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-2

$begingroup$

I have put together an equation that has other equations in it. When it is all condensed together it looks like this with a set of given information. I'm trying to proof it out, but can't seem to work out the math to what X would equal. Normally my math is pretty decent, but I can't seem to get the reverse of it to solve. Anybody able to assist?

$$1 = Bigl(0.003491 – fracX cdot 0.003366 + 200 cdot 0.003491 (X + 200) cdot 0.003491Bigr) cdot 100$$

algebra-precalculus

share|cite|improve this question

edited Mar 15 at 18:51

Vasya

4,1081618

asked Mar 15 at 18:32

Pantheon Engineering DesignsPantheon Engineering Designs

1

$endgroup$

closed as off-topic by Henrik, Vinyl_cape_jawa, Lee David Chung Lin, dantopa, Aweygan Mar 16 at 1:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Vinyl_cape_jawa, Lee David Chung Lin, dantopa, Aweygan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Multiply both sides of the = by the denominator of the fraction, particularly the $X+200$
  $endgroup$
  – Alan Baljeu
  Mar 15 at 18:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this what you want? wolframalpha.com/input/?i=1+%3D+(0.003491+%E2%80%93+%5B(%5B(X+*+0.003366)+%2B+(200+*+0.003491)%5D+%2F+(X+%2B+200))+%2F+0.003491%5D)+*+100
  $endgroup$
  – ensbana
  Mar 15 at 18:49
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you just multiply both sides by $(X+200) cdot 0.003491$ you will get a linear equation in $X$ which is easy to solve
  $endgroup$
  – Aditya Dua
  Mar 15 at 19:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you really just multiply both sides by ((X + 200) x 0.003491)? With it being within the parenthesis even?
  $endgroup$
  – Pantheon Engineering Designs
  Mar 15 at 19:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
  $endgroup$
  – dantopa
  Mar 16 at 0:34
  

  
  
  
  

add a comment | 

-2

$begingroup$

I have put together an equation that has other equations in it. When it is all condensed together it looks like this with a set of given information. I'm trying to proof it out, but can't seem to work out the math to what X would equal. Normally my math is pretty decent, but I can't seem to get the reverse of it to solve. Anybody able to assist?

$$1 = Bigl(0.003491 – fracX cdot 0.003366 + 200 cdot 0.003491 (X + 200) cdot 0.003491Bigr) cdot 100$$

algebra-precalculus

share|cite|improve this question

edited Mar 15 at 18:51

Vasya

4,1081618

asked Mar 15 at 18:32

Pantheon Engineering DesignsPantheon Engineering Designs

1

$endgroup$

closed as off-topic by Henrik, Vinyl_cape_jawa, Lee David Chung Lin, dantopa, Aweygan Mar 16 at 1:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Vinyl_cape_jawa, Lee David Chung Lin, dantopa, Aweygan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Multiply both sides of the = by the denominator of the fraction, particularly the $X+200$
  $endgroup$
  – Alan Baljeu
  Mar 15 at 18:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this what you want? wolframalpha.com/input/?i=1+%3D+(0.003491+%E2%80%93+%5B(%5B(X+*+0.003366)+%2B+(200+*+0.003491)%5D+%2F+(X+%2B+200))+%2F+0.003491%5D)+*+100
  $endgroup$
  – ensbana
  Mar 15 at 18:49
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you just multiply both sides by $(X+200) cdot 0.003491$ you will get a linear equation in $X$ which is easy to solve
  $endgroup$
  – Aditya Dua
  Mar 15 at 19:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you really just multiply both sides by ((X + 200) x 0.003491)? With it being within the parenthesis even?
  $endgroup$
  – Pantheon Engineering Designs
  Mar 15 at 19:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
  $endgroup$
  – dantopa
  Mar 16 at 0:34
  

  
  
  
  

add a comment | 

$begingroup$

I have put together an equation that has other equations in it. When it is all condensed together it looks like this with a set of given information. I'm trying to proof it out, but can't seem to work out the math to what X would equal. Normally my math is pretty decent, but I can't seem to get the reverse of it to solve. Anybody able to assist?

$$1 = Bigl(0.003491 – fracX cdot 0.003366 + 200 cdot 0.003491 (X + 200) cdot 0.003491Bigr) cdot 100$$

