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[SOLVED] Arranging Formula - Part 2

dipster307

New member
Jun 11, 2012
5
I an having problems trying arrange the formula below. I want to change it so the formula starts wuth "G13" equals to.

40981.10 = G13 - ( (((G13-B12-C8)*C3) + (B12*B3)) + (((C18-B9)*C5) + ((G13-C18)*C19)) )

The equation above is used in MS Excel, this is why you see the variable G13, B12 etc.
B12 = 34370
C8 = 8105
C3 = 0.4
B3 = 0.2
C18 = 42475
B9 = 7605
C5 = 0.12
C19 = 0.02
Can someone please help.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
First step: distribute the minus sign to obtain

40981.10 = G13 - (((G13-B12-C8)*C3) + (B12*B3)) - (((C18-B9)*C5) + ((G13-C18)*C19)).

Next, you have to scan the equation for appearances of G13. Where are they? Do they multiply other numbers? If so, can you gather all those expression in one place?
 

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
I an having problems trying arrange the formula below. I want to change it so the formula starts wuth "G13" equals to.

40981.10 = G13 - ( (((G13-B12-C8)*C3) + (B12*B3)) + (((C18-B9)*C5) + ((G13-C18)*C19)) )

The equation above is used in MS Excel, this is why you see the variable G13, B12 etc.
B12 = 34370
C8 = 8105
C3 = 0.4
B3 = 0.2
C18 = 42475
B9 = 7605
C5 = 0.12
C19 = 0.02
Can someone please help.
Let's define some constants to make them easier to work with. You can define these in excel if you like instead of back subbing.

$\alpha = B12+C8$
$\beta = B12 \cdot B3$
$\gamma = (C18 - B9) \cdot C5$

In case you're wondering I have used an addition sign in $\alpha$ because G13 - B12-C8 = G13 - (B12+C8)

We can now sub in the constants we've defined above:
40981.10 = G13 -( (((G13 - $\alpha$ )*C3) + $\beta$ + ( $\gamma $ + ((G13-C18)*C19)) )

After distributing the minus sign and clearing up superfluous brackets:
40981.10 = G13 - (G13 - $\alpha$ )*C3 - $\beta$ - ( $\gamma $ + (G13-C18)*C19)


You can distribute that minus sign in the last term too. From there it's a case of multiplying out and then combining those terms with G13 in them and those that don't.
 

dipster307

New member
Jun 11, 2012
5
Let's define some constants to make them easier to work with. You can define these in excel if you like instead of back subbing.

$\alpha = B12+C8$
$\beta = B12 \cdot B3$
$\gamma = (C18 - B9) \cdot C5$

In case you're wondering I have used an addition sign in $\alpha$ because G13 - B12-C8 = G13 - (B12+C8)

We can now sub in the constants we've defined above:
40981.10 = G13 -( (((G13 - $\alpha$ )*C3) + $\beta$ + ( $\gamma $ + ((G13-C18)*C19)) )

After distributing the minus sign and clearing up superfluous brackets:
40981.10 = G13 - (G13 - $\alpha$ )*C3 - $\beta$ - ( $\gamma $ + (G13-C18)*C19)


You can distribute that minus sign in the last term too. From there it's a case of multiplying out and then combining those terms with G13 in them and those that don't.

So the would the answer be:

40981.10 = G13 - (G13 - α )*C3 - β - ( γ + (G13-C18)*C19)

40981.10 = X

X = G13 – G13*C3 - α*C3 - β - γ + G13*C19 – C18*C19
X = G13 – G13*C3 + G13*C19 - α*C3 - β - γ – C18*C19
X = G13(1 – C3 + C19) - α*C3 - β - γ – C18*C19
X + α*C3 + β + γ + C18*C19 = G13(1 – C3 + C19)

(X + α*C3 + β + γ + C18*C19 ) / (1 – C3 + C19) = G13

However if I put the values in I get 71594.13 = G13.

The answer should be 58965.52, G13 must equal to 58965.52.
So where am I going wrong in the arrangement??
 

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
So the would the answer be:

40981.10 = G13 - (G13 - α )*C3 - β - ( γ + (G13-C18)*C19)

40981.10 = X

X = G13 – G13*C3 - α*C3 - β - γ + G13*C19 – C18*C19
X = G13 – G13*C3 + G13*C19 - α*C3 - β - γ – C18*C19
X = G13(1 – C3 + C19) - α*C3 - β - γ – C18*C19
X + α*C3 + β + γ + C18*C19 = G13(1 – C3 + C19)

(X + α*C3 + β + γ + C18*C19 ) / (1 – C3 + C19) = G13

However if I put the values in I get 71594.13 = G13.

The answer should be 58965.52, G13 must equal to 58965.52.
So where am I going wrong in the arrangement??
You've got some signs muddled up

40981.10 = G13 - (G13 - α )*C3 - β - ( γ + (G13-C18)*C19) -- original equation for reference.

