Need help!!!

G

Guest

Guest
I have a question and I have no clue how to get started. Here is the question:

Solve the Equations for the solutions, x and y

Ax + By = C
Dx + Ey = F

x= _______ / (ae-bd)
y= _______ / (ae-bd)

Any help would be great!!! :D :?
 
Here's a way, it's a bit roundabout. Can you tidy it up, while you are waiting for a better response.

Ax + By = C
Dx + Ey = F

Make y the subject of both equations.

y = (C - Ax)/B

y = (F - Dx)/E

Equate

(C - Ax)/B = (F - Dx)/E

solve for x.

E(C - Ax) = B(F - Dx)

CE - AEx = BF - BDx

CE - BF = -BDx + AEx

CE - BF = x(-BD + AE)

x = (CE - BF)/(AE - BD)


Try it for y?

Edit: Apologies for leaving out the parentheses.
 
Thanks for your help. Just wondering, does y= af-cd/(ae-bd)
 
You are welcome.

I got what you got for, y.

Don't forget the parentheses

y= (af-cd)/(ae-bd)


I'm very sorry about leaving out the parentheses, I trust you were doing it right on paper!

Thanks for pointing that out Gene.

You should substitute into the original equations to check.
 
Except it is
x = (CE - BF)/(AE - BD)
x = CE - BF/(AE - BD) means
divide BF by (AE-BD) then add CE to that. A completly different equation.
 
Hello, Billy-O!

We can also solve it by Elimination (or "Addition").

Solve the equations: \(\displaystyle \;\begin{array}{cc}(1)\;Ax\,+\,By\:=\:C\\(2)\;Dx\,+\,Ey\:=\:F\end{array}\)

\(\displaystyle x\:=\:\)______\(\displaystyle \;/(AE\,-\,BD)\)

\(\displaystyle y\:=\:\)______\(\displaystyle \;/(AE\,-\,BD)\)
Click to expand...
Multiply (1) by \(\displaystyle E:\;\;AEx\,+\,BEy\:=\:CE\)
Multiply (2) by -\(\displaystyle B:\;\)-\(\displaystyle BDx\,-\,BEy\:=\,\)-\(\displaystyle BF\)

Add: \(\displaystyle \;AEx\,-\,BDx\:=\:CE\,-\,BF\)

Factor:\(\displaystyle \;(AE\,-\,BD)x\:=\:CE\,-\,BF\)

. . Therefore: \(\displaystyle \L\;x\;=\;\frac{CE\,-\,BF}{AE\,-\,BD}\)


In a similar manner, solve for \(\displaystyle y\).
 
I'm very sorry about leaving out the parentheses, I trust you were doing it right on paper!

Thank you for pointing that out Gene.
 
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