algebra-precalculus

share|cite|improve this question

edited Mar 15 at 18:51

Vasya

4,1081618

asked Mar 15 at 18:32

Pantheon Engineering DesignsPantheon Engineering Designs

1

$endgroup$

share|cite|improve this question

edited Mar 15 at 18:51

Vasya

4,1081618

asked Mar 15 at 18:32

Pantheon Engineering DesignsPantheon Engineering Designs

1

share|cite|improve this question

edited Mar 15 at 18:51

Vasya

4,1081618

asked Mar 15 at 18:32

Pantheon Engineering DesignsPantheon Engineering Designs

1

edited Mar 15 at 18:51

Vasya

4,1081618

edited Mar 15 at 18:51

Vasya

4,1081618

asked Mar 15 at 18:32

Pantheon Engineering DesignsPantheon Engineering Designs

1

asked Mar 15 at 18:32

Pantheon Engineering DesignsPantheon Engineering Designs

1

2 Answers
2

active

oldest

votes

0

$begingroup$

The first thing I would do is multiply that 100 into the parentheses to reduce that a little:
$1= 0.3491- frac0.3366X+ 200(0.3491)0.003491X+ 0.6982$
$frac0.3366X+ 200(0.3491)0.003491X+ 0.6982= 0.3491- 1= -0.6509$
Now multiply on both sides by 0.003491X+ 0.6982. That gives
0.3366X+ 69.82= -0.0022722919X- 0.45445838.

Add 0.0022722919X to both sides and subtract 69.82 from both sides:
0.3388722919X= -70.27445838. Finally, divide both sides by 0.3388722919: X= 207.37741048695046760770587511112. To four significant figures, X= 207.4.

share|cite|improve this answer

answered Mar 15 at 19:12

user247327user247327

11.5k1516

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not sure how exactly, but it seems as if the bottom was not multiplied by 200 in there. So that number comes out wrong.
  $endgroup$
  – Pantheon Engineering Designs
  Mar 15 at 19:56
  

  
  
  
  

add a comment | 

0

$begingroup$

Yes, of course...

Ignoring all of the exact numbers, because they don't matter in the explanation, just call each number $a,b,c,dots$ for now, and keep $X$ written as a capital.

You started with the equation

$$1 = left(a - fracbX + cd(X+e)fright)g.$$

By multiplying both sides of the equation by $(X+e)f,$ you are left with

$(X+e)f = left(a(X+e)f - (bX+cd)right)g.$

After simplifying everything, this is simply going to be of the form

$hX + i = jX + k,$ which is trivial to solve.

It will be tedious to do, but the steps should be incredibly straightforward.

share|cite|improve this answer

edited Mar 15 at 19:13

J. W. Tanner

3,6631320

answered Mar 15 at 19:09

JMoravitzJMoravitz

48.8k43988

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you really just multiply both sides by ((X + 200) x 0.003491)? With it being within the parenthesis even?
  $endgroup$
  – Pantheon Engineering Designs
  Mar 15 at 19:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PantheonEngineeringDesigns Yes. Remember by distributivity that $(a+b)c = ac + bc$ and so $(a+fracbc)c = ac + b$. Further, $(a+fracbc)dc = (ac+b)d$. It is just repeated application of the distributive and commutative properties of multiplication and addition and choosing not to distribute the $d$ just yet.
  $endgroup$
  – JMoravitz
  Mar 15 at 20:00
  
  

  
  
  
  

add a comment | 

0

$begingroup$

The first thing I would do is multiply that 100 into the parentheses to reduce that a little:
$1= 0.3491- frac0.3366X+ 200(0.3491)0.003491X+ 0.6982$
$frac0.3366X+ 200(0.3491)0.003491X+ 0.6982= 0.3491- 1= -0.6509$
Now multiply on both sides by 0.003491X+ 0.6982. That gives
0.3366X+ 69.82= -0.0022722919X- 0.45445838.

Add 0.0022722919X to both sides and subtract 69.82 from both sides:
0.3388722919X= -70.27445838. Finally, divide both sides by 0.3388722919: X= 207.37741048695046760770587511112. To four significant figures, X= 207.4.

share|cite|improve this answer

answered Mar 15 at 19:12

user247327user247327

11.5k1516

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not sure how exactly, but it seems as if the bottom was not multiplied by 200 in there. So that number comes out wrong.
  $endgroup$
  – Pantheon Engineering Designs
  Mar 15 at 19:56
  

  
  
  
  

add a comment | 

0

$begingroup$

The first thing I would do is multiply that 100 into the parentheses to reduce that a little:
$1= 0.3491- frac0.3366X+ 200(0.3491)0.003491X+ 0.6982$
$frac0.3366X+ 200(0.3491)0.003491X+ 0.6982= 0.3491- 1= -0.6509$
Now multiply on both sides by 0.003491X+ 0.6982. That gives
0.3366X+ 69.82= -0.0022722919X- 0.45445838.

Add 0.0022722919X to both sides and subtract 69.82 from both sides:
0.3388722919X= -70.27445838. Finally, divide both sides by 0.3388722919: X= 207.37741048695046760770587511112. To four significant figures, X= 207.4.

share|cite|improve this answer

answered Mar 15 at 19:12

user247327user247327

11.5k1516

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not sure how exactly, but it seems as if the bottom was not multiplied by 200 in there. So that number comes out wrong.
  $endgroup$
  – Pantheon Engineering Designs
  Mar 15 at 19:56
  

  
  
  
  

add a comment | 

$begingroup$

The first thing I would do is multiply that 100 into the parentheses to reduce that a little:
$1= 0.3491- frac0.3366X+ 200(0.3491)0.003491X+ 0.6982$
$frac0.3366X+ 200(0.3491)0.003491X+ 0.6982= 0.3491- 1= -0.6509$
Now multiply on both sides by 0.003491X+ 0.6982. That gives
0.3366X+ 69.82= -0.0022722919X- 0.45445838.

