2/3(2-3x)=5: I have tried 3 times and got 3 different answers

zekham

New member
Joined
Oct 31, 2014
Messages
2
I have tried 3 times and got 3 different answers

2/3(2-3x)=5
 
Last edited by a moderator:
I have tried 3 times and got 3 different answers

2/3(2-3x)=5
Is the above meant to be one of the following?

. . . . .\(\displaystyle \mbox{a. }\, \dfrac{2}{3(2\, -\, 3x)}\, =\, 5\)

. . . . .\(\displaystyle \mbox{b. }\, \dfrac{2}{3}(2\, -\, 3x)\, =\, 5\)

When you reply, please include a clear listing of your efforts so far, so we can see where you're having trouble. Thank you! ;)
 
1.

\(\displaystyle \dfrac{2}{3(2 - 3x)} = 5 \)

\(\displaystyle \dfrac{2}{6 - 9x} = 5 \)

\(\displaystyle 2 = 5(6 - 9x)\)

\(\displaystyle 2 = 30 - 45x\)

\(\displaystyle -28 = -45x\)

\(\displaystyle \dfrac{-28}{-45} = x\)

\(\displaystyle \dfrac{28}{45} = x\)

Check

\(\displaystyle \dfrac{2}{3(2 - 3(28/45)) }= 5 \)

2.

\(\displaystyle \dfrac{2}{3}(2 - 3x) = 5\)

\(\displaystyle \dfrac{4}{3} - \dfrac{6x}{3} = 5\)

\(\displaystyle -\dfrac{6x}{3} = 5 - \dfrac{4}{3}\)

\(\displaystyle -\dfrac{6x}{3} = \dfrac{11}{3}\)

\(\displaystyle -6x = \dfrac{11}{3}(3)\)

\(\displaystyle -6x = \dfrac{33}{3}\)

\(\displaystyle -6x = 11\)

\(\displaystyle -x = \dfrac{11}{6}\)


\(\displaystyle x = -\dfrac{11}{6}\)

Check:

\(\displaystyle \dfrac{2}{3}(2 - 3(-\dfrac{11}{6})) = 5\)
 
Last edited:
Jason,

Please respect other poster's "wish" to "see" poster's effort! A complete ready-for-camera solution does not help the student.
 
2/3(2-3x)=5 \(\displaystyle \ \ \ \ \)To be clear, you should put the fraction in grouping symbols.

4/3-6x/3=5

4/3-2x=5

-2x=5-4/3

-2x=-11/3\(\displaystyle \ \ \ \ \)The sign of the fraction is wrong.

x=-11/3*1/2\(\displaystyle \ \ \ \ \)When you multiplied by the reciprocal, you didn't include the negative sign.

x=-11/6\(\displaystyle \ \ \ \ \) Your two errors above cancelled each out. This happens to be the correct answer.

zekham, I'm going to show my full solution, because you showed your attempt at a
full solution, and it's been more than a day and a half.


Let's look at the problem again, rewritten:


(2/3)(2 - 3x) = 5

zekham, instead of you and Jason76 distributing the 2/3, why not multiply each side by 3
and clear the fraction at the beginning?

3[(2/3)(2 - 3x)] = 3[5]

2(2 - 3x) = 15

4 - 6x = 15

-6x = 15 - 4

-6x = 11

x = 11/(-6)

x = -11/6
 
Top