X = G13 – G13*C3 - α*C3 - β - γ + G13*C19 C18*C19

^ I've put in red where you have the wrong sign.

For the first one you're distributing the minus sign across both terms.: -C3 * - $\alpha$ = C3*$\alpha$ .. I find it helps if you either imagine/put C3 at the front or act like you're distributing a -1 where there is just a minus sign

The second and third ones are a little trickier to spot. I started by eliminating the bracket inside (G13-C18) by expansion before applying the minus sign outside the ($\gamma$ + (G13-C18)*C19) brackets:


If we just concentrate on this bit: -($\gamma$ + (G13-C18)*C19)

Expanding out the inside brackets: -($\gamma$ + G13*C19 - C18*C19)

Now it's easier to distribute the minus sign: -$\gamma$ - G13*C19 + C18*C19 (because I am multiplying two negatives)


Brought back into the equation as a whole: X = G13 – G13*C3 + $\alpha$*C3 - $\beta$ - $\gamma$ - G13*C19 + C18*C19.

The rest of goes as you worked out but with the sign changes and you end up with: G13 = (X - $\alpha$*C3 + $\beta$ + $\gamma$ - C18*C19 ) / (1 – C3 - C19)

Unfortunately I am about 0.3 out (I get 58965.21) yet I cannot spot where I went wrong but hopefully someone else will be able to see it.

For reference the values I took:
$\alpha = 42475$
$\beta = 6874$
$\gamma = 4184.4$
C3 = 0.4
C18 = 42745
C19 = 0.02
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I an having problems trying arrange the formula below. I want to change it so the formula starts wuth "G13" equals to.

40981.10 = G13 - ( (((G13-B12-C8)*C3) + (B12*B3)) + (((C18-B9)*C5) + ((G13-C18)*C19)) )

The equation above is used in MS Excel, this is why you see the variable G13, B12 etc.
B12 = 34370
C8 = 8105
C3 = 0.4
B3 = 0.2
C18 = 42475
B9 = 7605
C5 = 0.12
C19 = 0.02
Can someone please help.
Hi dipster307, :)

Here is the result obtained using Maxima.

\[G13=\frac{10\,C3\,C8+\left( 10\,B9-10\,C18\right) \,C5+10\,B12\,C3+10\,C18\,C19-10\,B12\,B3-409811}{10\,C3+10\,C19-10}\]

Kind Regards,
Sudharaka.
 

Reckoner

Member
Jun 16, 2012
45
My working (open the spoiler if you want to look at a jumbled mess of variables):

\[40\,981.10 = \mathrm{G13} - \bigg[\Big[\big[(\mathrm{G13}-\mathrm{B12}-\mathrm{C8})\cdot\mathrm{C3}\big] + (\mathrm{B12}\cdot\mathrm{B3})\Big] + \Big[\big[(\mathrm{C18}-\mathrm{B9})\cdot\mathrm{C5}\big] + \big[(\mathrm{G13}-\mathrm{C18})\cdot\mathrm{C19}\big]\Big] \bigg]\]
\[\Rightarrow40\,981.10 = \mathrm{G13} - \Big[\big[(\mathrm{G13}-\mathrm{B12}-\mathrm{C8})\cdot\mathrm{C3}\big] + (\mathrm{B12}\cdot\mathrm{B3})\Big] - \Big[\big[(\mathrm{C18}-\mathrm{B9})\cdot\mathrm{C5}\big] + \big[(\mathrm{G13}-\mathrm{C18})\cdot\mathrm{C19}\big]\Big]\]
\[\Rightarrow40\,981.10 = \mathrm{G13} - \big[(\mathrm{G13}-\mathrm{B12}-\mathrm{C8})\cdot\mathrm{C3}\big] - (\mathrm{B12}\cdot\mathrm{B3}) - \big[(\mathrm{C18}-\mathrm{B9})\cdot\mathrm{C5}\big] - \big[(\mathrm{G13}-\mathrm{C18})\cdot\mathrm{C19}\big]\]
\[\Rightarrow40\,981.10 = \mathrm{G13} - \mathrm{G13}\cdot\mathrm{C3}+(\mathrm{B12}+ \mathrm{C8})\cdot\mathrm{C3} - (\mathrm{B12}\cdot\mathrm{B3}) - (\mathrm{C18}-\mathrm{B9})\cdot\mathrm{C5} - \mathrm{G13}\cdot\mathrm{C19}+\mathrm{C18}\cdot \mathrm{C19}\]
\[\Rightarrow40\,981.10 = \mathrm{G13} - \mathrm{G13} \cdot \mathrm{C3}+\mathrm{B12} \cdot \mathrm{C3} + \mathrm{C8}\cdot\mathrm{C3} - \mathrm{B12} \cdot \mathrm{B3} - \mathrm{C18} \cdot \mathrm{C5}+\mathrm{B9} \cdot \mathrm{C5} - \mathrm{G13} \cdot \mathrm{C19}+\mathrm{C18}\cdot \mathrm{C19}\]
\[\Rightarrow40\,981.10 = \mathrm{G13}(1 - \mathrm{C3} - \mathrm{C19})+\mathrm{B12} \cdot \mathrm{C3} + \mathrm{C8}\cdot\mathrm{C3} - \mathrm{B12} \cdot \mathrm{B3} - \mathrm{C18} \cdot \mathrm{C5}+\mathrm{B9} \cdot \mathrm{C5}+\mathrm{C18}\cdot \mathrm{C19}\]

