Kiss those Math Headaches GOODBYE!


Of course there’s a standard way to find the y-intercept of any line, and there’s nothing wrong with using that approach.

High-Octane Boost for Math
High-Octane Boost for Math Ed

But the method I’ll present here is a bit faster and therefore easer. And hey, if we can save time when doing math, it’s worth it … right?

So first let’s recall that the y-intercept of any function is the y-value of the function when the x-value = 0. That’s because the y-intercept is the y-value where the function crosses or touches the old, vertical y-axis, and of course all along the y-axis the x-value is always 0 (zero).

So the standard slope-intercept formula is y = mx + b. In a problem asking for the y-intercept, you’ll be given one point that the line passes through (that point’s coordinates will provide you with an x-value and a y-value), and you will also be told the slope of the line (the line’s m-value).
So then, to get the b-value, which is the value of the y-intercept, you just grab your y = mx + b equation (dust it off if you haven’t used it in a while), and plug in the three value you’ve been given: those for x, y and m. Then you solve the equation for the one variable that’s left: b, the value of the y-intercept.

Let’s look at an example: a line with a slope of 2 passes through the point (3, 10). What is this line’s y-intercept.

Now, according to the problem, the m-value = 2, the x-value = 3, and the y=value = 10. We just take these values and plug them into the equation:
y = mx + b, like this:

10 = (2)(3) + b

After doing these plug-ins, you just solve the equation for b, finding that
b = 4. That means that the y-intercept of the line = 4.

Now let’s see how you can do the same problem, but a little bit faster.
To do so, we first need to play around with the y = mx + b equation by subtracting the mx-term from both sides, like this:

y = mx + b [Standard equation.]
– mx = – mx [Subtracting mx from both sides.]
y – mx = b [Result after subtracting.]
b = y – mx [Result after flipping left & right sides
of the equation above.]

Aha! Look at that final, beautiful equation. This equation has b isolated on the left-hand side. So now if we want to solve for b, all we do is plug in the x, y and m values into the right-hand side of the equation and simplify the value, and the value we get will be the b-value.

For the problem we just solved, with x = 3, y = 10, m = 2, watch how easy it is to solve:

b = y – mx
b = 10 – (2)(3)
b = 10 – 6
b = 4

So notice that this technique, just like the first technique, reveals that the
y-intercept of the line is 4, or (0, 4). The techniques agree, they just get to the same end in slightly different ways.

Notice that with the second, quicker technique, you don’t need to add or subtract any terms. And that’s a key reason that this technique is faster and easier to use than the standard method. So try it out and stick with it if you like it.



Percents Defying Intuition


Question related to percents of increase and decrease … we know what happens if you increase an original quantity by adding x% of it, then subtracting that same x%: you’ll wind up with the quantity you started with. That is to say, the value of the original quantity will stay the same.

Now let’s pose the same question with regard to multiplication and division. Suppose you take an original quantity and increase it by x%. Then you turn around and decrease the new quantity by x%?

Will the new amount be: a) the same? b) greater? c) less? or d) might it depend on the value of x? What if you reverse the order … you first decrease the quantity by x%, and then you turn around and increase that new quantity by x%. Will the amount you wind up with be the same or different than what you got the other way around?

Many students intuitively think the final result will be the same as the original quantity. That makes this concept fun to teach; math lessons always are more fun when the results go against intuition.

We’ve already said that the amount of the increase and decrease is x%.

Now to keep the math as simple as possible, we’ll just use 1.00 as the original quantity. That way all values that we wind up with will automatically be in percent form.

When we increase our original quantity, 1.00, by x percent, we multiply it by 1.00 + x% = 1.00 + x/100 = 100/100 + x/100 = (100 + x)/100.

Similarly, when we decrease the original quantity, 1.00, by x percent, we multiply it by 1.00 – x% = 1.00 – x/100 = 100/100 – x/100 =(100 – x)/100.

So, then what are the steps performed in the problem? In the first way described, we 1st) increase the original quantity by x%, then 2nd) starting with the increased value, decrease it by x%. Let’s do that now.

