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Archive for the ‘Solving Equations’ Category

How to quickly find the y-intercept (b-value) of a line


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

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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.



“Algebra Survival” Program, v. 2.0, has just arrived!


The Second Edition of both the Algebra Survival Guide and its companion Workbook are officially here!

Check out this video for a full run-down on the new books, and see how — for a limited time — you can get them for a great discount at the Singing Turtle website.

 

Here’s the PDF with sample pages from the books: SAMPLER ASG2, ASW2.

And here’s the website where you can check out the books more fully and purchase the books.

 

 

 

 

 

 

 

How to Factor the Simplest Kind of Quadratic Trinomials (a = +1)


Yep, factoring quadratic trinomials is a key skill for Algebra 1. And the process can seem intimidating, especially at first.

But it’s actually surprisingly easy if taught in a certain way. And of course, that’s what I’m going to do here … teach it in the easiest and fastest way possible.

Believe it or not, there’s a reason teachers make you factor trinomials. They may not have told you yet, but they do this so you can solve equations with quadratic trinomials. Once you can factor one of these little beasts, solving an equation that contains one becomes amazingly simple. But without the ability to factor the trinomial, solving it is much more difficult.

You’ll notice that this video starts with four preliminary concepts. These are pretty simple concepts, and for most of you these will feel like review. But make sure you know all of those concepts before you go on, especially the concept of absolute value.

With these preliminaries “under your belt,” factoring trinomials will be rather easy.

To put this video into perspective, it shows how to factor two of the four kinds of quadratic trinomials, those with the pattern of + + + and + – +. After this video, I will post another that shows how to factor quadratic trinomials with the patterns of + + – and + – –.

Also, my first two videos on factoring trinomials are for trinomials whose a-value = + 1. There’s a different, more complicated process for factoring quadratic trinomials whose a-value is not = + 1.  I’ll go over that in a few later videos.

In any case, this will get you started in a way that shouldn’t feel too painful. Follow along and good luck.

 

 

 

 

How to Transform from Standard Form to Slope-Intercept Form


Is there any point to doing something the long way when you can just as well do it in a shorter, much more efficient way? I say, Heck no! We can do things in the “Triple-F” way:  Fast, Fun & Friendly, and with deep understanding, to boot.

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So in that spirit, today I’ll get us started on quickly and effortlessly converting a linear equation from what’s called standard form (Ax + By = C) to what we know as the good-old slope-intercept form (y = mx + b).

To better grasp standard form, let’s replace its mysterious A, B, and C with actual numbers:  4 for A, 2 for B and 8 for C. That gives us the more typical looking equation of an actual line:  4x + 2y = 8. Do you recall seeing this kind of equation in your algebra text and class? Sure you do. You get this kind of equation in the chapter(s) on the coordinate plane and in other spots, too.

Now usually when books teach us how to convert from this “standard” form to slope-intercept form, they tell us to solve the equation for y. Of course that works, but it takes too darn long.

To understand the quicker way, let’s have a little fun with the standard form of the equation: Ax + By = C

We’re going to start with this standard form and solve that for y. And as you’ll see, we’ll learn some useful things from the result.

To kick things off, we start with Ax + By = C, and we subtract the Ax term from both sides. That leaves us with this equation:

By = – Ax + C

Now take this new equation and divide both sides by B. That gives us this little gem of an equation:

y = (– A/B) x + C/B

I’m going to call this the magic equation both to give us a way to refer to it and to show us what’s so useful about it.

The big insight is that this magic equation is actually, believe-it-or-not, in slope-intercept form; we just need to SEE it that way. Here’s how.

In slope-intercept form (y = mx + b), notice that the y variable is all by its lonesome on the left side. Do we have that in the magic equation? Yes we do. So … CHECK!

In slope-intercept form, there’s a value called m (aka, the slope) that is multiplying the x variable. Do we have something in the magic equation that’s multiplying the x variable? Why yes, and it happens to be
(–A/B). So do we have the slope showing in the magic equation? Yes, the slope is:  (–A/B). So … CHECK!

Finally, in slope-intercept form, there’s a constant (i.e., a number term, not a variable term) that appears after the mx term. So do we have a constant after the mx term in the magic equation? Yes, indeed. We have C/B. Note that in any actual linear equation, B and C will be actual numbers, not variables. So the value you get when you divide B by C (the quotient B/C), also must be a real number, just as surely as the real numbers 8 and 2 gives us the real number 4 when we divide 8 by 2.

So, to address the final question, do we have a b-value in the magic equation? Yes, it’s (C/B). C/B is the y-intercept, the real number we call b in slope-intercept form. So once again … CHECK.

