Sove simple equation doubt

• Sep 21st 2011, 07:06 AM
FernandoBasso
Sove simple equation doubt
I'm solving this:
Code:

3 – 7 * (1 - 2x) = 5 – (x + 9)
And am trying this way:
Code:

3- 7 * (1 - 2x) = 5 - (x + 9) 3 - 7 ..?? ??  = 5 - x - 9)
What should be done in the part I put those question marks?

I'm really lost.

Thanks in advance for any help.
• Sep 21st 2011, 07:11 AM
Siron
Re: Sove simple equation doubt
Do you know the distributive law?
In general: $a\cdot (b+c)=a\cdot b+a\cdot c$

Use it to expand $-7\cdot (1-2x)=...$
• Sep 21st 2011, 07:31 AM
FernandoBasso
Re: Sove simple equation doubt
Thanks for the reply.

So, I should do:

$3 - 7 . (1 - 2x) = 5 - (x + 9)$
$3 - 7 . 1 - 7 . 2x = 5 - x - 9)$
$3 - (-7) - 14x = 5 - x - 9$
$-10x = ...$

I'm lost...
• Sep 21st 2011, 07:47 AM
Siron
Re: Sove simple equation doubt
You have to pay attention with the signs:
$3-7\cdot (1-2x)=5-(x+9)$
$\Leftrightarrow 3-7\cdot 1 -7\cdot (-2x)=5-x-9$
$\Leftrightarrow 3-7+14x=5-x-9$
$\Leftrightarrow 3-7-5+9=-x-14x$
$\Leftrightarrow 0=-15x$
$\Leftrightarrow x=0$
• Sep 21st 2011, 09:43 AM
FernandoBasso
Re: Sove simple equation doubt
I was doing:
Code:

3 - 7 * (1 - 2x) = 5 - (x + 9) 3 - 7 * 1 - 7 * (-2x) = ...  3 - (-7) - 14x = ...
When I should be doing:
Code:

3 - 7 * (1 - 2x) = 5 - (x + 9) 3 - 7 * 1 - 7 * (-2x) = ...  3 - 7 - 14x = ...
Why don't I put -7 inside parentheses?
• Sep 21st 2011, 10:31 AM
Siron
Re: Sove simple equation doubt
If you expand:
$7(1-2x)=7-14x$
And the LHS is:
$3-[7(1-2x)]=3-(7-14x)=3-7+14x$

If you expand:
$-7(1-2x)=-7+14x$
Now, the LHS is:
$3+[-7(1-2x)]=3-7+14x$