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|3x-1|=|2x+6|

 

How?

Guest Mar 6, 2017

Best Answer 

 #1
avatar+17614 
+10

|3x - 1|  =  |2x + 6|

 

Either     3x - 1  =  2x + 6     or     3x - 1  =  -(2x + 6)

 

Can you finish it from here?

geno3141  Mar 6, 2017
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7+0 Answers

 #1
avatar+17614 
+10
Best Answer

|3x - 1|  =  |2x + 6|

 

Either     3x - 1  =  2x + 6     or     3x - 1  =  -(2x + 6)

 

Can you finish it from here?

geno3141  Mar 6, 2017
 #5
avatar+90633 
0

Hi Geno,

It is really good to see you again   laugh laugh laugh

Melody  Mar 6, 2017
 #2
avatar
+5

In response to geno3141 (Idk how to reply to your comment. Sorry)

 

I do know how to finish from there, but as of my understanding, I needed to do:

 

+ and +   

+ and -

- and -

- and +

 

You gave me only these solutions:

+ and +             

+ and -

 

Could you explain why?

Guest Mar 6, 2017
 #3
avatar+77149 
+5

Note, guest that  the  - /-  solution is

 

- (3x - 1)  = - (2x + 6)

 

But multiplying through by -1 would produce the first solution that geno presented

 

And the  -/+   solution is

 

- (3x - 1)  = (2x + 6)

 

Again, mutiplying through by -1 produces geno's second solution

 

So.....these other two are superfluous........

 

 

cool cool cool

CPhill  Mar 6, 2017
 #4
avatar+90633 
+5

I'll give the explanation a go :)

 

|3x-1|=|2x+6|

 

+ and +           3x-1 = 2x+6      (1)

+ and -           3x-1 = - ( 2x+2)    (2) 

- and -           -  (3x-1) = - ( 2x+2)    (3) 

- and +           - (3x-1) = + ( 2x+2)    (4) 

 

 

If you look at these 4 options you can see that  

1 and 3 are really the same

and

2 and 4 are really the same.   :)

 

So there are really only 2 distict options :)

 

Does that explain it good enough for you?

Melody  Mar 6, 2017
 #6
avatar+18629 
+5

|3x-1|=|2x+6|

 

\(\begin{array}{|rcll|} \hline |3x-1| &=& |2x+6| \quad & | \quad \text{square both sides} \\ (3x-1)^2 &=& (2x+6)^2 \\ 9x^2-6x+1 &=& 4x^2+24x+36 \\ 9x^2-4x^2-6x-24x+1-36 &=& 0 \\ 5x^2-30x-35 &=& 0 \\ 5x^2-30x-35 &=& 0 \quad & | \quad : 5 \\ x^2-6x-7 &=& 0 \\ (x-7)(x+1) &=& 0 \\ \hline \end{array} \)

 

x = 7 or x = -1

 

check:

\(\begin{array}{|rcll|} \hline |3x-1| &=& |2x+6| \quad & | \quad x=-1 \\ |3\cdot(-1)-1| &=& |2\cdot(-1)+6| \\ |-4| &=& |4| \\ 4 &=& 4 \ \checkmark \\\\ |3x-1| &=& |2x+6| \quad & | \quad x=7 \\ |3\cdot7-1| &=& |2\cdot7+6| \\ |20| &=& |20| \\ 20 &=& 20 \ \checkmark \\ \hline \end{array}\)

 

 

laugh

heureka  Mar 6, 2017
 #7
avatar
+5

Thank you so much everyone for answering. I'm very grateful.

Guest Mar 6, 2017

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