Expressions, when to add and when not to!

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Probability

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I've been wondering and reached an idea that still might be incorrect!

I have say an expression;

[MATH]{-2a}-({-5a^2})+({-4a})[/MATH]
I'm asked to simplify the expression, thus;

[MATH]{5a^2}-{6a}[/MATH]
But I've been getting confused over when it is correct to square the 'a' terms and when not to!

As in this example, when I say [MATH]{-2a}-({-4a})={-6a}[/MATH] seems correct to me, but if the expression said multiply the terms, this is the time that [MATH]{a}[/MATH] would become [MATH]{a^2}[/MATH] is that correct!
 
Probability said:
I've been wondering and reached an idea that still might be incorrect!

I have say an expression;

[MATH]{-2a}-({-5a^2})+({-4a})[/MATH]
I'm asked to simplify the expression, thus;

[MATH]{5a^2}-{6a}[/MATH]
But I've been getting confused over when it is correct to square the 'a' terms and when not to!

As in this example, when I say [MATH]{-2a}-({-4a})={-6a}[/MATH] seems correct to me, but if the expression said multiply the terms, this is the time that [MATH]{a}[/MATH] would become [MATH]{a^2}[/MATH] is that correct!
Click to expand...
Now sure what you mean. Squaring would change the value of the expression.
 
Probability said:
[MATH]{-2a}-({-5a^2})+({-4a})[/MATH]
I'm asked to simplify the expression, thus;
[MATH]{5a^2}-{6a}[/MATH]
But I've been getting confused over when it is correct to square the 'a' terms and when not to!

As in this example, when I say [MATH]{-2a}-({-4a})={-6a}[/MATH] seems correct to me, but if the expression said multiply the terms, this is the time that [MATH]{a}[/MATH] would become [MATH]{a^2}[/MATH] is that correct!
Click to expand...

Yes.

If I add 2 inches + 3 inches, I get 5 inches (the same kind of quantity).

But if I multiply 2 inches by 3 inches, I get 6 square inches (a new kind of quantity).

But it's not that [MATH]a[/MATH] "becomes" [MATH]a^2[/MATH]; rather, [MATH]a[/MATH] times [MATH]a[/MATH] is [MATH]a^2[/MATH]. You are not "squaring the a terms", but multiplying them, which results in a square.
 
Probability said:
I've been wondering and reached an idea that still might be incorrect!

I have say an expression;

[MATH]{-2a}-({-5a^2})+({-4a})[/MATH]
I'm asked to simplify the expression, thus;

[MATH]{5a^2}-{6a}[/MATH]
But I've been getting confused over when it is correct to square the 'a' terms and when not to!

As in this example, when I say [MATH]{-2a}-({-4a})={-6a}[/MATH] seems correct to me, IT'S NOT but if the expression said multiply the terms, this is the time that [MATH]{a}[/MATH] would become [MATH]{a^2}[/MATH] is that correct!
Click to expand...
[MATH](-2a) - (-4a) = (-2a) + 4a = 4a - 2a = a(4 - 2) = a * 2 = 2a \ne (-6a).[/MATH]
We took advantage of a general rule called the distributive law

[MATH]a(b + c) \equiv (ab) + (bc).[/MATH]
It is a generalization of the numeric fact that

[MATH]7(4 + 5) = 7 * 9 = 63 = 28 + 35 = (7 * 4) + (7 * 5).[/MATH]
Now what does [MATH]-5a^2[/MATH] MEAN?

[MATH](- 5a^2) = (-5)(a^2) = (-5)(a * a) = (-5a)(a).[/MATH]
So it is perfectly valid to say

[MATH]{-2a}-({-5a^2})+({-4a}) = (-2)a - (-5a)a + (-4)a = \\ a\{(-2) - (-5a) + (-4)\} = a(5a - 2 - 4) = a(5a - 6).[/MATH]
Now whether that is SIMPLER than [MATH]5a^2 - 6a[/MATH]
is a matter of opinion or purpose because the two mean the exact same thing.
 
Simplifying expressions containing fractions. I've no written examples of expressions containing fractions so the work below is purely trial and error. I think I can get to the correct solution but not sure the method how I get there is correct!

[MATH]{4s}\times\frac{1}{2}rst{-2}[-\frac{1}{2}]s=[/MATH]
I'm thinking that [MATH]{4s}-{2}={2}s[/MATH] and that [MATH]\frac{1}{2}{rst}-\frac{1}{2}s=\frac{2}{4}-\frac{2}{4}={2}r{s^2}t+s[/MATH]
I'm thinking I'm not too far off on a tangent here but could somebody please offer some guidance in my techniques where I may have gone wrong.
 
JeffM said:
[MATH](-2a) - (-4a) = (-2a) + 4a = 4a - 2a = a(4 - 2) = a * 2 = 2a \ne (-6a).[/MATH]
We took advantage of a general rule called the distributive law

[MATH]a(b + c) \equiv (ab) + (bc).[/MATH]
It is a generalization of the numeric fact that

[MATH]7(4 + 5) = 7 * 9 = 63 = 28 + 35 = (7 * 4) + (7 * 5).[/MATH]
Now what does [MATH]-5a^2[/MATH] MEAN?

