## SAT Calculator Strategies – SAT Test Prep #5

The two best SAT prep books of all time? We think so.

Filed under Difficulty: Easy, SAT, Test Prep, TI-83 Plus, TI-84 Plus, TI-89 | Leave a comment

## How to Round Numbers on the TI-84 Plus and TI-89

Sometimes it’s useful to work in units with only a certain number of decimals. For example, if you’re doing money calculations, you probably don’t want to think about fractions of cents. If you want to change the number of decimals your calculator uses, do the following:

TI-83 Plus and TI-84 Plus

1. Press to enter the mode menu.
2. Use the arrow keys to navigate to the line which says “FLOAT” and lists digits from zero to nine.
3. Select the number from the list corresponding to how many digits you would like. FLOAT corresponds to the full number of digits.
4. Press to confirm.
5. Return to the home screen by pressing 2nd .

TI-89 Series

1. Press to enter the mode menu.
2. Use the arrow keys to select the “Display Digits” menu and press .
3. Select the option you would like. Float before a number in this case means that the number is rounded to that number of digits. Use FIX to determine the number of decimal points to use.
4. Press to confirm.
5. Return to the home screen by pressing 2nd .

There is also a round( function on both the TI-84 and TI-89 series calculators which can be reached through the MATH menu. In this case, first enter the number you want to round (or the variable that contains it), press , then enter the number of digits. Finish off with a closing parenthesis and then press to evaluate. Happy rounding!

Filed under Algebra, Difficulty: Easy, TI-83 Plus, TI-84 Plus, TI-89, TI-92 Plus | Leave a comment

## TuesDAZED / Vintage Calcblog, Reader Questions, and Some Musings

I usually use Calcblog’s Twitter for less-official communication, but as I always have lots of thoughts on education and how to improve Calcblog, I thought I’d write a post which is less of a tutorial and more of a true blog entry. If you like this, let me know, and I’ll start doing it more often.

As an introductory statement, as always, I love to hear from readers via our contact form, Twitter account, or Facebook page. I especially think Facebook is a great platform for discussion, though most of our users follow us on Twitter. Once we get more followers on Facebook, I’d like to open up a math help discussion there or maybe hold some contests. Of course, if you haven’t already followed us on either of these networks, we’d love to have you join us!

Occasionally, readers send in questions related to topics not covered in our tutorials. Of course, some of these questions are simpler than what a normal-length article would warrant, but I’d like to share the answers with all of our readers.

Question: How do I enter sec (or csc and cot) on my TI-89?

Answer: There is no secant, cosecant, or cotangent function on the TI-89 (the same goes for the TI-84 Plus and most other calculators), but you can easily enter these functions as the inverses of regular trigonometric functions:

secant(x) = 1/cosine(x)

cosecant(x) = 1/sine(x)

cotangent(x) = 1/tangent(x)

If you’d like to learn more about entering logarithms of different bases on your calculator, a similar question, check out our quick tip on this.

Question: I’m trying to graph the equation r=2-2*sin(θ) on my TI-84 calculator, and it looks all jagged, which isn’t what we learned in class. What’s going wrong?

Answer: Most likely, you need to set your calculator’s “theta step” to a smaller value, which can be reached by pressing the button. Typically, the TI-84 Plus and TI-84 plus use π/24, which is about .13, and graph over the range of θ = 0 to 2*π, so you have (2*π – 0)/(π/24) = 48 points plotted on your graph. (Remember, your calculator graphs by plotting points just as if you were graphing by hand.) If your theta step value is larger, you could be looking at angles on the order of 10-100 degrees between points, which is going to give you a jagged graph which in some cases may not even look like the function it’s supposed to represent.

Window Menu for Polar Graphing on the TI-84 with standard settings for θmin, θmax, and θstep.

## Thoughts

Polar graphing can be somewhat difficult to learn at first, but it’s incredibly powerful, and it’s useful for students to grasp the idea that there is more than one way of representing a point in space. (Many students only think about Cartesian coordinates.) Graphing, as well, is a very useful way of looking at data, and it’s an important skill to understand what the graph of a function tells us. This has applications in science, statistics, and many aspects of daily life, so it isn’t simply some pedantic idea used to make students suffer.

