The true result for A is -0.1762.

  • Thread starter Andreii
  • Start date
  • Tags
    Mathematic
In summary, the conversation is about solving a basic algebraic equation for the variable A. The equation involves several terms with different letters and numbers, and the person is having trouble finding the mistake they made in their calculations. They have tried multiple approaches and are seeking help to figure out where they went wrong. One suggestion is to use a computer program to check their work, but it is also important to learn algebra by practicing it. The conversation ends with a suggestion to rearrange the terms in the equation to make it easier to solve.
  • #1
Andreii
53
0
Hi, here is my equation of basic mathematic.. I know it looks a bit huge but actually on the paper it isn't. In the equation is one mathematic mistake, I checked several times alone and I am not able to find it, I hope i can get some help and thank you:

If I write like this '' L1 '' (without the space between letter and number) then this means L with index 1. I mean the same for each other letter and number where there is no space between them. Datas are the following: L is 0.25, L1 is 1480, L2 is 1680, L3 is 1660, L4 is 590, F is 21000, A1 is 1661.06, A2 is 1074.665, A3 is 132.665, E1 is 110000, E2 is 110000, E3 is 210000, Z1 is 7 * 10^-6, Z2 is 30 * 10^-6, Z3 is 12 * 10^-6 and T is 63.8

Step Number 1: L = ((A * L1) / (A1 * E1)) + Z1 * T * L1 + ((A * L2) / (A2 * E2)) + Z2 * T * L2 + ((A + F) * L4 / (A2 * E2)) + Z2 * T * L4 + ((A + F) * L3 / (A3 * E3)) + Z3 * T * L3
Step Number 2: A ((L1 / (A1 * E1)) + Z1 * T * L1 + (L2 / (A2 * E2)) + Z2 * T * L2 + (L4 / (A2 * E2)) + Z2 * T * L4 + (L3 / (A3 * E3)) + Z3 * T * L3) + ((F * L4) / (A2 * E2)) + ((F * L3) / (A3 * E3)) = L
Step Number 3: A = (L – ((F * L4) / (A2 * E2)) – ((F * L3) / (A3 * E3))) / (L1 / (A1 * E1)) + Z1 * T * L1 + (L2 / (A2 * E2)) + Z2 * T * L2 + (L4 / (A2 * E2)) + Z2 * T * L4 + (L3 / (A3 * E3)) + Z3 * T * L3)
Step Number 4: A = (0.25 – ((21000 * 590) / (1074.665 * 110000)) – ((21000 * 1660) / (132.665 * 210000))) / (1480 / (1661.06 * 110000)) + 7 * 10^-6 * 63.8 * 1480 + (1680 / (1074.665 * 110000)) + 30 * 10^-6 * 63.8 * 1680 + (590 / (1074.665 * 110000)) + 30 * 10^-6 * 63.8 * 590 + (1660 / (132.665 * 210000)) + 12 * 10^-6 * 63.8 * 1660
Step Number 5: A = (0.25 – 0.104811 – 1.25127) / ( 8.099 * 10^-6 + 0.660968 + 1.42 * 10^-5 + 3.21552 + 4.99 * 10^-6 + 1.12926 + 5.96 * 10^-5 + 1.270896)
Step Number 6: A = (-1.106081) / (6.27673)
Step Number 7: A = - 0.1762

In Step Number 4, I also tried: 7 * 10^6 instead of 7 * 10^-6 and 30 * 10^6 instead of 30 * 10^-6 (in both 30 * 10^-6) and 12 * 10^6 instead of 12 * 10^-6 but I still don't get the true result (for A) – mistake is still there, I just can't find it :frown:
 
Physics news on Phys.org
  • #2
I have no idea of what you are trying to do, or even what your question is.

What are you doing?
 
  • #3
I don't understand where you are having difficulties.

If I understand this correctly it's just a basic algebraic equation, in which you are solving for A.
 
  • #4
I am trying to solve A from this equation... It is beginner's one, so I wonder where have I done the mistake because I can't find it and this is why I made a topic here.

If I understand this correctly it's just a basic algebraic equation, in which you are solving for A.

Yes, I meant this.
 
  • #5
The first step is to get your algebra correct. Putting the numbers in is meaningless without a correct starting point.
is this your 1st step ?


[tex] L = \frac {A L_1} {A_1 E_1} + T Z_1 L_1 + \frac {A L_2} {A_2 E_2} + T Z_2 L_2 + \frac {(A+ F)} {L_4}} {A_2 E_2} + T Z_2 L_4 + \frac {(A+ F)}{ L_3}} {A_3 E_3} + T Z_3 L_3 [/tex]

I question the term [itex] T Z_2 L_4 [/itex] since it does not follow the pattern established by the similar terms.

Anything you can tell us about where you got this expression will also help.