Add 0.0022722919X to both sides and subtract 69.82 from both sides:
0.3388722919X= -70.27445838. Finally, divide both sides by 0.3388722919: X= 207.37741048695046760770587511112. To four significant figures, X= 207.4.

share|cite|improve this answer

answered Mar 15 at 19:12

user247327user247327

11.5k1516

$endgroup$

share|cite|improve this answer

answered Mar 15 at 19:12

user247327user247327

11.5k1516

answered Mar 15 at 19:12

user247327user247327

11.5k1516

answered Mar 15 at 19:12

user247327user247327

11.5k1516

0

$begingroup$

Yes, of course...

Ignoring all of the exact numbers, because they don't matter in the explanation, just call each number $a,b,c,dots$ for now, and keep $X$ written as a capital.

You started with the equation

$$1 = left(a - fracbX + cd(X+e)fright)g.$$

By multiplying both sides of the equation by $(X+e)f,$ you are left with

$(X+e)f = left(a(X+e)f - (bX+cd)right)g.$

After simplifying everything, this is simply going to be of the form

$hX + i = jX + k,$ which is trivial to solve.

It will be tedious to do, but the steps should be incredibly straightforward.

share|cite|improve this answer

edited Mar 15 at 19:13

J. W. Tanner

3,6631320

answered Mar 15 at 19:09

JMoravitzJMoravitz

48.8k43988

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you really just multiply both sides by ((X + 200) x 0.003491)? With it being within the parenthesis even?
  $endgroup$
  – Pantheon Engineering Designs
  Mar 15 at 19:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PantheonEngineeringDesigns Yes. Remember by distributivity that $(a+b)c = ac + bc$ and so $(a+fracbc)c = ac + b$. Further, $(a+fracbc)dc = (ac+b)d$. It is just repeated application of the distributive and commutative properties of multiplication and addition and choosing not to distribute the $d$ just yet.
  $endgroup$
  – JMoravitz
  Mar 15 at 20:00
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Yes, of course...

Ignoring all of the exact numbers, because they don't matter in the explanation, just call each number $a,b,c,dots$ for now, and keep $X$ written as a capital.

You started with the equation

$$1 = left(a - fracbX + cd(X+e)fright)g.$$

By multiplying both sides of the equation by $(X+e)f,$ you are left with

$(X+e)f = left(a(X+e)f - (bX+cd)right)g.$

After simplifying everything, this is simply going to be of the form

$hX + i = jX + k,$ which is trivial to solve.

It will be tedious to do, but the steps should be incredibly straightforward.

share|cite|improve this answer

edited Mar 15 at 19:13

J. W. Tanner

3,6631320

answered Mar 15 at 19:09

JMoravitzJMoravitz

48.8k43988

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you really just multiply both sides by ((X + 200) x 0.003491)? With it being within the parenthesis even?
  $endgroup$
  – Pantheon Engineering Designs
  Mar 15 at 19:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PantheonEngineeringDesigns Yes. Remember by distributivity that $(a+b)c = ac + bc$ and so $(a+fracbc)c = ac + b$. Further, $(a+fracbc)dc = (ac+b)d$. It is just repeated application of the distributive and commutative properties of multiplication and addition and choosing not to distribute the $d$ just yet.
  $endgroup$
  – JMoravitz
  Mar 15 at 20:00
  
  

  
  
  
  

add a comment | 

$begingroup$

Yes, of course...

Ignoring all of the exact numbers, because they don't matter in the explanation, just call each number $a,b,c,dots$ for now, and keep $X$ written as a capital.

You started with the equation

$$1 = left(a - fracbX + cd(X+e)fright)g.$$

By multiplying both sides of the equation by $(X+e)f,$ you are left with

$(X+e)f = left(a(X+e)f - (bX+cd)right)g.$

After simplifying everything, this is simply going to be of the form

$hX + i = jX + k,$ which is trivial to solve.

It will be tedious to do, but the steps should be incredibly straightforward.

share|cite|improve this answer

edited Mar 15 at 19:13

J. W. Tanner

3,6631320

answered Mar 15 at 19:09

JMoravitzJMoravitz

48.8k43988

$endgroup$

share|cite|improve this answer

edited Mar 15 at 19:13

J. W. Tanner

3,6631320

answered Mar 15 at 19:09

JMoravitzJMoravitz

48.8k43988

edited Mar 15 at 19:13

J. W. Tanner

3,6631320

edited Mar 15 at 19:13

J. W. Tanner

3,6631320

answered Mar 15 at 19:09

JMoravitzJMoravitz

48.8k43988

answered Mar 15 at 19:09

JMoravitzJMoravitz

48.8k43988