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Find $a+b+c$, given that $x+y\neq -1$ and \begin{align*} ax+by+c&=x+7,\\ a+bx+cy&=2x+6y,\\ ay+b+cx&=4x+y. \end{align*}

Guest Oct 1, 2017

Best Answer 

 #3
avatar+4806 
+2

I think I got it!!!!!!! laugh

 

Adding these three equations together will give us another true equation.

Like this...

 

ax + a + ay + by + bx + b + c + cy + cx  =  x + 2x + 4x + 6y + y + 7     Then factor it like this...

 

a(x + 1 + y) + b(y + x + 1) + c(1 + y + x)  =  7x + 7y + 7      Divide through by  (x + y + 1) .

 

a + b + c  =  \(\frac{7x+7y+7}{x+y+1}\)

 

a + b + c  =  \(\frac{7(x+y+1)}{x+y+1}\)

 

a + b + c  =  7

 

 

*edit*

 

Also... to show that it produces a true equation when you add them together...let's say...

 

A = B  ,   C = D ,  and  E = F                          (And  A, B, C, D, E, and F  can be expressions.)

 

A + C + E  =  A + C + E      Now we can substitute  B  in for  A ,  D  in for  C , and  F  in for  E .

 

A + C + E  =  B + D + F

hectictar  Oct 2, 2017
edited by hectictar  Oct 2, 2017
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5+0 Answers

 #1
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0

Does not translate into LaTex !!.

Guest Oct 1, 2017
 #2
avatar+732 
+1

The editing label depicting LaTex is a misnomer. This forum doesn’t use LaTeX . It uses TeX

 

Despite the French sounding name, the “La” in LaTex (probably) comes from (Leslie B. Lamport), the initial developer who wrote advanced scripted algorithms that augmented Donald Knuth’s TeX typesetting program to a more advanced markup and layout program used for publishing.

 

Translating LaTex into TeX, requires only a slightly higher intelligence than the machine you are using. 

 

\(\text{Find } a+b+c, \text{given that } x+y\neq -1 \\ \text{ and }\\ \begin{align*} ax+by+c&=x+7,\\ a+bx+cy&=2x+6y,\\ ay+b+cx&=4x+y. \end{align*} \)

 

 

GA

GingerAle  Oct 1, 2017
 #3
avatar+4806 
+2
Best Answer

I think I got it!!!!!!! laugh

 

Adding these three equations together will give us another true equation.

Like this...

 

ax + a + ay + by + bx + b + c + cy + cx  =  x + 2x + 4x + 6y + y + 7     Then factor it like this...

 

a(x + 1 + y) + b(y + x + 1) + c(1 + y + x)  =  7x + 7y + 7      Divide through by  (x + y + 1) .

 

a + b + c  =  \(\frac{7x+7y+7}{x+y+1}\)

 

a + b + c  =  \(\frac{7(x+y+1)}{x+y+1}\)

 

a + b + c  =  7

 

 

*edit*

 

Also... to show that it produces a true equation when you add them together...let's say...

 

A = B  ,   C = D ,  and  E = F                          (And  A, B, C, D, E, and F  can be expressions.)

 

A + C + E  =  A + C + E      Now we can substitute  B  in for  A ,  D  in for  C , and  F  in for  E .

 

A + C + E  =  B + D + F

hectictar  Oct 2, 2017
edited by hectictar  Oct 2, 2017
 #4
avatar+77221 
+1

Impressive, hectictar........!!!!!!

 

Even Columbo would have a hard time figuring this one out....!!!!!

 

 

 

cool cool cool

CPhill  Oct 2, 2017
 #5
avatar+4806 
+1

Hahaha!! It was easy once I tried adding them together!

hectictar  Oct 2, 2017
edited by hectictar  Oct 2, 2017
edited by hectictar  Oct 2, 2017

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