Linear equation

**sinners** · February 2nd 2008, 06:15 PM

Hey hopefully some one can help me with this cuase i cant get it for the life of me

solve for 'x'

b-cx over a + a -cx over b + 2 = 0

also im new so i dont know how to get all signs and stuff lol

Thanks for the help.

**Jhevon** · February 2nd 2008, 06:23 PM

Originally Posted by sinners

Hey hopefully some one can help me with this cuase i cant get it for the life of me

solve for 'x'

b-cx over a + a -cx over b + 2 = 0

also im new so i dont know how to get all signs and stuff lol

Thanks for the help.

$\frac {b - cx}a + \frac {a - cx}b + 2 = 0$ ..........split up the fractions

$\Rightarrow \frac ba - \frac {cx}a + \frac ab - \frac {cx}b + 2 = 0$ .............group like terms and factor out the x

$\Rightarrow \left( - \frac ca - \frac cb \right)x + \frac ba + \frac ab + 2 = 0$

can you continue?

i have not come to help the math geniuses, but to help sinners do his problem...

...did anyone but me get that?

**sinners** · February 2nd 2008, 06:44 PM

this is what i did next i dont know if it is right or not.

(-cb over ab - ca over ba) x + b^2 over ab + a^2 over ba + 2ab = 0 so there is a common denominator then crossed all the common denominators out. So its..

(-cb - ca)x + b^2 + a^2 + 2ab = 0 i then moved the x part over to the other side so its...

b^2 + a^2 + 2ab = 0 over ac + bc then i moved the rest of equation over to the other side so it is now...

b^2 + a^2 + 2ab over ac + bc is this the correct answer? i am really confused because the book had the answer as x = a^2 + b^2 + 2ab over ac + bc = a+b over c. Is the book right or am i? also can some one show me how to actually show up as a over b instead of typing it

thanks in advance

**Jhevon** · February 2nd 2008, 07:12 PM

Originally Posted by sinners

this is what i did next i dont know if it is right or not.

(-cb over ab - ca over ba) x + b^2 over ab + a^2 over ba + 2ab = 0 so there is a common denominator then crossed all the common denominators out. So its..

(-cb - ca)x + b^2 + a^2 + 2ab = 0 i then moved the x part over to the other side so its...

b^2 + a^2 + 2ab = 0 over ac + bc then i moved the rest of equation over to the other side so it is now...

b^2 + a^2 + 2ab over ac + bc is this the correct answer? i am really confused because the book had the answer as x = a^2 + b^2 + 2ab over ac + bc = a+b over c. Is the book right or am i?

the book is right, it hard to read through your work though because of how you typed it, so i will show you the solution.

so we were at:

$\left( - \frac ca - \frac cb \right)x + \frac ba + \frac ab + 2 = 0$ ............bring the x term on the other side, we are solving for x after all

$\Rightarrow \left( \frac ca + \frac cb \right)x = \frac ba + \frac ab + 2$ .............combine all the fractions

$\Rightarrow \left( \frac {ac + bc}{ab}\right)x = \frac {a^2 + b^2 + 2ab}{ab}$ .........divide both sides by (ac + bc)/ab

$\Rightarrow x = \frac {a^2 + b^2 + 2ab}{ab} \cdot \frac {ab}{ac + bc}$ .............the ab's cancel, now factorize the other terms

$\Rightarrow x = \frac {(a + b)^2}{ab} \cdot \frac {ab}{c(a + b)}$

$\Rightarrow x = \frac {(a + b)^2}{c(a + b)}$

$\Rightarrow x = \frac {a + b}{c}$

also can some one show me how to actually show up as a over b instead of typing it

use LaTex, learn how to here

or simply type a/b to mean a over b. but please, use parentheses if you have a lot of terms. that is, type (ac - ab)/(b + c) to mean $\frac {ac - ab}{b + c}$

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