1st) 1.00 [times] (100 + x)/100 = (100 + x)/100

2nd) (100 + x)/100 [times] (100 – x)/100 = (100 + x)(100 – x)/100^2 = (100^2 – x^2) / 10,000 = (10,000 – x^2) / 10,000.

We can draw a number of conclusions from this final algebraic expression, boldfaced.

1) It represents the quantity we are left with after increasing, then decreasing a quantity of 1.00, by x%.

2) Since this was done in a general way, this expression serves as a formula to predict the new value for any situation where an original quantity gets increased or decreased by the same percent.

Example: if $2,000 gets increased by 3%, then decreased by 3%, the amount left would be given by this:
2000 [times] (10,000 – 3^2) / 10,000 = 2000(10,000 – 9)/10,000 =
2000(9991/10,000) = 1998.2

3) Since this expression by which the original value gets multiplied does not changed, no matter the original value, no matter the percent, it behooves us to deeply understand this expression:
(10,000 – x^2) / 10,000

4) First, let’s notice that this expression does not change whether we increase the original value first, then decrease — or if we switch the order by decreasing the original value first, then increasing the result. We can see this is true because the numerator of the expression: (10,000 – x^2) arises as the product of (100 + x) and (100 – x). The commutative property guarantees that the order of the factors does not change their product.

So the point here is that it doesn’t matter whether we first increase the original quantity, then decrease it, or if we first decrease the original quantity, then increase it; we’ll get the same result each time.

5) Now let’s look at the relationship between the value of the percent, x, and the outcome, the change in the original quantity.

From the expression we see that as x^2 gets larger, the numerator,
10,000 – x^2, will get smaller. And of course, the value of x^2 increases as the value of x increases.

So in a transitive way of thinking, as the percent, x, that we increase and decrease by gets larger, the ultimate decrease of the original value gets larger. Or, slightly differently, as the percent of change we put the value through increases, the original quantity will end up decreasing more at the end.

Here’s a table that shows the relationship between the x-value, the percent of increase and decrease, and the percent of the original value you’re left with after the increase and decrease are carried out.

This is the first time I’ve explored this topic. So feel free to share your thoughts, insights on it.




There are many ways to teach about linear equations, right. You can (a) use the old t-table approach; (b) you can draw a line and then figure out its slope and y-intercept; or you can (c) first explain the slope-intercept formula and then explain how the equation aligns with the formula.

However I would contend that all three of those approaches leaves students somewhat baffled.

Always Several Options on The Math Cafe’s Menu

The (a) t-table approach makes it appear that points just appear because there’s some machine (the equation) that spits out points in a rather random way, and then these points (the ones on the lines) are just special points that we have found, and perhaps the only ones that are on the line.

The (b) technique gives the impression that — I don’t know —every line just has these two weird properties, slope and y-intercept, and you are simply finding them for this line. While this is true, it does not feel inherently interesting to students, in my experience, partly because slope and y-intercept seem so different from each other. Why would a line necessarily have these two properties? There’s nothing obvious about why a line would have these two properties and no other properties.

The (c) technique is little better than the (b) technique. Students are still left wondering why a line would have these two properties: slope and y-intercept. The two properties seem unrelated to each other and not necessarily essential to the nature of a line.

So is any approach better? Here’s an idea. Get kids thinking about lines by thinking about the relationship between the x- and y-values of the points on the lines.

Start out with this line: y = x. Point out that y = x is in a sense an equation that sends out a call, much as a town squire might shout out a call in a village square for certain people to assemble. While the town squire might be calling for certain men (or women) over a certain age to assemble, the equation y = x, on the other hand, sends out a call for certain pairs of points to assemble, specifically, the pairs of points whose y-values are equal to their x-values, since y = x. Ask students what kind of points those would be. So if the x-value of such a point is 3, what would its y-value have to be? Answer: it would have to be the y-value that is the same as the x-value, so the y-value would also need to be 3. So the point called to assemble would be this point: (3, 3).