If I read it correctly, your 2nd step does not follow. I would appreciate it if you would click on the above formula to see how to use our LaTex feature. You may also want to look in the Latex thread in the tutorial section. Using this format makes it much easier for us to help you. TIA
 
Last edited:
  • #6
The easiest way to check your work in problem where the math is not difficult but rather tedious; is to use a program such as MathCAD or Matlab. I suggest MathCAD because it is displayed in standard mathematical notation, not like a programming language. Check in your school's computer lab or ask in your math department. Often MathCAD is already loaded on these PCs.
 
Last edited:
  • #7
Oh boy, I am not sure that I can condone using computerised math to avoid learning simple algebra. The only way to learn algebra is to do algebra.
 
  • #8
[tex] L = \frac {A L_1} {A_1 E_1} + T Z_1 L_1 + \frac {A L_2} {A_2 E_2} + T Z_2 L_2 + \frac {(A+ F)} {L_4}} {A_2 E_2} + T Z_2 L_4 + \frac {(A+ F)}{ L_3}} {A_3 E_3} + T Z_3 L_3 [/tex]

Assuming that this is correct (and Integral has a good argument that it doesn't look right) then You need to separte terms that have an "A" from those that don't:
[tex] L = A\left(\frac {L_1} {A_1 E_1}+ \frac {L_2} {A_2 E_2}+ \frac {1} {L_4}} {A_2 E_2}+ \frac {1}{ L_3}} {A_3 E_3} \right)+ T Z_1 L_1 + T Z_2 L_2 + \frac {(F)} {L_4}} {A_2 E_2} + T Z_2 L_4 + \frac {(F)}{ L_3}} {A_3 E_3} + T Z_3 L_3 [/tex]
Now move all that does not involve "A" to the left side (i.e. subtract it from both sides:
[tex] L- \left(T Z_1 L_1 + T Z_2 L_2 + \frac {(F)} {L_4}} {A_2 E_2} + T Z_2 L_4 + \frac {(F)}{ L_3}} {A_3 E_3} + T Z_3 L_3\right) = A\left(\frac {L_1} {A_1 E_1}+ \frac {L_2} {A_2 E_2}+ \frac {1} {L_4}} {A_2 E_2}+ \frac {1}{ L_3}} {A_3 E_3} \right) [/tex]
Finally solve for A by dividing both sides by that number (in parentheses) multiplying A.
 
  • #9
HallsofIvy and Integral: This is not my first step. I tried LaTeX type of writing but it keep saying to me '' LaTeX graphic is being generated. Reload this page in a moment. ''... In my first step is difference (im talking about the one which you typed Integral) in the following thing:
near first '' (A + F) '' (from left to right) is * L4 and under it is '' A2 * E2 ''
and near it is '' + Z2 * T * L4 '' which is the same as '' + Z2 T L4 ''
near second '' (A + F) '' (from left to right) is * L3 and under it is '' A3 * E3 '' and near it is it '' + Z3 * T * L3 '' which is the same as '' + Z3 T L3 ''

Anything you can tell us about where you got this expression will also help.

I got this from mechanic construction - I am trying to figure out tensions in specific cuts ''at'' the summer when the weather is more hot and this means tensions inside the mechanic construction will change, also size of it might change too. But, from the mechanic vision is everything good - i haven't done any mistake but the mistake is in mathematical part of the task. Actually those Z's are temperature's extenses (marked as symbol alpha) but to avoid higher confusion, I marked them as Z's.
 
  • #10
Do the problem in portions!
Assuming the expression in Step 1 is correct, we have:
[tex] L = \frac{A * L1}{A1 * E1} + Z1 * T * L1 + \frac{A * L2}{A2 * E2} + Z2 * T * L2 + \frac{(A + F) * L4}{A2 * E2} + Z2 * T * L4 + \frac{(A + F) * L3}{A3 * E3} + Z3 * T * L3=[/tex]
[tex]Af_{1}+t_{1}+Af_{2}+t_{2}+Af_{3}+t_{3}+t_{4}+Af_{4}+t_{5}+t_{6}\to{A}=\frac{L-\sum_{i=1}^{6}t_{i}}{f_{1}+f_{2}+f_{3}+f_{4}}[/tex]
With the relations:
[tex]f_{1}=\frac{L_{1}}{A_{1}E_{1}}, f_{2}=\frac{L_{2}}{A_{2}E_{2}}, f_{3}=\frac{L_{4}}{A_{2}E_{2}}, f_{4}=\frac{L_{3}}{A_{3}E_{3}}[/tex]
And:
[tex]t_{1}=Z_{1}TL_{1},t_{2}=Z_{2}TL_{2}, t_{3}=\frac{FL_{4}}{A_{2}E_{2}}, t_{4}=Z_{2}TL_{4}, t_{5}=\frac{FL_{3}}{A_{3}E_{3}}, t_{6}=Z_{3}TL_{3}[/tex]

Calculate the t's and f's separately..
Alternatively, note for example, that
[tex]t_{1}+t_{2}+t_{4}+t_{6}=T*10^{-6}(7*L_{1}+30*L_{2}+30*L_{4}+12*L_{3})[/tex]
 
Last edited:
  • #11
Assuming the expression in Step 1 is correct, we have:

Yes, I meant this equation.