[MATH](- 5a^2) = (-5)(a^2) = (-5)(a * a) = (-5a)(a).[/MATH]
So it is perfectly valid to say

[MATH]{-2a}-({-5a^2})+({-4a}) = (-2)a - (-5a)a + (-4)a = \\ a\{(-2) - (-5a) + (-4)\} = a(5a - 2 - 4) = a(5a - 6).[/MATH]
Now whether that is SIMPLER than [MATH]5a^2 - 6a[/MATH]
is a matter of opinion or purpose because the two mean the exact same thing.
Click to expand...
JeffM,

Thanks for the above explanation, however I'm not up to the distributive law just yet as that is next in the work book I think now they are starting to refer to brackets.
 
WHOA

Imagine s is 10. So

[MATH]4 * 10 - 2 = 40 - 2 = 38 \ne 20 = 2 * 10.[/MATH]
The hardest thing for beginning students in elementary algebra to grasp is that the letters stand for numbers. And you know all about numbers. You are just seeing them in a more general and abstract light.

If you ever wonder whether some relationship is generally true, substitute in two numbers (never zero or one) and see whether it is true for both numbers. If it is not, your algebraic manipulation is wrong.

Taking 2 away from four times a number does not always result in twice the number. In this simple case, if you say it in English, you will realize that it is wrong. In more complex cases, just do an experiment with at least two distinct numbers (minus two and plus three are my favorite candidates). If your experiment is wrong for either number, your hypothesis is wrong.

Students typically find fractions in algebra a pain. I do myself. So one thing you can do is to use the distributive law.

[MATH]4s * \dfrac{1}{2} * rst - 2 * \left ( - \dfrac{1}{2} \right ) * s.[/MATH]
I see that both summands contain 1/2, s and a multiple of 2 so all three can be factored out.

[MATH]4s * \dfrac{1}{2} * rst - 2 * \left (- \dfrac{1}{2} \right ) * s = \\ \dfrac{1}{2} * 2 s * \{2 * rst - 1 * (- 1)\} = \dfrac{\cancel 2 s}{\cancel 2} * (2rst + 1) = s(2rst + 1).[/MATH]
That is one way to handle the fractions. Another way is to perform the indicated multiplications and divisions and see whether that eliminates some or all of the fractions. That is the quicker method in THIS case.

[MATH]4s * \dfrac{1}{2} * rst - 2 * \left (- \dfrac{1}{2} \right ) * s =\\ \left (\dfrac{1}{2} * 4 \right ) * s * rst - \left ( \dfrac{1}{2} * 2 \right ) * (-1) * s =\\ 2s^2rt + s = s(2rst + 1).[/MATH]
 
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I can see based on my limited experience with these expressions I need more practice. So I'll take it that the method I used would be considered wrong then.
 
Probability said:
Simplifying expressions containing fractions. I've no written examples of expressions containing fractions so the work below is purely trial and error. I think I can get to the correct solution but not sure the method how I get there is correct!

[MATH]{4s}\times\frac{1}{2}rst{-2}[-\frac{1}{2}]s=[/MATH]
I'm thinking that [MATH]{4s}-{2}={2}s[/MATH] and that [MATH]\frac{1}{2}{rst}-\frac{1}{2}s=\frac{2}{4}-\frac{2}{4}={2}r{s^2}t+s[/MATH]
I'm thinking I'm not too far off on a tangent here but could somebody please offer some guidance in my techniques where I may have gone wrong.
Click to expand...
You write:

[MATH]{4s}-{2}={2}s[/MATH]

How do you get that?
 
Probability said:
Simplifying expressions containing fractions. I've no written examples of expressions containing fractions so the work below is purely trial and error. I think I can get to the correct solution but not sure the method how I get there is correct!

[MATH]{4s}\times\frac{1}{2}rst{-2}[-\frac{1}{2}]s=[/MATH]
I'm thinking that [MATH]{4s}-{2}={2}s[/MATH] and that [MATH]\frac{1}{2}{rst}-\frac{1}{2}s=\frac{2}{4}-\frac{2}{4}={2}r{s^2}t+s[/MATH]
I'm thinking I'm not too far off on a tangent here but could somebody please offer some guidance in my techniques where I may have gone wrong.
Click to expand...
You wrote: 4s - 2 = 2s. This is incorrect. 4s - 2s = 2s but 4s - 2 is only equal to 4s - 2. And I have no idea why you are pointing this out. It doesn't come up in the problem.

Also, you wrote
[MATH]\frac{1}{2}{rst}-\frac{1}{2}s=\frac{2}{4}-\dfrac{2}{4}={2}r{s^2}t+s[/MATH]
[math]\dfrac{1}{2} rst - \dfrac{1}{2}s[/math] is not [math]\dfrac{2}{4} - \dfrac{2}{4} = 0[/math]. Where did the rst go?
and
[math]\frac{2}{4} - \frac{2}{4} = 2 r s^2t + s[/math] is not correct, either. Where did the rst come back from?