Similarly, if x = -3, the associated y-value would be what? Answer: -3. So this point called to assemble would be: (-3, -3). And if x = 0, the associated y-value would be what? Answer: 0. So this point would be: (0, 0). So when these three points are assembled, you would have these points standing bravely on the cold, coordinate plane: (3, 3), (-3, -3), (0, 0). Tell students to graph these three points. When they do so carefully, using a ruler, ask them to name some other points that appear to stand on this same line. They should report that this line also contains points such as (–5, -5), (6, 6), (9 9), (5,5), even (0.5, 0.5), etc.

Students should gradually come to see that when they graph y = x, they wind up graphing the infinitely many points whose y-values are exactly equal to their x-values. And thus they should see that this is NOT a coincidence. They should make the connection that the line, y = x, calls forth the ordered pairs whose y-values are equal to their x-values! This insight is the fundamental concept behind this lesson.

What is so amazing is how many students do NOT realize this most fundamental fact. I have been tutoring students in algebra for 25 years, and I find that when I explain this rather fundamental fact to students, 9 times out of 10 students are amazed.

So I cannot state how important it is that we help students make this connection: there is a relationship between an equation and the coordinate points that the equation generates.

Once we help students see this for y = x, we should move forward in small steps. I will show how I do this in the next post.


“Oh my god, I’m getting these!”

Sierra (not her real name) lay down her phone on the tutoring desk and shot me a quick glance, eyes wide.

“Two minutes ago I would’ve had no way to do these in my head,” she said.

Sierra had just completed a round of mental math addition problems through a phone app*. She was adding two two-digit numbers and had just gotten four problems correct in a row.

The app just showed Sierra how to do addition from left to right, not right to left, thus “chunking” the problem.

For example, take the problem 58 + 79. The app encouraged Sierra to add the tens digits first, helping her see that the 5 and 7, respectively, actually represent 50 and 70, which sum to 120.

The app instructed Sierra to hold that 120 in her short-term memory, then sum the digits in the ones place, 8 and 9, to get 17.

Finally, the app told Sierra to add the two subtotals to find the answer:
120 + 17 = 137.

Sierra said she had never realized that an addition problem could be tackled this way, even though she’s now 16 years old. She left the tutoring session saying she was excited to continue working with the app. The best feature of the app, she said, is that it provides tips on how to do each problem quickly and efficiently.

As a longtime tutor, I’ve long noticed that teaching mental math strategies often helps boost students’ “math self-esteem.” Today’s session with Sierra reinforced that point in a big way.

Before today’s session, Sierra had been voicing a range of negative feelings about her relationship to math. She said she can’t be “a good logical thinker” because she struggles with math. She said that she has “always” been bad at math. When I explored that statement, I found that its roots trace back to 4th grade when she had a teacher who forced students to do problems on the board, whether or not they were prepared. On a number of occasions Sierra was called to the board and confused. Rather than helping Sierra through her confusion, the teacher allowed her to wallow in it publicly. I know … terrible, right?!

But, back to the healing side of things … how does the development of mental math skills make such a big difference for many students? I think the answer relates largely to two concepts: labor/efficiency, and power.

If a student lacks good mental math skills, s/he will find the process of doing arithmetic laborious, i.e., like drudgery. If you doubt that, remind yourself of what it felt like to do a long division problem of the sort: 25,682 ÷ 479. By contrast, a student with strong mental math skills finds the process of arithmetic easy and somewhat fun. Knowing those shortcuts is actually kind of exhilarating, especially if you know A LOT of shortcuts.

Mental math skills also make a student feel powerful. For example, a student who has the rules of divisibility solidly locked down will have no problem when learning the skill of prime factorizing numbers; those divisibility rules will make the factoring process fast and fun. On the other hand, a student who doesn’t know divisibility rules will need to rely on multiplication facts (a rather backward way of working), or on a calculator (slow for this kind of task) — just to perform the division aspect of prime factorizing. As a result, this student will not enjoy prime factorizing numbers, and won’t even know how it might be possible to enjoy the process.

