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 February 14th, 2012, 01:54 AM #1 Member   Joined: Nov 2011 Posts: 72 Thanks: 0 Algebra Question ((Lf1-2*t3*b2*c2)/L2)+((Lf1*a1*e1)/L1) = a1e1 + b2c2 Solve for Lf1
February 14th, 2012, 04:28 AM   #2
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Re: Algebra Question

Hello, hatchelhoff!

I hope I interpreted the problem correctly . . .

Quote:
 $\frac{L\!f_1\,-\,2t_3b_2c_2}{L_2}\,+\,\frac{L\!f_1a_1e_1}{L_1} \:=\:a_1e_1\,+\,b_2c_2$ $\text{Solve for }L\!f_1.$

$\text{Let: }\:\begin{Bmatrix}x=&L\!f_1 \\ \\ T=&t_3 \\ \\ A=&a_1e_1 \\ \\ B=&b_2c_2 \end{Bmatrix}=$

$\text{W\!e have: }\:\frac{x\,-\,TB}{L_2}\,+\,\frac{xA}{L_1} \:=\:A\,+\,B$

[color=beige]. . . . . . .[/color]$\frac{x}{L_2}\,-\,\frac{2TB}{L_2}\,+\,\frac{Ax}{L_1} \:=\:A\,+\,B$

[color=beige].n. . . . . . . . . . .[/color]$\frac{x}{L_2}\,+\,\frac{ax}{L_1} \:=\:A\,+\,B\,+\,\frac{2TB}{L_2}$

[color=beige]n . . . . . . . .[/color]$\left(\frac{1}{L_2}\,+\,\frac{A}{L_1}\right)x \:=\:\frac{AL_2\,+\,BL_2\,+\,2TB}{{L_2}$

[color=beige]n . . . . . . .[/color]$\left(\frac{L_1\,+\,AL_2}{L_1L_2}\right)x \:=\:\frac{AL_2\,+\,BL_2\,+\,2TB}{L_2}$

[color=beige]n . . . . . . . . . . . . . . . [/color]$x \:=\:\frac{L_1(AL_2\,+\,BL_2\,+\,2TB)}{L_1\,+\,AL_ 2}$

$\text{Back-substitute: }\; L\!f_{1} \;=\;\frac{L_1(a_1e_1L_2\,+\,b_2c_2L_2\,+\,2t_3b_2 c_2)}{L_1\,+\,a_1e_1L_2}$

 February 15th, 2012, 03:32 AM #3 Member   Joined: Nov 2011 Posts: 72 Thanks: 0 Re: Algebra Question Thats great. Thanks very much Soroban
February 15th, 2012, 06:33 AM   #4
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From: Ottawa Ontario, Canada

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Re: Algebra Question

Quote:
 Originally Posted by hatchelhoff ((Lf1-2*t3*b2*c2)/L2)+((Lf1*a1*e1)/L1) = a1e1 + b2c2 Solve for Lf1
(Lf1-2*t3*b2*c2)/L2 : (x - tbc) / v
(Lf1*a1*e1)/L1 : xae / u
a1e1 + b2c2 : ae + bc

(x - tbc) / v + xae / u = ae + bc
Solve for x

Multipy through by uv, do some backflips, and end with:

x = u(2bct + aev + bcv) / (u + aev)

Looks like same as Soroban's....no guarantee!

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