|3x - 1| = |2x + 6|
Either 3x - 1 = 2x + 6 or 3x - 1 = -(2x + 6)
Can you finish it from here?
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?
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........
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?
|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:
\(\begin{array}{|rcll|} \hline |3x-1| &=& |2x+6| \quad & | \quad x=-1 \\ |3\cdot(-1)-1| &=& |2\cdot(-1)+6| \\ |-4| &=& |4| \\ 4 &=& 4 \ \checkmark \\\\ |3x-1| &=& |2x+6| \quad & | \quad x=7 \\ |3\cdot7-1| &=& |2\cdot7+6| \\ |20| &=& |20| \\ 20 &=& 20 \ \checkmark \\ \hline \end{array}\)