So it’s critically important that we help students nurture and maintain a solid toolbox of mental math skills The internet is loaded with resources for mental math. Just go to YouTube and search for this topic. This blog, also, has many articles and videos on the topic. Just search it for mental math, and see what pops up.


* the phone app Sierra was using was Magoosh’s Mental Math Flashcards app.






Coffee, Pi and More

Coffee, Pi and More

Full confession: I’m a word-nerd. I love language. When I was a kid I would read the dictionary and explore the etymologies of words in a beat-up edition of the American-Heritage dictionary. The other side of this is that I’m highly tuned in to words, the power they have to draw people in, or push people away.

So in my tutoring I pay a lot of attention to the words I use when I talk to my students about math. And I think this has paid off over the years, as I’ve developed a reputation as a person parents bring their kid to when the child is frozen with math anxiety.

Since this is the start of a school year I’d like to make a few suggestions on how all of us who are math instructors — teachers, tutors, homeschoolers, parents — phrase our “math talk” so we keep the level of math anxiety as low as possible to keep our students as open to math learning as possible.

  1. Avoid closed-ended questions.  Wait! you’re thinking. This is math class. There have to be closed-ended questions! Math problems have just one answer. Perhaps, but you don’t have to phrase every question in a closed-ended way. Suppose you have the diagram of a rectangle that is 12″ x 7″, and you want your children to figure out its area.
    You could simply ask: What is its area?
    Or, you could ask:  Does anyone have an idea about its area?
    The first question makes children feel that there is only one way to answer, and their answer will be either wrong or right. The second question allows a child to say a variety of things that could be right, such as: a) the area is base times height, or b) the area is in square inches, or c) the area is more than 12 square inches. Because the question is open-ended, it allows for a variety of answers, not just one answer, therefore it invites participation from more children, not just from the child who knows the answer in the precise number of square inches.
  2. Encourage exploration, and let the learning come as a natural result. Example, suppose that you’re teaching the concept of equivalent fraction and you’re using fraction circles to do so. Imagine that you’ve just helped your child understand the barest beginning of the concept of equivalent fractions by helping him see that 1/3 = 2/6 by placing two 1/6 pieces on top of one 1/3 piece.
    You might next want the child to follow this up by finding another kind of fraction that is equivalent to 1/3.
    The closed-ended way to do this would be to tell the child that such a fraction exists and to ask him to find it.
    The more open-ended way would be to ask the child if s/he thinks there might be another piece, other than sixths, that would fit perfectly on top of the 1/3 piece.
    “Do you want to explore if there’s another piece that might fit on top of the 1/3 piece?” you could say to the child. That’s all you’d need to say, and the child will be off and carrying out the challenge.
  3. Allow for multiple approaches to the same answer. Let’s say you’re teaching an algebra lesson on simplifying numerical square roots. It’s a nice idea to teach the process in two ways and let the students know that both ways are equally valid. The first way involves prime factorizing the number under the square root, then taking out one of every pair of prime factors. The second way involves breaking the number into a perfect square times another number, then taking out the square root of the perfect square. You demonstrate both techniques and have students learn both techniques. Then you let each student decide on which technique s/he wishes to use. Doing this models the idea that there is more than one way to do math procedures, and that shows students that they can develop their own ways to do math procedures as long as their way leads to the correct answer.
  4. So these are a few ideas on ways to talk about math to students. I know from my own experience that the way we talk about math has a big impact on how comfortable students feel around the subject of math.

ACT Math Test: Myth #1


I’ve been coaching many students on the ACT math test lately, and I’ve heard a myth from them and their parents that I’d like to dispel.

The myth is this: If you’re an “A” math student, you’ll ace the ACT math test, and you won’t need long to study for it.

This notion is absolutely FALSE,  for four key reasons.