So all in all, do we now have the equation in slope-intercept form? Yes, indeed. You just need to realize that
(–A/B) is the slope, and (C/B) is the y-intercept.

In my next post I’ll show you how you use these results to quickly transform the equation from standard form to slope-intercept form. It will be amazing.

The Log Blog


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Tales from the Tutoring Experience

It’s that time of year again when Algebra 2 students are all studying logs — not the lumberjack things, but ‘logarithms’ — so I’d like to present a concept that helps students work with logs.

I call it the “Whole-Part-Equality Principle” (as I’ve never seen it named by anyone else … is there a name for it? Anyone know?) But I prefer to call it the “Peter-Paul-Pennies-in-the-Pocket Principle.”

Here’s how it works. There’s Peter, and there’s Paul. We are told that Peter and Paul have no money except pennies, and they transport their pennies only in their right and left pants pockets (if anyone can think of a way to pack this story with even more p’s, please let me know).

Anyhow, we know three additional facts:

1)  The number of pennies that Peter is transporting equals the number of pennies that Paul is transporting.

2) Peter and Paul each have three pennies in their right pants pockets.

3)  Peter and Paul transport their pennies NOWHERE but in their pants pockets.

QUESTION:  What can we conclude about the number of pennies that Peter and Paul have in their left pants pockets?

ANSWER:  It’s obvious, right? While we don’t know how many pennies Peter and Paul could be transporting in their left pants pockets (it could be any number, right?), it is nevertheless clear that they must have the same number of pennies in their left pants pockets.

WHY?  View it like this … If the wholes are the same (the total number of pennies that Peter and Paul each has), and if one of two key parts are the same (the number of pennies that Peter and Paul have in their right pants pockets), then the other parts must also be equal (the number of pennies they have in their left pants pockets).

Why am I bringing this up? To point out an important principle.

This same principle — if the wholes are equal, and if one of their two parts are equal, then the other parts must also be equal — can be used to solve many log and exponent problems.

EXAMPLE 1:  Suppose you have this equation:  log x = log 7.2. What can we conclude? Well, the wholes are equal (meaning the left and right sides of this equation are equal), and the bases of the logs are equal (logs are always base 10 unless another base is given), therefore the remaining parts, the ‘arguments,’ also must be equal. The ‘argument’ is the term after the word ‘log,’ so for this equation the arguments are x and 7.2, and they must be equal … meaning that  x = 7.2.

EXAMPLE 2:   Suppose you have the equation:  log 2^x = log 16. Again, the wholes are equal, and the logs have the same base, so the arguments must be equal. That means that 2^x = 16. Since 2^4 = 16, x = 4, and that’s the answer.

EXAMPLE 3:  Suppose you have the equation:   a^log x = a^log 12.9. Since the wholes (the left and right sides of the equation) are equal, and since the bases are equal as they are both ‘a,’ therefore the only remaining parts, the exponents, must also be equal. So this means that log x = log 12.9. Following the same logic as we used in Examples 1 and 2, this means that x = 12.9.

Any questions? If so, please post as a comment. If not, please use this principle, and enjoy its profound practicality. (OK, I’m done.)

Josh Rappaport is the author of the Algebra Survival Guide and Workbook, which together comprise an award-winning program that makes algebra do-able! Josh also is the author of PreAlgebra Blastoff!, an engaging, hands-on approach to working with integers. All of Josh’s books, published by Singing Turtle Press, are available on Amazon.com

The “Unknown” Order of Operations


Talk about a major point that’s usually unspoken …

We make such a big deal out of the Order of Operations in Algebra, and yet there’s a second order of operations, equally important but seldom mentioned.

First, to clarify, the standard Order of Operations (caps on the two O’s to indicate this one) helps us simplify mathematical expressions. It tells us how to take a group of math terms and boil them down to a simpler expression. And it works great for that, as it should, as that’s what it’s designed for.

EXAMPLE:  this Order of Operations tells us that, given an expression like:  – 2 – 3(4 – 10), we’d first do the operations inside PARENTHESES to get – 6, then we’d MULTIPLY the 3 by that – 6 to get – 18. Then we would SUBTRACT the – 18 from the – 2, to get 16. You know, PEMDAS.

But it turns out that there’s another order of operations, the one used for solving equations. And students need to know this order as well.

In fact, a confusing thing is that the PEMDAS order is in a sense the very opposite of the order for solving equations. And yet, FEW people hear about this. In fact, I have yet to see any textbook make this critical point.  That’s why I’m making it here and now: so none of you  suffer the confusion.