If there’s enough interest, I’ll make a video about polar graphing on either the TI-84 or TI-89 (or both). Making videos like this is time consuming, but can be very useful since it shows students exactly how to go through the process. Speaking of which, if you haven’t already seen it, check out and share our video on graphing on the TI-84 Plus. It’s amazing to think about all the articles Calcblog has after only being around for about 10 months, and this was certainly one of my favorites:

This made me think about some of the other great articles we’ve posted…here are some of my favorites:

Quadratic Formula Programming Tutorial on the TI-83 Plus and TI-84 Plus or on the TI-89

Linear Regression on the TI-83 Plus and TI-84 Plus or on the TI-89

Z-Testing on the TI-83 Plus, TI-84 Plus, and TI-89

I like these articles because they teach a new concept from beginning to finish, and these concepts are ones which are used in many different disciplines. Of course, we have lots of other great articles, and I can’t even remember them all! If you’d like, take your own journey through the Calcblog archives and let me know which articles you like the most or have used personally.

## Other News

I’d like to thank everyone who has purchased our book, Ace the SAT Using Your TI Calculator, and especially those who have given us so much positive feedback. I feel extremely blessed that I am able to help people learn and excel and hope to continue this into the feature. If you haven’t read Ace the SAT yet and would like to order a copy for you or as a gift for someone else, you can find a link to its Amazon.com listing on our sidebar.

Thank you for following Calcblog!

Eric

## Algebra Mini-Series #5: Fraction Mistakes, Cross Multiplying, and Solving Rational Equations

This is the final installment of our weeklong series on improving your algebra skills. Our goal for this series was to provide a useful resource for both students and teachers, so that this article can be used in the classroom, for test prep, or to help yourself practice and master skills you never learned. We hope you’ve found this helpful, and that we can do more series like this one in the future.

Back to: Part 4: How and When to Multiply in Algebra

## Five Tricks to Improve Your Algebra Skills

### Part 5: Common Fraction Mistakes and Clearing Denominators

An old joke reads “4/3 of people have trouble with fractions.” Of course, were this really true then nobody would get the joke, but there is unfortunately a lot of truth in this witticism nonetheless. If I had a dollar for every time I’ve seen a student in precalculus and beyond mess up a problem because they don’t know how to deal with fractions, I would be a very rich man.

Which this in mind, I urge all teachers of math to spend their first class each term “re-teaching” students how to use fractions. Chances are, at least a third of your class is not really comfortable with fractions at all, and most of the other students could at least use a refresher. In this light, this article isn’t really about learning anything new per se. Instead, it focuses on a few of the things about fractions many students aren’t familiar with (even though they should be), and then on clearing denominators to solve equations with fractions.

## Common Fractions Mistakes

### Mistake: Not Multiplying Through

Problem:
3 * (x+2)/(x-4) = ?

Incorrect Method:

Correct Method:

I’ve often seen this mistake in cases where students understand perfectly how to distribute, but somehow dealing with fractions scares them to the point of not knowing what to do with the factor of 3.

### Mistake: Splitting Up Fractions Incorrectly

Problem:
(3x + 2)/(3x + 3) = ?

Incorrect Method:

Correct method:

This mistake is also extremely common. The cause is simply unfamiliarity with fractions so that rules of multiplication and addition become confused.

### Mistake: Dividing Fractions by Nested Fractions Incorrectly

Problem:
2/(4/3) = ?

Incorrect Method:

Correct Method:

This one is also often missed by students, usually because it isn’t in the typical form of (a/b) / (c/d), which more students solve correctly. It’s best to review both of these, especially if students are going to be working with rational equations.

### Mistake: Not Knowing How to Deal With Negative Signs in Fractions

Problem:
2/-4, -3/7, etc…

Correct Method:

This is absolutely fundamental, though many students still don’t understand this. If you’re a student, make sure you know this one—it’s as often missed as the ones above, but is absolutely crucial to good algebra.

Other important topics to review include factoring and dividing numbers and expressions and rationalizing denominators using conjugates, since these too are often missed despite their usefulness.

