Box Method

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(C)Copyright 2020, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

I used to have a coworker who was so convinced that Ken was based on him that he was shocked when I pointed out that Ken had neither a beard nor mustache.

But give him some time off, and who knows what will happen.

I have been keeping busy in my house, so I didn't get to address the issue of using the Area model (aka, the Box Method) for multiply 2- and 3-digit numbers. In the past couple of weeks, an example floated around social media of a front-line worker who couldn't do the "new math" with their child, and just did it the "old way". This, I understand. The outpouring of ridicule for the new math ... well, I guess I understand that, too, because people haven't learned it, or More Importantly, learned WHY it's being taught this way.

Side note: an issue I did bring up on Twitter is that, when first rolled out, some instructors didn't know why they were doing this either. They just did it because they were told to, and they might not have even done that much well because of resistance. Learning it that way would make the core knowledge behind it difficult to scale up.

If I ever do write an article, I'd likely reference Mr. Incredible from The Incredibles 2, when he ponders "Why did they change math?" The superhero reference is not accidental since I would be addressing an essential worker who is giving their all out there while many of us aren't leaving our homes. But, he did eventually get it, and understand it.

If a student learns rules, they may be able to recite them from memory, and maybe they can apply them through rote. On the other hand, if a student develops the Number Sense and makes the connections, the extra steps in the first exercise can be used over and over through high school into college with more advanced topics. Along the way, they can build layers upon layers on their foundation.

Odds are I won't get around to writing that blog post, because I never seem to. But at least there is this entry, so my plans aren't totally FOILed.

Come back often for more funny math and geeky comics.

Daily Regents: Rectangular Garden

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams begin next month. At least, that is the plan.

June 2014, Questions 34

34. A rectangular garden measuring 12 meters by 16 meters is to have a walkway installed around it with a width of x meters, as shown in the diagram below. Together the walkway and the garden have an area of 396 square meters.

Write an equation that can be used to find x, the width of the walkway.

Describe how your equation models the situation.

Determine and state the width of the walkway, in meters.

Unlike the previous question, this was a majorly complicated problem, with more binomial multiplication.

The length of the rectangle is x + 16 + x or (2x + 16).
The width of the rectangle is x + 12 + x or (2x + 12).
The area of the rectangle is 396.

So (2x + 16)(2x + 12) = 396
4x² + 24x + 32x + 192 = 396
4x² + 56x - 204 = 0
Any time you have a coefficient in front of x², check if you can factor it out. In this case, all the coefficients are multiples of 4, so we can just divide each term on the left by 4. Dividing 0 by 4 will leave 0 on the right side.
x² + 14x - 51 = 0 Factor the trinomial back into two binomials and apply the Zero Product Property. (One of the binomials MUST equal zero.)
(x + 17)(x - 3) = 0
x + 17 = 0 or x - 3 = 0
x = -17 or x = 3

We can discard the negative value not because we made a mistake, but because a negative value makes no sense in the context of length and width.
You are left with the width of the walkway being 3 meters wide.

The equation models the situation because the Area is the product of the length times the width. The length of the garden plus the walkway on either side is 2x + 16. The width of the garden plus the walkway is 2x + 12. The Area of the garden is 396 square meters.

Any questions?

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If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Daily Regents: Trinomials

I'll be reviewing a New York State Regents Exam Question every day from now until the Regents exams begin next month. At least, that is the plan.

June 2014, Questions 33

33. Write an equation that defines m(x) as a trinomial where m(x) = (3x - 1)(3 - x) + 4x² + 19. Solve for x when m(x) = 0.

You need to multiply the binomials and then combine all the like terms.
(3x - 1)(3 - x) = 9x - 3x² - 3 + x = -3x² + 10x - 3
To that, add 4x² + 19.
You get m(x) = x² + 10x + 16. Don't forget to write it as an equation!

You needed to find when m(x) = 0, so x² + 10x + 16 = 0. That's a simple quadratic equation to solve. What two factors of 16 add up to 10? That's 8 and 2.
So (x + 8)(x + 2) = 0
x + 8 = 0 or x + 2 = 0
Then x = -8 or x = -2.

Any questions?

---

If anyone in Brooklyn is looking for an Algebra or Geometry Regents Prep tutor, send me a note. I have a couple of weekly spots available between now and June.

Five Things You Need to Know About Algebra, Part IV

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(C)Copyright 2015, C. Burke.

I'd like to take an EPEE to FOIL. Or maybe a SABER.

Multiply everything inside the second set of parentheses by what's outside the parentheses, which in the case of two binomials is the first binomial. And then do it again with the other set of parentheses. This method will work with trinomials as well without leaving you scratching your head.

I've had students who could multiply a trinomial by a binomial because they didn't know what to do with the extra term because they'd learned a short-cut but not the concept.

Interestingly, when I try to demonstrate using integers instead of variables, the point gets lost because they finally start using the Order of Operations, correctly, and add what's inside the parentheses first. Sigh.

June 2014 Common Core Algebra 1 Regents Exam, Part 3

Update: I now have a Common Core Regents Review books available on Amazon.

This is the second thread dealing with the Common Core Algebra test, and probably the last for a while, as I'll be away from the weekend. The first thread was about Part 2 of the test. This post deals with part 3 and 4.

Algebra 1 (Common Core), Part 3

Questions is Part 3 are worth four points. Again, I won't assume the point distribution for partial credit.

33. Write an equation that defines m(x) as a trinomial where m(x) = (3x - 1)(3 - x) + 4x² + 19. Solve for x when m(x) = 0.

