Solving for a Specified Value

holmes1172

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P=2l+2w, I am trying to solve for 'w'

I'm really wanting help clarifying if I'm on the right track or not.
So, P=2l+2w, do I combine the 2l+2w to get 4lw and then divide both sides by 'w'?

[FONT=&quot][/FONT]P/w=4lw/w
then getting P/w+4l?

I'm confused on exactly how to do this problem b/c of the example I am given isn't exactly like this problem.

One thing that is making this class difficult is it is online and sometimes (as now) I cannot find examples that are comparable with whatever problem I'm needing to find a solution to and I want to learn it so I will actually do well on the exams. Any help I would appreciate. I don't ness. want anybody doing the problem for me, but just helping me understand (and for future questions as well) so I can do well in this course. Thanks!

J. Holmes
 
P=2l+2w, I am trying to solve for 'w'

I'm really wanting help clarifying if I'm on the right track or not.
So, P=2l+2w, do I combine the 2l+2w to get 4lw and then divide both sides by 'w'?

P/w=4lw/w
then getting P/w+4l?

I'm confused on exactly how to do this problem b/c of the example I am given isn't exactly like this problem.

One thing that is making this class difficult is it is online and sometimes (as now) I cannot find examples that are comparable with whatever problem I'm needing to find a solution to and I want to learn it so I will actually do well on the exams. Any help I would appreciate. I don't ness. want anybody doing the problem for me, but just helping me understand (and for future questions as well) so I can do well in this course. Thanks!

J. Holmes
You have an equation in three unknowns. It cannot be solved numerically. In general, a numerical solution is only possible (though not guaranteed) if you have n unknowns and n equations. If you are asked to "solve" something, that generally means to find a numeric solution, which in this case is impossible. If you are asked to "solve in terms of w," that can be done and means to restate the equation so that w appears once on only one side of the equation.

Let's try a different example.

\(\displaystyle c = \dfrac{5(f - 32)}{9}\) is the equation for turning fahrenheit temperature to celsius temperature. Solve in terms of f.

\(\displaystyle c = \dfrac{5(f - 32)}{9} \implies 9c = 5(f - 32) = 5f - 160 \implies 5f = 9c + 160 \implies f = \dfrac{9c}{5} + 32.\)

This is the equation for turning celsius into fahrenheit.

So what is the specific question actually asked. Please be complete and exact.
 
This is something I have seen when dealing with polynomials too and it confuses me a bit. How do you know when it is okay to switch items around? To me, it seems like you can do anything to get the answer and it causes me pause because without know when you can move things around I may get the wrong answer. What is the rule on this?
 
This is something I have seen when dealing with polynomials too and it confuses me a bit. How do you know when it is okay to switch items around? To me, it seems like you can do anything to get the answer and it causes me pause because without know when you can move things around I may get the wrong answer. What is the rule on this?
Addition and multiplication of numbers are commutative, meaning

\(\displaystyle a + b = b + a\ and\ a * b = b * a.\)

So any rearrangement involving the terms in a sum or a product is permissible. With me to here?

The use of the minus sign in algebra is admittedly confusing to beginners.

\(\displaystyle 6 - 4 = 2 \ne - 2 = 4 - 6.\)

In the line above, the minus sign is being used to indicate the operation of subtraction, which is not commutative, and also the sign of a number.

However, the minus sign is also used in algebra to indicate the additive inverse of a number.

\(\displaystyle 6 + (- 4)\) means to add 6 and the additive inverse of 4, and addition is commutative.

\(\displaystyle 6 + (- 4) = (- 4) + 6 = - 4 + 6.\)

In any expression of the form \(\displaystyle a - b\), then \(\displaystyle a - b = - b + a\), but \(\displaystyle a - b \ne b - a\) unless \(\displaystyle a = b.\)
 
I can see that now. But then in a problem like the one I have above, how do we know it's okay to separate the 2 from w and divide by it? If I'm making it more complicated than it has to be or reading too much into the method, please let me know. Just want my head around it in case I see problems like this in the future.
 
teeter-totter perspective

This is something I have seen when dealing with polynomials too and it confuses me a bit. How do you know when it is okay to switch items around? To me, it seems like you can do anything to get the answer and it causes me pause because without know when you can move things around I may get the wrong answer. What is the rule on this?

scale.jpgteeter.jpgteeter_totter.jpg

You seem to be asking a very general and interesting question M. Holmes. Offering you another perspective.

