Help with fairly simple, quick problem.

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August 26th 2008, 05:03 PM
alkali112

Help with fairly simple, quick problem.

Can someone solve this and show the work for me, please?

3(x-y)=2(y-x) + 8
-(2x - y) = 3(y + 3x - 1)

Thanks very much. i\I just need to know how to work it.
August 26th 2008, 05:26 PM
Quick

Quote:

Originally Posted by alkali112

3(x-y)=2(y-x) + 8
-(2x - y) = 3(y + 3x - 1)

hmm...

first let's distribute out the equations to get:

$3(x-y)=2(y-x) + 8\quad\rightarrow\quad 3x-3y=2y-2x + 8$
$-(2x - y) = 3(y + 3x - 1)\quad\rightarrow\quad -2x+y=3y+9x-3$

Alright, now let's solve for x (or y, it's your choice) for the first problem:

$3x-3y=2y-2x + 8\quad\rightarrow\quad 5x=5y+8\quad\rightarrow\quad x=\frac{5y+8}{5}$

now take the second equation:

$-2x+y=3y+9x-3$

and also solve for x:

$-2x+y=3y+9x-3\quad\rightarrow\quad -11x=2y-3\quad\rightarrow\quad x=\frac{2y-3}{-11}$

and now we can start solving stuff. First things first, you have to write down that:

$x=x$

So now we substitute:

$\overbrace{\frac{5y+8}{5}}^{\text{from the first equation}}=\underbrace{\frac{2y-3}{-11}}_{\text{from the second equation}}$

so now multiply both sides by -11 and 5 to get:

$-11(5y+8)=5(2y-3)$

now use the distributive property:

$-55y-88=10y-15$

now move things around:

$-65y=73$

then divide:

$y=\frac{-73}{65}$

Now you can solve for x.

On a side note, $\frac{-73}{65}$ is not a nice number, you should double check my work and see if my arithmetic is right.
August 26th 2008, 05:44 PM
Jhevon

a "quick" problem indeed :D

Quote:

Originally Posted by Quick

hmm...

first let's distribute out the equations to get:

$3(x-y)=2(y-x) + 8\quad\rightarrow\quad 3x-3y=2y-2x + 8$
$-(2x - y) = 3(y + 3x - 1)\quad\rightarrow\quad -2x+y=3y+9x-3$

Alright, now let's solve for x (or y, it's your choice) for the first problem:

$3x-3y=2y-2x + 8\quad\rightarrow\quad 5x=5y+8\quad\rightarrow\quad x=\frac{5y+8}{5}$

now take the second equation:

$-2x+y=3y+9x-3$

and also solve for x:

$-2x+y=3y+9x-3\quad\rightarrow\quad -11x=2y-3\quad\rightarrow\quad x=\frac{2y-3}{-11}$

and now we can start solving stuff. First things first, you have to write down that:

$x=x$

So now we substitute:

$\overbrace{\frac{5y+8}{5}}^{\text{from the first equation}}=\underbrace{\frac{2y-3}{-11}}_{\text{from the second equation}}$

so now multiply both sides by -11 and 5 to get:

$-11(5y+8)=5(2y-3)$

now use the distributive property:

$-55y-88=10y-15$

now move things around:

$-65y=73$

then divide:

$y=\frac{-73}{65}$

Now you can solve for x.

On a side note, $\frac{-73}{65}$ is not a nice number, you should double check my work and see if my arithmetic is right.