## Solving Rational Equations by Clearing Denominators

The easiest way to teach solving rational equations is by clearing denominators. Take the following for example:

Solve for x:
1/(x-1) – 1/(x+1) = 1/(x)

Solution: Multiply all terms by (x-1)(x+1)(x) and expand. This gets rid of all equations and turns the problem into a polynomial equation:

(x+1)(x) – (x-1)(x) = (x-1)(x+1)
x² + x – x² + x = x² -1
2x = x² – 1
x² – 2x – 1 = 0

By the quadratic formula, x = 1 +/- √(2)

For simple equations consisting of only two terms, clearing denominators can easily be condensed into the mnemonic of “cross multiplication:”

While it’s useful as a student to understand this technique, the ultimate goal of course is not to learn to solve such equations by rote memorization. The ultimate goal should be the same as always: to develop a problem solving skillset aimed at isolating relevant quantities to get equations into a form easy to solve.

Filed under Algebra, Difficulty: Easy | 2 Comments

## Algebra Mini-Series #4: Use Multiplication in Equations Wisely

This is part four of our weeklong series on improving your algebra skills. Our goal for this series is to provide a useful resource for both students and teachers, so that this article can be used in the classroom, for test prep, or to help yourself practice and master skills you never learned. With these tips, we hope you can make your algebra cleaner, faster, and more intuitive. Check back tomorrow for another informative article on some aspect of elementary algebra!

Back to: Part 3: Using Substitution to Solve Equations

Forward to: Part 5: How to Use Fractions with Algebra Correctly

## Five Tricks to Improve Your Algebra Skills

### Part 4: Know When (and how) To Multiply Large Numbers Together

Oftentimes big numbers can make a simple algebra problem nasty. This, of course, should never be something to hold you back, but students aren’t usually taught how to deal with numbers in equations. Rather it’s assumed that this is the easy part and that the algebra involved is the meat of the problem. If you’re ever faced with a situation where it’s the numbers that are tripping you up, here are two easy ways which can help you to deal with large numbers more quickly and effectively.

### A Quick Trick to Multiply Large Numbers

Quick! What’s 19*32? Did you guess 608? Or did you grab a sheet of paper and start working it out by hand? Typically guessing isn’t going to get you anywhere, and you don’t want to start writing out long multiplication in the middle of a math test. Here’s an easier way to do this: rewrite the numbers as (20-1)(30+2) and distribute (or FOIL) them out. 19*32 = 20*30 + 2*20 – 1*30 – 1*2 = 600 + 40 – 30 – 2 = 608. I bet you never thought you’d use the distributive property this way! (This is actually a fault of our educational system—most students are taught so much abstraction that they forget all the X and Ys they’re dealing with are actually numbers.)

After breaking the number up, we didn’t have to do any long multiplication at all, but mostly addition and subtraction. The reason that I chose 20 – 1 for 19 is that we would like our digits to all be as small as possible—you could also try 10 + 9 and this would work too. In fact, this way prevents you from having to deal with negatives, so it’s really your choice. One advantage of the first method is that you’ll never have to deal with multiplying digits larger than 5.

Let’s try this on the following problem:

Solve the following for y:
22x = (y + x)/17 – 1

First, move the 1 to the left side:

22x + 1 = (y + x)/17

Next, clear the denominator by multiplying through:

22x*17 + 1*17 = y + x

At this point, we unfortunately have to multiply out 22*17 since we’re going to have to subtract x from the left side. So let’s do it the “easy” way: (20 + 2)(20 – 3) = 400 – 60 + 40 – 6 = 374.

374x + 17 = y + x

373x + 17 = y

Unfortunately, this was the easiest way to do this problem, as we can see from the fact that 373 is prime. But sometimes it’s possible to avoid having to do this multiplication altogether. Take the following for example:

If y = 17/21, then (y*60 + 17/7)/3 =?

Plugging in y, we have (17*60/21 + 17/7)/3. But let’s not multiply 17*60 just yet..rather, distribute the /3 to get (17*60/21)/3 + (17/7)/3 = 17*20/21 + 17/21. Again, don’t multiply 17*20 and add 17—this is just 17*20 + 17 = 17*21. (Don’t forget what multiplication means!) So our answer is 17*21/21 = 17.