An equation is a math "statement" with an "=" sign separating two "expressions". The equal sign means that the numerical weight of the quantity of each expression is the same; it is then said that the "equation is balanced" and that the statement is "true". If that were not the case then the equation would be declared "unequal", "unbalanced" and "false", a daunting denunciation, no?. 1 =1 is true and balanced, 1 = 2 is false and unbalanced.

The four fundamental operations of math are addition, subtraction, multiplication, and division each of these "operations" can be applied to an equation but you must apply them to both side of the equation in order for the equation to remain true and balanced.

For example you can 1 to both side of 1=1 and get 1+1 = 1+1 => 2 = 2, a fair and balanced reporting of the news.

You can can subtract 1 from both sides: 1=1 and get 1-1 = 1-1 => 0=0, fair and balanced

You can can multiply both sides by some number "x": 1=1 and get 1*x = 1*x => x=x, fair and balanced

You can divide both sides by some number "x": 1=1 and get 1(1/x) = 1(1/x) =>1/ x=1/x, fair and balanced BUT ...
only if you publicly announce that x cannot be zero, dividing by zero is "undefined", i.e has no meaning.

So you cannot "do anything" to get the answer (hey, where are your ethics), in fact, you can do only a very few things. Think of an equation as a balance scale, to keep the scale balanced you must add, subtract, multiply, or divide the weight on each pan equally or the balance and thus the truthfulness of the equation is lost.

Final note, besides the four fundamental operations, other operations may be applied to an equation (like the square root operator, or the Log operator), but as always they must be applied equally to both sides of the equation (both expressions), also, each new operation probably comes with restrictions (like the division operation disallowing zero) which must be minded (hate that but what are you going to do?).

hmmm ... in retrospect I may have missed the point of your question, if so sorry.

If p=2l+2w is balanced then

p -2w = 2I +2w -2w =>
p -2w = 2I is balanced

If
p -2w = 2I is balanced then

p -2w -p = 2I-p -2w = 2I-p is balanced

If -2w = 2I-p is balanced then

-2w(1/-2) = (2I-p)(1/-2) => w = -(2I-p)/2 or w = (p-2I)/2, when you distrbute the (-) = (-1)

Checking tip: Find a set of numbers that balance the original equation, oddball numbers that are unlikely to cancel one another out (are incommensurate with one another) , like

I = 1.123, w = 2.2123, so p = 2I + 2w = 2(1.123)+2(2.123) = 6.492, p = 6.492

checking to see if these number still work in the final equation:

w = (p-2I)/2 ,

2.2123 =? (6.492-2*1.123)/2

2.2123 = 2.2123 Hooray !!!! (lights, bells)

If the result had been negative then one would have go back to the middle of the chain of the statements to start closing in on where the imbalance had occurred.




 
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No, no! I appreciate that explanation. I don't like assuming, so it's just good you stated all that. And yes, it does help. Thank you!
 
let a = 1.123, b = 2.123, s = 3.123, t = 4.123, z = 5.123 ... or any other crazy numbers (it is best to use crazy numbers to prevent accidental cancellations)

Hi Dale, is the set of Crazy numbers what we get after removing the Integers from the set of Rational numbers? ;)
 
Updated post.

Hi Dale, is the set of Crazy numbers what we get after removing the Integers from the set of Rational numbers? ;)

Yikes, amended post, because my checking procedure was incomplete. "Crazy numbers" are ones you pull from your imagination because you need numbers that are overwhelmingly, most likely incommensurate. 1.123, 2.123 etc are nice because they are easy to punch into the calculator. Nuts, there is a name for this type of checking algorithm in software, the term floats just beyond reach.

True, many are now spending hours googling "crazy numbers", millions of dollars in lost time. Manifold apologizes to those misled.:(
 
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