  1.  The ACT test has a challenging time limitation, and until you learn how to master the time factor, you will struggle with the ACT math test.
  2.  The ACT math test requires you to remember a large quantity of math material, stretching all  the way back from PreAlgebra to Trigonometry. It’s quite possible to be an “A” math student who memorizes a lot of info before each big test, then forgets that info right after each test. If you’re that kind of “A” math student, you’ll struggle with the ACT because you’ll need to re-learn all of that math that you have forgotten.
  3. In school you’re taught how to do the problems that come up on the test. And then you take the test on those skills. So as long as you study, you should have a good idea how to do those problems correctly. But on the ACT math section, the problems require you to figure out what to do right then and there. You have to “think on your feet” because the problems are often non-traditional in nature. So the ACT math section is testing your overall ability to think mathematically, not just your ability to regurgitate a bunch of stuff you have memorized.
  4. On school math tests you focus on one topic at a time. For example, a test might be on three methods of factoring, and that’s all that you’re tested on at that time. But on the ACT math section, any given problem might require you to use math skills from seemingly unrelated areas of math. Example: one problem might require you to use geometry’s Pythagorean Theorem and also require you to use algebra’s rule for factoring quadratic trinomials. So you have to work in a more fluid, flexible way.For all of these reasons I always cringe when a parent calls me up and tells me that his/er child needs to bring up an ACT math score up by 5 points by studying “really intensely” in the last week before the test.

    In fact, my recommendation is that students start preparing for the ACT no later than the summer before their junior year of high school. That way students get an early feel for the challenges of this particular test, and they can make a plan for re-learning all of the info they need to re-learn. They also have time to learn the strategies they’ll need for this test.

In a future post I will share some examples of actual ACT math problems so you can get a better sense of the situations I’m describing. In the meantime, consider starting out with that idea of summer before junior year as the ideal time to get your child started studying for the ACT math section.

 

 

 

 


If you’re staring at two terms you need to factor, but feel like a deer looking at the headlights of an oncoming semi, here’s a way to leap to safety!

It’s called the “Difference of Two Squares” trick.

High-Octane Boost for Math

It requires four simple steps.

  1. Figure out if each of the terms is a “perfect square.”
  2. If so, take the square root of each term.
  3. Put each square root in its proper place inside two (    ).
  4. Put a + sign inside the first (   ), and put a – sign inside the second (   ).

Let’s do an easy example. Suppose the terms you’re looking at are these:
x^2  – 9

Let’s go through the 4 steps together.

  1. Figure out if each term is a “perfect square.”

    So, what does it mean for a number or term to be a “perfect square”?  It means that you get the number or term by multiplying a number or term by itself. For example, 16 is a perfect square because you can get 16 by multiplying 4 by itself:  4 x 4 = 16.

    So when we look at our two terms, x^2 and 9, we notice that both
    are perfect squares.
    9 is just 3 times 3.
    And in the same way, x^2 is just x times x.

  2.  Take the square root of each term.
    The square root of x^2 is just x.
    And the square root of 9 is just 3.

  3. Put each square root in the proper place inside two sets of (    ).
    We put the square root of the term that was positive first, and the square root of the term that was negative second.Since the x^2 was the positive term, we put its square root, x, first inside each
    (   ).  So far, that gives us:  (x    ) (x     )

    Since the 9 was the negative term because it had the negative sign in front of it: – 9, we put its square root, 3, second inside each (   ). So our (   )s now look like this:  (x   3) (x   3)

  4. Finally, we just need to put in signs that connect the terms inside
    the (    )s.

    That’s easy. We put a + sign inside one (    ), and we put a – sign
    inside the other (    ).
    I prefer to put the + inside the first (   ), but it really doesn’t matter.The final factored form, then, looks like this:  (x + 3) (x – 3)
    That’s all there is to it.

Now try these problems for practice.

           a)  x^2 – 16
           b)  x^2 – 100
           c)   x^2 – 121
           d)   x^4 –  16x^2
           e)   49x^8 – 144y^12

Answers:

           a)   x^2 – 16   =  (x + 4) (x – 4)
           b)  x^2 – 100  = (x + 10) (x – 10)
           c)   x^2 – 121  = (x + 11) ( x – 11)
           d)   x^4 –  16x^2  = (x^2 + 4x) (x^2 – 4x)
           e)   49x^8 – 144y^12  = (7x^4 + 12y^6)(7x^4 – 12y^6)