While the cases where this is useful may at first seem limited, waiting to multiply numbers until the end can greatly reduce the amount of mental grinding you have to do to solve a problem and will often result in cancelled factors. Using this in conjunction with the above technique should make dealing with numbers in your equations a lot simpler.

Challenge problem: 211*39=?

Solution: 211 = 200 + 10 + 1, 39 = 40 – 1, so 211*39 = (200 + 10 + 1)*(40 – 1) = (200 + 10 + 1)*40 – (200 + 10 + 1) = 8000 + 400 + 40 – 200 – 10 – 1 = 8000 + 200 + 30 – 1 = 8229.

Filed under Algebra, Difficulty: Easy | 2 Comments

## Algebra Mini-Series #3: Using Substitution To Solve Equations

This is part three of our weeklong series on improving your algebra skills. Our goal for this series is to provide a useful resource for both students and teachers, so that this article can be used in the classroom, for test prep, or to help yourself practice and master skills you never learned. With these tips, we hope you can make your algebra cleaner, faster, and more intuitive. Check back tomorrow for another informative article on some aspect of elementary algebra!

Back to: Part 2: Moving Quantities Left and Right in Equations

Forward to: Part 4: How and When to Multiply in Algebra

## Five Tricks to Improve Your Algebra Skills

### Part 3: Solving Equations Using Substitution

As we discussed in our first article, substitution is a great way to make complicated equations simpler. However, it’s also a great technique to actually bring difficult and even “unsolvable” equations well within reach. Consider the following for example:

Solve for x:
(x²+2x+3)² – 4*(x²+2x+3) + 4 = 0

Were we to distribute out the terms of this equation, we would get an ugly quartic polynomial. Instead, let’s try the following. Let y = x²+2x+3, then:

 Step Reason y² – 4y + 4 = 0 Using Substitution (y-2)² = 0 Factoring the above expression y = 2 By the zero product property x²+2x+3 = 2 Back substitute for y x²+2x+1 = 0 Moving the 2 to the left (x+1)² = 0 Factoring the above x = -1 By the zero product property

This allowed us to solve a nasty expression using elementary algebra. This technique can also be used to help solve equations of the form a*f(x)² + b*f(x) + c = 0 by replacing f(x) with another variable and using the quadratic formula. Let’s try this on the following classic problem:

Solve for x:

First, multiply through by e^x

Next, let y=e^x

Then just factor or use the quadratic formula.

Back substitute for y.

This technique is one which is often forgotten in favor of simply covering more new material, though it is extremely useful in solving a variety of difficult equations, in the same way that differentiating under the integral sign is ignored despite being useful for solving many otherwise near-impossible calculus problems. If you’re teaching algebra, try introducing it to your students by first asking them to solve an equation like the following:

To them, this equation will probably seem difficult, but substitution is certainly the most natural way to solve such a problem since it does not immediately lend itself to factoring.

Filed under Algebra, Difficulty: Easy | 1 Comment

## Algebra Mini-Series #2: Moving Quantities Left and Right in Equations

This is part two of our weeklong series on improving your algebra skills. Our goal for this series is to provide a useful resource for both students and teachers, so that this article can be used in the classroom, for test prep, or to help yourself practice and master skills you never learned. With these tips, we hope you can make your algebra cleaner, faster, and more intuitive. Check back tomorrow for another informative article on some aspect of elementary algebra!

Back to: Part 1: Replacing Large Expressions With a Single Variable

Forward to: Part 3: Using Substitution to Solve Equations

## Five Tricks to Improve Your Algebra Skills

### Part 2: Move Quantities Left and Right Instead of Adding to Both Sides

What you’re about to learn is one of the most fundamental ways many effective students differentiate themselves from average students. This technique might take some getting used to, but it is so important that it can mean the difference between being enjoying learning math and having to put in many frustrated hours just to keep up with your peers. So what is it? Let’s begin by thinking about the following equation:

x + 5 = 7

How would you solve this? Most students initially learn to solve this equation by “adding” -5 to both sides to cancel the five on the left and preserve overall equality. Of course, this is mathematically sound, but just as though you have to take off your training wheels before you can learn to bike up a mountain, it’s likewise necessary to move on from this technique before attempting any mathematical Tour de France.