\[\Rightarrow\mathrm{G13} = \frac{40\,981.10 - \mathrm{B12} \cdot \mathrm{C3} - \mathrm{C8}\cdot\mathrm{C3} + \mathrm{B12} \cdot \mathrm{B3} + \mathrm{C18} \cdot \mathrm{C5}-\mathrm{B9} \cdot \mathrm{C5}-\mathrm{C18}\cdot \mathrm{C19}}{1 - \mathrm{C3} - \mathrm{C19}}\]

And after comparing my solution with SuperSonic4's, they appear to be identical. And I get ~58965.52 in both cases, so I'm guessing that some number got entered incorrectly somewhere.
 

dipster307

New member
Jun 11, 2012
5
My working (open the spoiler if you want to look at a jumbled mess of variables):

\[40\,981.10 = \mathrm{G13} - \bigg[\Big[\big[(\mathrm{G13}-\mathrm{B12}-\mathrm{C8})\cdot\mathrm{C3}\big] + (\mathrm{B12}\cdot\mathrm{B3})\Big] + \Big[\big[(\mathrm{C18}-\mathrm{B9})\cdot\mathrm{C5}\big] + \big[(\mathrm{G13}-\mathrm{C18})\cdot\mathrm{C19}\big]\Big] \bigg]\]
\[\Rightarrow40\,981.10 = \mathrm{G13} - \Big[\big[(\mathrm{G13}-\mathrm{B12}-\mathrm{C8})\cdot\mathrm{C3}\big] + (\mathrm{B12}\cdot\mathrm{B3})\Big] - \Big[\big[(\mathrm{C18}-\mathrm{B9})\cdot\mathrm{C5}\big] + \big[(\mathrm{G13}-\mathrm{C18})\cdot\mathrm{C19}\big]\Big]\]
\[\Rightarrow40\,981.10 = \mathrm{G13} - \big[(\mathrm{G13}-\mathrm{B12}-\mathrm{C8})\cdot\mathrm{C3}\big] - (\mathrm{B12}\cdot\mathrm{B3}) - \big[(\mathrm{C18}-\mathrm{B9})\cdot\mathrm{C5}\big] - \big[(\mathrm{G13}-\mathrm{C18})\cdot\mathrm{C19}\big]\]
\[\Rightarrow40\,981.10 = \mathrm{G13} - \mathrm{G13}\cdot\mathrm{C3}+(\mathrm{B12}+ \mathrm{C8})\cdot\mathrm{C3} - (\mathrm{B12}\cdot\mathrm{B3}) - (\mathrm{C18}-\mathrm{B9})\cdot\mathrm{C5} - \mathrm{G13}\cdot\mathrm{C19}+\mathrm{C18}\cdot \mathrm{C19}\]
\[\Rightarrow40\,981.10 = \mathrm{G13} - \mathrm{G13} \cdot \mathrm{C3}+\mathrm{B12} \cdot \mathrm{C3} + \mathrm{C8}\cdot\mathrm{C3} - \mathrm{B12} \cdot \mathrm{B3} - \mathrm{C18} \cdot \mathrm{C5}+\mathrm{B9} \cdot \mathrm{C5} - \mathrm{G13} \cdot \mathrm{C19}+\mathrm{C18}\cdot \mathrm{C19}\]
\[\Rightarrow40\,981.10 = \mathrm{G13}(1 - \mathrm{C3} - \mathrm{C19})+\mathrm{B12} \cdot \mathrm{C3} + \mathrm{C8}\cdot\mathrm{C3} - \mathrm{B12} \cdot \mathrm{B3} - \mathrm{C18} \cdot \mathrm{C5}+\mathrm{B9} \cdot \mathrm{C5}+\mathrm{C18}\cdot \mathrm{C19}\]

\[\Rightarrow\mathrm{G13} = \frac{40\,981.10 - \mathrm{B12} \cdot \mathrm{C3} - \mathrm{C8}\cdot\mathrm{C3} + \mathrm{B12} \cdot \mathrm{B3} + \mathrm{C18} \cdot \mathrm{C5}-\mathrm{B9} \cdot \mathrm{C5}-\mathrm{C18}\cdot \mathrm{C19}}{1 - \mathrm{C3} - \mathrm{C19}}\]

And after comparing my solution with SuperSonic4's, they appear to be identical. And I get ~58965.52 in both cases, so I'm guessing that some number got entered incorrectly somewhere.
Thanks everyone for helping out, I just need to keeping practicing my maths skills a bit more :)