Is Using a Calculator Necessary for Solving Precalculus Problems?

[ECHOSIDE]

I am making my way through a Precalculus text. It is not for a class, just independent study. I am doing every problem in the book. So far, I have done all of the problems without a calculator to better my arithmetic skills. This is frequently time consuming because the book isn't giving me "nice" numbers. I frequently need to calculate 8 and 10 digit numbers with long division and long multiplication.
I am finding that some of the problems require 10 minutes or more of arithmetic. When I am finished, I frequently end up with the wrong answer because of human error. When this happens, I re-work the problem from the beginning but will frequently make the same arithmetic error(s) the second and third times I work the problem.
I have a bit of a dichotomy.
If my arithmetic is as bad as it seems to be, I need more practice with it, not less. Thus it seems I should continue to work the problems out longhand. (Although it is frustrating and not fun.)
On the other hand, if the risk of human error introduced to the problem is unreasonable, perhaps I should approach the problems like the rest of my buddies and use a calculator.

I assume I am not the only person who has run into this decision. Do any of you have thoughts, experiences, or input?

Thanks!

 

[Jack21222]

Never in real life will you ever have to do 8 digit long division by hand. On the other hand, you should be worried that it takes 10 minutes and you keep getting wrong answers. I'm pretty sure 4th graders can do it more quickly.

 

[olivermsun]

Example problems?

Calculating long multiplication and division for 8 and 10 digit numbers sounds a little excessive if you're trying to polish your precalc concepts.

 

[ECHOSIDE]

Never in real life will you ever have to do 8 digit long division by hand. On the other hand, you should be worried that it takes 10 minutes and you keep getting wrong answers. I'm pretty sure 4th graders can do it more quickly.

Jack21222, perhaps I was unclear. I'm sorry about that.

Example problems?

Calculating long multiplication and division for 8 and 10 digit numbers sounds a little excessive if you're trying to polish your precalc concepts.

Olivermsun, I can give you an example.
This problem took me at least 15 minutes to solve and check.
0.03x-0.02=0.2(x+0.03)

Maybe it shouldn't have taken me so long. I admit that I'm not great at math.

 

[micromass]

I'd say: YES, use a calculator! Calculating 8 digit long divisions by hand is madness, there are better things to put your time in.

Try to judge for yourself, if it's just a calculation like 7²-4.5.2, then don't use a calculator, but if the calculation gets ugly, then there's nothing much to learn by doing evertything by hand.

I do respect your courage, though...

 

[olivermsun]

Olivermsun, I can give you an example.
This problem took me at least 15 minutes to solve and check.
0.03x-0.02=0.2(x+0.03)

A lot of "hand" math depends on making the problem easier.
On this one, for example, you can eliminate some decimals by multiplying everything by 100. You could of course do it keeping the decimals in until the end, but sometimes doing it this way helps to minimize decimal place mistakes:

3x-2 = 20(x+0.03)
3x-2 = 20x+0.6
-2.6 = 17x
And at this point I don't see any reason not to use your calculator, although you could also learn to recognize that you can multiply by 5 to get rid of the remaining decimals :
-13 = 85x
-13/85 = x
which is a perfectly valid answer.

Maybe it shouldn't have taken me so long. I admit that I'm not great at math.

Nothing that can't be fixed up with a little practice. But again, make it easier not harder for yourself.

 

[priquas]

I like to do all my calculations by hand, also when it is frustrating, boring and too easy to make errors, although occasionally I make a check with a calculator. The effort is balanced with a better concentration while working on a problem, the search for better ways in doing calculations and in obtaining better precisions. There is always something to learn while doing hand calculations. Usually I compute also sines, cosines, tangents, square and cubic roots (!).

Strangely, after many years of this practice I found more mechanical to push buttons on a calculator then in doing computations in my head or on paper.

 

[ECHOSIDE]

A lot of "hand" math depends on making the problem easier.
On this one, for example, you can eliminate some decimals by multiplying everything by 100. You could of course do it keeping the decimals in until the end, but sometimes doing it this way helps to minimize decimal place mistakes:

3x-2 = 20(x+0.03)
3x-2 = 20x+0.6
-2.6 = 17x
And at this point I don't see any reason not to use your calculator, although you could also learn to recognize that you can multiply by 5 to get rid of the remaining decimals :
-13 = 85x
-13/85 = x
which is a perfectly valid answer.

Olivermsun, your insight is brilliant - particularly your ability to get rid of the decimals in -2.6=17x! I will keep this in mind.
However, I'm afraid that in this case it has produced an erroneous result. x=-13/85 does not check. The only real solution to this equation is x=-103/85. Perhaps we could identify what caused your method to fail in this case, because I like your method a lot better.

 

[BloodyFrozen]

Olivermsun, your insight is brilliant - particularly your ability to get rid of the decimals in -2.6=17x! I will keep this in mind.
However, I'm afraid that in this case it has produced an erroneous result. x=-13/85 does not check. The only real solution to this equation is x=-103/85. Perhaps we could identify what caused your method to fail in this case, because I like your method a lot better.

Try and show the work, if possible.

Olivermsun's answer is correct, maybe you checked wrong -or- the book, etc. is wrong.

I'm providing a simple check

LHS

3x-2
3(-13/85)-2
-209/85

RHS

20(x+0.03)
20(-13/85+0.03)
20(-209/1700)
-209/85

LHS = RHS
-209/85 = -209/85

For the sake of typing out your answer---

LHS--> -479/85
RHS--> -2009/85

As you can see, they are not equal.

 

[chronon]

-103/85 looks like the answer to 0.03x-0.2=0.2(x+0.03) not 0.03x-0.02=0.2(x+0.03)

 

[priquas]

My approach to the problem:

0.03x - 0.02 = 0.2( x + 0.03 )

0.03x - 0.02 = 0.2x + 0.006

30x -20 = 200x + 6

(200 - 30)x = - 20 - 6

170x = - 26

x = - 26/170 = -13/85

 

[ECHOSIDE]

I'm sorry, Olivermsun, you were correct!
I did not transcribe the problem properly.
Chronon, the error you suspect is the error I made when transcribing.

(embarrassment!)

 

[pwsnafu]

I assume I am not the only person who has run into this decision. Do any of you have thoughts, experiences, or input?
Thanks!

You need to learn http://en.wikipedia.org/wiki/Trachtenberg_system" because the long multiplication and division are just too inefficient.