You need to multiply the binonials and then combine all the like terms.
(3x - 1)(3 - x) = 9x - 3x² - 3 + x = -3x² + 10x - 3
To that, add 4x² + 19.
You get m(x) = x² + 10x + 16. Don't forget to write it as an equation!

For the second part of the question, take the equation you found in the first part (they allow for consistent errors) and substitute 0 for x. The answer should be obvious, but I would still show the work -- especially because it says solve

m(0) = (0)² + 10(0) + 16 = 0 + 0 + 16 = 16.

Frankly, I thought this problem was easier (or at least less complicated) than some of the Part 2 problems.

EDIT: As you may have seen in the comments below, in my rush to complete this section before leaving for the weekend, I misread my notes in solving this problem, despite having typed it correctly above. My original instinct was that this problem was too easy, and now I see why. Generally, when I get the feeling that something is too easy, there's usually something I missed, and I go look for it.

You needed to find when m(x) = 0, so x² + 10x + 16 = 0. That's a simple quadratic equation to solve. What two factors of 16 add up to 10? That's 8 and 2.
So (x + 8)(x + 2) = 0
x + 8 = 0 or x + 2 = 0
Then x = -8 or x = -2.

So that would be a Conceptual error for this part of the question, which is more points off than a simple Computational error. It could have been worse. I might have been incredibly wrong and arrogantly displayed a lack of understanding of how functions work, but that would be entirely overstating the mistake made. In any event, I am told that some people like to make damning accusations under the guise of anonymity rather than step out into the light of day. But we don't talk about such things in polite company.

Now to continue.... End of EDITED SECTION

34. A rectangular garden measuring 12 meters by 16 meters is to have a walkway installed around it with a width of x meters, as shown in the diagram below (Note: This will be added later) Together the walkway and the garden have an area of 396 meters.

Write an equation that can be used to find x, the width of the walkway.

Describe how your equation models the situation.

Determine and state the width of the walkway, in meters.

Unlike the previous question, this was a majorly complicated problem, with more binomial multiplication.

The length of the rectangle is (2x + 16). The width of the rectangle is (2x + 12). The area of the rectangle is 396.

So (2x + 16)(2x + 12) = 396
4x² + 24x + 32x + 192 = 396
4x² + 56x - 204 = 0
x² + 14x - 51 = 0
(x + 17)(x - 3) = 0
x + 17 = 0 or x - 3 = 0
x = -17 or x = 3
Discard the negative value and you are left with the width of the walkway being 3 meters wide.

Hopefully, that's enough.

35. Caitlin has a movie rental card worth $175. After she rents the first movie, the card's value is$172.25. After she rents the second movie, its value is $169.50. After she rents the third movie, the car is worth $166.75. Assuming the pattern continues, write an equation to define A(n), the amount of money on the rental card after n rentals.

Caitlin rents a movie every Friday night. How many weeks in a row can she afford to rent a movie using the rental card only? Explain how you arrived at your answer.

First, "Explain" means in words; equations will NOT be enough information. I am not kidding.

Subtract 175-172.25 and you get $2.75. Check the next and the next to be sure, but you see that each rental is $2.75
That makes the equation A(n) = -2.75n + 175.

To answer the second part, make an inequality greater than 0, and find the largest whole number which makes it true. (Or make an equation equal to 0 and drop everything after the decimal.)
-2.75n + 175 > 0
-2.75n > -175
n Each movie costs $2.75 to rent. If you divide $175 by $2.75, you can get 63 rentals. There is money left over, but not enough for another rental using the card only.

36. An animal shelter spends $2.35 per day to care for each cat and $5.50 per day to care for each dog. Pat noticed that the shelter spent $89.50 caring for cats and dogs on Wednesday.

Write an equation to represent to represent the possible numbers of cats and dogs that could have been at the shelter on Wednesday. Pat said that there might have been 8 cats and 14 dogs at the shelter on Wednesday. Are Pat's numbers possible? Use your equation to justify your answer. Later, Pat found a record sowing that there were a total of 22 cats and dogs at the shelter on Wednesday. How many cats were at the shelter on Wednesday?

This is a lot of work for a system of equations. I heard some teachers complaining about it (one of whom has seen the scoring rubric), and I heard students tell me that this was so easy. The students explained to me how they did it, so they might be correct in how "easy" they found it.

We'll use C for the number of cats and D for the number of dogs. That makes 2.35C the amount spent on cats and 5.50D is the amount spent on dogs.

The equation for Wednesday is 2.35C + 5.50D = 89.50.

For the second part, substitute 8 for C and 14 for D, and see if it's a true statement.

2.35(8) + 5.50(14) = 89.50 ?
18.80 + 77.00 = 89.50 ?
95.80 =/= 89.50 X
This is NOT a possible combination.

The final part states that C + D = 22. We now have a system of equations. Multiply this equation by -2.35

-2.35C + -2.35D = -51.70
2.35C + 5.50D = 89.50
combine: 3.15D = 37.80
divide: D = 12
There are 12 Dogs, so there are 22 - 12 = 10 = C, 10 Cats.

* * * * * * * * * * *

Okay, that's enough for me. This is taking a lot out of me.

What are your opinions of all this?

FOIL (Again)

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(C)Copyright 2009, C. Burke. All rights reserved.


The *nsync version is "Bi- by Bi-".



I was tempted to do a Formula 409 comic (for the 409th comic).
The temptation passed.