

$-11=x$

$\frac{n+5}{-16}=-1$

$\frac{n+5}{-16}\cdot-16=-1\cdot-16$

$n+5=16$

$n+5-5=16-5$

$n=11$

$\frac{m}{9}-1=-2$

$\frac{m}{9}-1+1=-2+1$

$\frac{m}{9}=-1$

$\frac{m}{9}\cdot9=-1\cdot9$

$m=-9$