The unfortunate truth is that solving equations this way is an unnatural contrivance, a convention taught to make elementary teachers’ lives easier while eventually causing headaches for secondary teachers who wonder why their students can’t handle anything beyond linear single-variable algebra. If you ask a bright child who has not yet had the experience of learning classroom algebra the equivalent question “five plus what number is seven,” they will almost certainly think for a minute and then give you one of the following two responses:

1. 2, because 2 + 5 is 7
2. 2, because that’s 7 – 5.

Of course, the first answer simply demonstrates guessing, which isn’t going to help a student to find the roots of a cubic polynomial with irrational coefficients, but the second one, on the other hand, provides insight into how most people’s brains naturally work. The child will most likely not be able to tell you how they arrived at the second answer, but rather that that’s just the way it works. Of course, this is equivalent to subtracting 5 from both sides, but it skips this step altogether, and as a result saves time and effort.

What that child intuitively demonstrates is this: when a quantity is added or subtracted to one side of the equation, it’s easiest to isolate the variable by moving quantities from one side to the other and flipping their sign. In a sense, all we’re doing is putting the things we don’t care about in one box and leaving everything we do care about in another. While this is inherent in the “adding a number to both sides” method as well, the means does not make intuitive sense to many students since we’re adding and subtracting things which “aren’t there” with no natural end in mind. Solving our equation using our new method proceeds as such:

x + 5 = 7
x = 7 – 5
x = 2

The end result that this method is both more natural and generally more efficient. The key is twofold. First, for this problem, it doesn’t seem much faster (if at all) to use this method, but in fact, with practice, the intermediary step is completely eliminated, whereas when adding to both sides, students will never get away from writing “-5″ beneath both sides of the equation, costing time and causing clutter. Second, for more complicated problems this becomes significantly faster, both for mental math and for reducing the amount of work needed on paper to solve a problem. Consider the following problem (which has only one degree of additional difficulty over the above equation) and the difference in solving it using both methods.

Old Method:
2x + 5 = x + 7
-x    -x
x + 5 = 7
-5   -5
x = 2

New Method:
2x + 5 = x + 7
x + 5 = 7
x = 2

It might not seem like much, but we’ve ultimately reduced the amount of work necessary by a factor of about one half. The more notable expression of this comes from personal experience: I have never had a student ask me for help with basic algebra who does NOT use the “old” method, and conversely I almost never have a student who DOES use this “moving left and right” method need help with algebra. If you’re trying to practice this technique for yourself, the best thing you can do is sit down with a list of simple algebra problems and run through them. With time, it will become almost automatic and will greatly increase your equation solving efficiency.

Bonus: The same is also true of multiplication and division, except instead of switching positive negative with switch numerators and denominators. However, you need to be a bit more careful. Take the following for example:

Solve for y:
3x = y/5 – 1

Clearly, y = 15x + 5, but if we’re going to move a fraction from one side to another, it needs to be the only fraction on one side, so first we have to move the -1 to the left side:

3x + 1 = y/5
15x + 5 = y

Filed under Algebra, Difficulty: Easy | 1 Comment

## Algebra Mini-Series #1: Replacing Large Expressions with a Single Variable

This is part one of our weeklong series on improving your algebra skills. Our goal for this series is to provide a useful resource for both students and teachers, so that this article can be used in the classroom, for test prep, or to help yourself practice and master skills you never learned. With these tips, we hope you can make your algebra cleaner, faster, and more intuitive. Check back tomorrow for another informative article on some aspect of elementary algebra!

## Five Tricks to Improve Your Algebra Skills

### Part 1: Replace Large Expressions with a Single Variable When Possible

Sometimes you might have to solve a particularly nasty looking equation which doesn’t actually require much difficult algebra, but is nonetheless difficult due to its complexity. This is true in practically all high school and college math, as well as in chemistry, physics, and other disciplines. If you’re going to be copying down ugly numbers again and again, why not replace them with a simpler placeholder until you aren’t ready to calculate everything together at the end? Consider the following example:

Solve for y:
(x+√(1/x+x²))*y + (x³+2)/y = y

If we replace everything besides our y variables with the following, the equation takes a much nicer form.