In the Order of Operations, we learn that we work the operations of multiplication and division before the operations of addition and subtraction. But when solving equations we do the exact opposite: we work with terms connected by addition and subtraction before we work with the terms connected by multiplication and division.

Example: Suppose we need to solve the equation,
4x – 10 = 22

What to do first? Recalling that our goal is to get the ‘x’ term alone, we see that two numbers stand in the way: the 4 and the 10. We might  think of them as x’s bodyguards, and our job is to get x alone so we can have a private chat with him.

To do this, we need to ask how each of those numbers is connected to the equation’s left side. The 4 is connected by multiplication, and the 10 is connected by subtraction. A key rule comes into play here. To undo a number from an equation, we use the opposite operation to how it’s connected.

So to undo the 4 — connected by multiplication — we do division since division is the opposite of multiplication. And to undo the 10 — connected by subtraction — we do addition since addition is the  opposite of subtraction.

So far, so good. But here’s “the rub.” If we were relying on the PEMDAS Order of Operations, it would be logical to undo the 4 by division BEFORE we undo the 10 with addition … because that Order of Operations says you do division before addition.

But the polar opposite is the truth when solving equations!

WHEN SOLVING EQUATIONS, WE UNDO TERMS CONNECTED BY ADDITION AND SUBTRACTION BEFORE WE UNDO TERMS CONNECTED BY MULTIPLICATION OR DIVISION.

Just take a look at how crazy things would get if we followed PEMDAS here.

We have:  4x – 10 = 22

Undoing the 4 by division, we would have to divide all of the equation’s terms by 4, getting this:

x – 10/4 = 22/4

What a mess! In fact, now we can no longer even see the 10 we were going to deal with. The mess this creates impels us to undo the terms connected by addition or subtraction before we undo those connected by multiplication or division.

For many, the “Aunt Sally” memory trick works for PEMDAS. I suggest that for solving equations order of operations, we use a different memory trick.

I just remind students that in elementary school, they learned how to do addition and subtraction before multiplication and division. So I tell them that when solving equations, they go back to the elementary school order and UNDO terms connected by addition/subtraction BEFORE they UNDO terms connected by multiplication/division.

And this works quite well for most students. Try it and see if it works for you as well.

Josh Rappaport is the author of the Algebra Survival Guide and Workbook, which together comprise an award-winning program that makes algebra do-able! Josh also is the author of PreAlgebra Blastoff!, an engaging, hands-on approach to working with integers. All of Josh’s books, published by Singing Turtle Press, are available on Amazon.com

How to Decrease Algebraic Mistakes – Part 5


This is the fifth in a series of posts on how to help students make fewer mistakes in algebra.

No Mistakes

Let's Reduce Mistakes in Algebra!

So far I have introduced a form of notation I have developed, the double-slash, which looks like this:

//

and I have described some of the ways that students can use it.

I’ll continue the conversation by showing how this notation can help students combine like terms with greater care.

(more…)

How to decrease Algebraic mistakes – Part 4


Combining integers … does any early algebraic skill cause more problems?

If so, I can’t think of one.

Fortunately, though, using the double-slash notation that I’ve been talking about this week helps students make sense of this tricky topic.

No Mistakes

Let's Reduce Mistakes in Algebra!


Even a problem as simple as the following can be made easier with the double-slash:

– 2 + 5 – 3 + 7 – 9

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How to Decrease Mistakes in Algebra – Part 3


When we left off, we were talking about the double-slash, a form of notation I’ve developed that helps students attain greater focus when simplifying algebraic expressions.

With greater focus, students make fewer mistakes. With the double-slash at their disposal, students avoid the mistake of combining terms that should not be combined. In the following example, students use the double-slash twice to simplify an algebraic expression:

     + 8 – 2(3x – 7)

=            + 8   //  – 2(3x – 7)

=            + 8  //  – 6x + 14

=            – 6x  //  + 8 + 14

=            – 6x  + 22

No Mistakes

Let's Reduce Mistakes in Algebra!

By cordoning off the section with the distributive property:  – 2(3x – 7), the double-slash allows students to see it distraction-free. With this heightened level of focus, students are more likely to work out the distributive property correctly, then continue on, simplifying the whole expression with no mistakes.

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How to Decrease Mistakes in Algebra – Part 2


When I left off yesterday, I pointed out a certain kind of algebra expression that tends to lead students to make mistakes. It was an expression like this:

 8 – 2(3x – 7)

I pointed out that students often mistakenly combine the 8 and the 2, to get this:

=            6(3x – 7)

=            18x – 42

The theme of these posts is:  how to help students avoid mistakes in algebra.

No Mistakes

Let's Reduce Mistakes in Algebra!

(more…)