Solving problems that have been modeled with equations.

Objective: When done with this lesson, you will have demonstrated how to solve problems using equations.

Approximate completion time: 4 hours

Perhaps you noticed that using tables and graphs can take a lot of time to do well. Could you imagine making a table with 300 rows? What if you didn't know how large you needed to make a graph and found out only too late that your graph was not large enough?

The most compelling reason to solve problems using equations is that solutions can be found relatively quickly--once you have some skills under your belt.

Let's revisit some problems from the last lessons and see the power of solving with equations.

Example 1

"Mike's Mowing Co. will mow lawns for a yearly fee of $50 and $12 per mowing. Carl's Cuts charges $120 for a yearly fee and then $6 per mowing."

Recreating the table in horizontal fashion to help find variable expressions:

# of mowings	0	1	2	3	4	5	...
Mike's Mowing Charge ($)	50	62	74	86	98	110	...
Carl's Cuts Charge ($)	120	126	132	138	144	150	...

Let n = the # of lawn mowings

Then the total cost from Mike's Mowing is 50 + 12·n since it starts with 50 and increases $12 for each additional mowing
and the total cost from Carl's Cuts is 120 + 6 ·n since it starts with 120 and increases $6 for each additional mowing.

# of mowings	0	1	2	3	4	5	...	n
Mike's Mowing Charge ($)	50	62	74	86	98	110	...	50 + 12·n
Carl's Cuts Charge ($)	120	126	132	138	144	150	...	120 + 6·n

 

Now for nearly any question that could be asked about total charges, the expressions 50 + 12·n and 120 + 6·n will be readily used.
(You will get faster and faster with writing variable expressions. Some people can write them with even creating a table--that's nice!)

Let's see these variable expressions in action. Here are some questions and the solutions with equations.

 

Question:	How much would Mike's Mowing charge for 12 mowings?
Solution:	n = 15 50 + 12·(15) = 230 It would cost $230.

Question:	How many mowings could be done with Carl's Cuts for $400?
Solution:	120 + 6·n = 400 6·n = 400 − 120 6·n = 280 n = 280/6 n = 46.6666... To stay under $400 only 46 mowings could be done. The charge would be over $400 for 47 mowings.

Question:

How many mowings until the charges from both companies would be the same?

Solution:

Algebra Steps Reasoning 50 + 12·n = 120 + 6·n The two companies charges would be the same. 50 + 12n = 120 + 6n Rewriting since "touching" implies multiplication. 50 + 6n = 120 Subtracted 6n from both sides 6n = 70 Subtracted 50 from both sides (Do you notice what this equation represents? The companies differ by $6 per lawn and $70 in their start-up charges.) n = 70/6 Divided both sides by 6 n = 11.666... Simplifying The charges would be the same at about 11.667 mowings. Yet the companies would likely only charge for complete mowings. Therefore the charges from both companies would never be the same. However, going from 11 to 12 mowings would make Carl's Cuts less expensive. (Mike's started lower but will be higher after 12 mowings.)

 

If one were quick with writing variable expressions, this solving by equations would be super fast! (By the way, there are some fast lawn mowers out there...)

 

Example 2

Assignment 5.03 - Solving problems that have been modeled with equations.

Now it's time for you to practice putting it all together.

Download your practice problems here. Take time to do a good job with these.