Your line is right if you leave out the 2/4s and multiply the first term by 4s.
[MATH]{4s}\times\frac{1}{2}rst{-2}[-\frac{1}{2}]s= 2rs^2t + s[/MATH]is correct.

You really need to proof-read your posts.

I don't know what to call your method here other than order of operations.

-Dan
 
Probability said:
Simplifying expressions containing fractions. I've no written examples of expressions containing fractions so the work below is purely trial and error. I think I can get to the correct solution but not sure the method how I get there is correct!

[MATH]{4s}\times\frac{1}{2}rst{-2}[-\frac{1}{2}]s=[/MATH]
I'm thinking that [MATH]{4s}-{2}={2}s[/MATH] and that [MATH]\frac{1}{2}{rst}-\frac{1}{2}s=\frac{2}{4}-\frac{2}{4}={2}r{s^2}t+s[/MATH]
I'm thinking I'm not too far off on a tangent here but could somebody please offer some guidance in my techniques where I may have gone wrong.
Click to expand...
The "method" above is so misguided - it is not even wrong (to paraphrase a famous physicist - "That is not only not right; it is not even wrong". - Wolfgang Pauli )
 
Subhotosh Khan said:
The "method" above is so misguided - it is not even wrong (to paraphrase a famous physicist - "That is not only not right; it is not even wrong". - Wolfgang Pauli )
Click to expand...
Some years ago I worked with a young lad who was at technical college and about to do his exams. I asked him what exams are you taking! He replied I'm doing my maths exam Thursday. I asked him, are you good at maths and he replied YES. I said that is good then. When he got his results through from college I asked him what grade he got for his maths exam! He replied E. I said to him, E you say, yes the young lad replied, but I then said to him, but you said you are good at maths?

The young lad replied, yes I am, its a different kind of good :LOL:
 
Probability said:
… when I say [MATH]\; {-2a} - ({-4a}) = {-6a} \;[/MATH] [it] seems correct to me …
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Hi Probability. That subtraction looks like a typo. I think you meant to type:

[MATH]{-2a} + ({-4a})={-6a}[/MATH]
?
 
OK guys you should not be so hard on me. I did point out at the beginning that I'd not had any examples of how to do these types of problems, and I did provide an example of how I did the problem, even though it was wrong, I did ask for advice. Please remember also I'm learning this on my own.

So here I have had another go at it now based on your replies above. Please advise how I'm doing this time...

We have;

[MATH]{4s}\times\frac{1}{2}\times rst-{2}(-\frac{1}{2}s)[/MATH]
Now if I look at the expression in two sections and work out each section one at a time, hopefully this will be correct, so here goes;

[MATH]{4s}\times\frac{1}{2}\times rst[/MATH]
I see this as [MATH]({4s}\times rst\times\frac{1}{2})=\frac{4}{2}r{s^2}t=2r{s^2}t[/MATH]
Now if I look at;

[MATH]{-2}(-\frac{1}{2}s)={s}[/MATH]
[MATH]={2}{r}{s^2}{t}+{s}[/MATH]
 
Probability said:
OK guys you should not be so hard on me. I did point out at the beginning that I'd not had any examples of how to do these types of problems, and I did provide an example of how I did the problem, even though it was wrong, I did ask for advice. Please remember also I'm learning this on my own.

So here I have had another go at it now based on your replies above. Please advise how I'm doing this time...

We have;

[MATH]{4s}\times\frac{1}{2}\times rst-{2}(-\frac{1}{2}s)[/MATH]
Now if I look at the expression in two sections and work out each section one at a time, hopefully this will be correct, so here goes;

[MATH]{4s}\times\frac{1}{2}\times rst[/MATH]
I see this as [MATH]({4s}\times rst\times\frac{1}{2})=\frac{4}{2}r{s^2}t=2r{s^2}t[/MATH]
Now if I look at;

[MATH]{-2}(-\frac{1}{2}s)={s}[/MATH]
[MATH]={2}{r}{s^2}{t}+{s}[/MATH]
Click to expand...
Looks good.
 
Otis said:
Hi Probability. That subtraction looks like a typo. I think you meant to type:

[MATH]{-2a} + ({-4a})={-6a}[/MATH]
?
Click to expand...
Otis said:
Hi Probability. That subtraction looks like a typo. I think you meant to type:

[MATH]{-2a} + ({-4a})={-6a}[/MATH]
?
Click to expand...
Hi, if you refer back to my first post you'll see its actually correct, however someone else along the way might of miss typed it?
 
Probability said:
Hi, if you refer back to my first post you'll see its actually correct …
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Hi. I quoted your first post, so that's your typing. (In this forum, one can use the link at the top of a quote to be taken to the source.)

You wrote -2a - (-4a) = -6a

That is not correct. Subtracting a negated quantity is the same as adding the opposite quantity. In other words:

-2 - (-4) = -2 + 4

I'm thinking the subtraction of -4a is a typo because your first expression adds -4a instead of subtracting -4a.

?
 
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