Arildno, I tried what you suggested twice (in case if I did a mistake in calculating first time) but A still isn't the true result. If I understood you right then I should make A = (L - T) / F where:
T equals the sum: t1 + t2 + t3 + t4 + t5 + t6
F equals the sum: f1 + f2 + f3 + f4

A = (L - 7.632727) / 0.000086881
A = - 84975.16 :frown:
 
  • #12
Calculate EACH of your terms SEPARATELY!
Let each term have a line of its own; it is totally impossible to see where you might have gone wrong elsewise.
 
  • #13
Arildno I have done exsactly this - I calculated each one separately but I got:

t1 = 0.660968
t2 = 3.21552
t3 = 0.104811
t4 = 1.12926
t5 = 1.251272
t6 = 1.270896
f1 = 0.000008099
f2 = 0.000014212
f3 = 0.00000499
f4 = 0.00005958
 
  • #14
Could you please insert into each line the appropriate form of this:
[tex]t_{1}=Z_{1}*T*L_{1}=7*10^{-6}*63.8*1480=(answer)[/tex]

If you could do this, I'll go over the individual calculations when I have a calculator ready.
 
Last edited:
  • #15
t1 = Z1 * T * L1 = 7 * 10^-6 * 63.8 * 1480 = 0.660968

t2 = Z2 * T * L2 = 30 * 10^-6 * 63.8 * 1680 = 3.21552

t3 = (F * L4) / (A2 * E2) = 12390000 / 118213150 = 0.104811

t4 = Z2 * T * L4 = 30 * 10^-6 * 63.8 * 590 = 1.12926

t5 = (F * L3) / (A3 * E3) = 34860000 / 27859650 = 1.251272

t6 = Z3 * T * L3 = 12 * 10^-6 * 63.8 * 1660 = 1.270896

f1 = (L1) / (A1 * E1) = 1480 / 182716600 = 0.000008099

f2 = (L2) / (A2 * E2) = 1680 / 118213150 = 0.000014212

f3 = (L4) / (A2 * E2) = 590 / 118213150 = 0.00000499

f4 = (L3) / (A3 * E3) = 1660 / 27859650 = 0.00005958
 
  • #16
I use MathCAD as a check, when solving long sets of interdependent equations. It is easy to make small errors (either in the algebra or in plugging in the numbers) that can carry over to the rest of the calculations. In using MathCAD as a tool (like a calculator) for checking tedious calculations, it comes down to the same idea as what alrildno and andreii are working on. Breaking down the calculation into smaller terms and checking those results (analogous to using breakpoints in debugging a program). Computerized math should not be used to circumvent learning algebra.
 
Last edited:
  • #17
Andreii said:
A = (L - 7.632727) / 0.000086881
A = - 84975.16 :frown:
My derivation agree's with Arildno's algebra. I too solved each term (in MathCAD) to compare with you and they agree. I also tried the final calculation and obtain an answer similar to yours A = -84969.33, so your method looks correct. The other things to double check is that you didn't make any typos in transcribing the original equations or constants.

Andreii said:
I still don't get the true result (for A)
what is the true result?
 
Last edited:

FAQ: The true result for A is -0.1762.

What is a basic mathematic equation?

A basic mathematic equation is a mathematical statement that uses symbols and numbers to show the relationship between two or more quantities. It can be solved to find the value of one or more variables.

What are the different types of basic mathematic equations?

Some common types of basic mathematic equations include linear equations, quadratic equations, exponential equations, and trigonometric equations. Each type has its own set of rules and methods for solving.

How do I solve a basic mathematic equation?

The first step in solving a basic mathematic equation is to identify the variables and constants. Then, use the appropriate method or formula to solve for the unknown variable. It is important to follow the correct order of operations when solving equations.

What is the importance of basic mathematic equations in science?

Basic mathematic equations are essential in science as they help describe and predict physical phenomena. They are used in fields such as physics, chemistry, and biology to model and understand natural processes.

Can I use basic mathematic equations to solve real-world problems?

Yes, basic mathematic equations can be applied to solve real-world problems such as calculating distances, determining rates, and predicting outcomes. They provide a powerful tool for making accurate and precise calculations in various fields.

Similar threads

Replies
2
Views
2K
Replies
2
Views
2K
Replies
1
Views
2K
Replies
3
Views
8K
Replies
7
Views
5K
Back
Top