Question

I need help on this. I'm stuck on this one because I don't know what to do with the "6".
Solve the equation. Check for extraneous solutions.
6|6-4x|=8x+4

Answer 1

Well, you basically 'get rid of it' by dividing both sides by 6 :)

Answer 2

I figured so. Wasn't sure. So that leaves me with |6-4x|=8x+ 2/3?

Answer 3

You also have to divide the 8x :)

Answer 4

\[\Large |6-4x |= \frac43x+\frac23\]

Answer 5

Savvy? ^_^

Answer 6

Or, there's actually another way...
You can just distribute the 6 to the absolute value...

Answer 7

It makes no difference really, I'm just delaying dealing with fractions until there's absolutely no choice XD

Answer 8

So, either way...
\[\Large |6-4x|= \frac43x +\frac23\]
or\[\Large |36-24x|= 8x + 4\]

Answer 9

They'll work out just fine ^_^

Answer 10

Thank you! Fractions do make it harder but the answers include some fractions so it would be better to do the fractions :)

Answer 11

Fractions inevitable in this question.
Can you take it from here?

Answer 12

I'm working it out to see if I get stuck again. I am a little stuck though. I'm not sure what to do next

Answer 13

Subtract 6 from both sides?

Answer 14

Well, yes and no.
When you have an absolute value equation, it basically gives you two equations to solve..
one which simply removes the absolute value bars...
\[\Large 6-4x = \frac43x+\frac23\]
and one which removes the absolute value bars BUT multiplies whatever was inside it by -1.
\[\Large \color{blue}{(-1)}(6-4x )= \frac43x + \frac23\]\[\Large4x - 6 = \frac43x + \frac23\]

Just solve both of these, you'll get two *possible* values for x.

Answer 15

Ohh okay. I was thinking \[6-4x = \frac{ 4}{ 3 }x + \frac{2}{3} \]
and
\[6-4x = -(\frac{4}{3}x - \frac{2}{3}) \]

Answer 16

No parentheses on the second. unless you have a + sign inside
\[\Large 6 - 4x = -\left(\frac43x +\frac23\right)\]or\[\Large 6-4x = -\frac43x - \frac23\]

But not\[\Large 6 -4x = -\left(\frac43x - \frac23\right)\color{red}{=-\frac43x + \frac23}\]

Answer 17

Ohhh okay thank you.
For the first one I came out with \[\frac{16}{3}x = -\frac{16}{3}\]

Answer 18

Something's wrong here...

Answer 19

Could you show your steps?

Answer 20

\[6|6-4x|=8x+4\]
Divided both sides by 6 and got \[6-4x=\frac{4}{3}x + \frac{2}{3}\]
then subtracted 6 from both sides and got
\[\frac{16}{3}x=-\frac{16}{3}\]
then divided both sides by \[\frac{16}{3}\] and got -1.

Answer 21

Tsk... that fateful 'wrong' sign...
if in doubt, don't skip steps... patience is a virtue :D
\[6-4x=\frac{4}{3}x + \frac{2}{3}\]
Subtract 6 from both sides
\[6-4x\color{red}{-6}=\frac{4}{3}x + \frac{2}{3}\color{red}{-6}\]\[\Large -4x = \frac43x\color{blue}{-\frac{16}3}\]
Right?

Answer 22

Ohhh I did skip a step!

Answer 23

Now, if we subtract \(\Large \frac43 x\) from both sides
\[\Large -4x\color{red}{-\frac43x} = \frac43x -\frac{16}3\color{red}{-\frac43x}\]
We, in fact, arrive at\[\Large \color{blue}{-\frac{16}3x }=-\frac{16}3\]

So... x = ?

And yes, one wrong side could be tragic :D

Answer 24

Err, one wrong sign*
sorry...

Answer 25

So x=1 :)

Answer 26

Yes, and that's one of the two solutions.
Now solve the other one, and for the love of ******* don't skip steps (yet) XD

Answer 27

LOL I won't this time! :)