Let:
a = (x+√(1/x+x²))
b = (x³+2)

Then we just need to solve:

ay+b/y=y
ay-y=-b/y
(a-1)y = -b/y
(a-1)y² = -b
y² = -b/(a-1)
y = ±√(-b/(a-1))
y = ±√(-(x³+2)/(x+√(1/x+x²)-1))

Of course, this answer isn’t nice at all, though if we hadn’t swapped out its components for simpler variables, it would have been even harder to solve by hand without making a mistake. As an added bonus, we don’t have to copy down a and b each time we write down a step, which saves time—and on standardized tests such as the SAT, or any other math test as well, time is invaluable.

Of course, the one down side of this technique is that it might be confusing at first to have so many variables in your equation, but this isn’t really an issue for a variety of reasons:

• As math classes get more advanced, this happens anyway, so if you’re not already used to this, it’s best to prepare early, which will give you a leg up on your peers.
• Relatively speaking, single letters are much nicer than the nasty expressions they replace.
• They’re a lot quicker to write, and it’s identical to algebra with normal variables.

Just remember the basic technique—replace coefficients and expressions which do not include the variable of interest (in the above case, y, but often x) with letter variables, solve, and then back substitute. If you’ve taken integral calculus before, you’ll recognize this as a simpler but just as powerful version of u-du substitution.

Filed under Algebra, Difficulty: Easy | 1 Comment

## SAT Prep #4: Using Your Calculator’s Fraction Features

Both the TI-83/TI-84 Plus and the TI-89 have built-in features for dealing with fractions, which can help to solve many of the fractions problems on the SAT. For example, on the TI-84, you can convert a decimal into a fraction by using the ▶Frac function. Simply type in a number, or else use 2nd to paste the Ans token and select your previous answer, then press to select the ▶Frac function, and again to evaluate.

On the TI-89, almost all math is done symbolically, so if you evaluate an equation which does not involve decimals, you will receive an exact answer. If you want to convert a fraction into a decimal, enter the fraction, press , then to evaluate. Try the following problem:

Sample Problem: Student Produced Response. When the fraction 441/333 is reduced to lowest terms, what is the sum of the numerator and denominator?

This problem can take a fair amount of effort to work out by hand unless you immediately see how the numbers factor, and it’s easy to make a mistake. However, using your calculator makes it quick and painless. On the TI-84, type 441/333 then press to evaluate it as a fraction in lowest terms. On the TI-89, just type the fraction in and press , and it’s automatically simplified for you. 441/333 = 49/37, so the answer is 49+37 = 86.

One more tip: if you would like to convert a decimal to a fraction on the TI-84, you can use the exact() function, which converts a decimal into its “exact” fractional form. This function can be reached by pressing 2nd to select the math menu, then pressing . To learn more about the exact function and other techniques for dealing with fractions on your calculator and for more practice, check out our book!

Filed under Difficulty: Easy, SAT, Test Prep, TI-83 Plus, TI-84 Plus, TI-89 | Leave a comment

## Ace the SAT – Our New Test Prep Book!

We’re very proud to announce Calcblog’s test preparation book, Ace the SAT Using Your TI Calculator: Using the TI-84 Plus and TI-89 to Crack the SAT Math Section! This book is packed with tips, tricks and test taking strategies on how to use your graphing calculator as a test taking tool to boost your SAT score on the Math section of the SAT exam.

The SAT is practically a rite of passage for college-bound teens, though many people do not realize its importance. The SAT is used not only for college acceptance, but also for determining scholarship cutoffs—if you spend 30 hours studying for the SAT and get a \$120,000 scholarship, that’s a salary of \$4,000 an hour. This book could be worth 10,000 times its cover price! Of course, nobody from College Board is going to tell you that the best way to do well on the SAT is to use your calculator when it’s supposed to be a test that evaluates your so-called “mathematical reasoning.” This well-kept secret can help you to vastly improve your confidence and your score.

To learn more about how this book can benefit your test prep regimen, check out our article, or order Ace the SAT from Amazon.com.

Filed under SAT, Test Prep, TI-83 Plus, TI-84 Plus, TI-89 | Leave a comment