How do I solve this?

May 2014
6
0
Bangkok
A little stuck on this. I eventually got the
answer by working through what x could
be, but it took a long time. So what's the quick way?
Thanks

X over 9 - X - 2 over 3 = 4
 

v8archie

Math Team
Dec 2013
7,712
2,682
Colombia
\begin{align*}
&&\frac{x}{9} - x - \frac{2}{3} &= 4 \\
&\text{Multiply by $9$ to remove quotients} & x - 9x -6 &= 36 \\
&\text{Collect similar terms} & -8x &= 42 \\
&\text{Divide by $-8$} & \boxed{x = \frac{-21}{4}} \\
\end{align*}
\begin{align*}
& & \frac{x}{9} - \frac{x - 2}{3} &= 4 \\
&\text{Multiply by $9$ to remove quotients} & x - 3(x - 2) &= 36 \\
&\text{Expand the brackets} & x - 3x + 6 &= 36 \\
&\text{Collect similar terms} & -2x &= 30 \\
&\text{Divide by $-2$} & \boxed{x = -15}
\end{align*}
 
Last edited:
May 2014
6
0
Bangkok
Thanks. If you have time, how would you solve this?
I've used the same problem with a small change.

9 over x - x+2 over 3 = 4
 
Last edited by a moderator:

v8archie

Math Team
Dec 2013
7,712
2,682
Colombia
I'd do it in exactly the same way.

\begin{align*}
& & \frac{x}{9} - \frac{x + 2}{3} &= 4 \\
&\text{Multiply by $9$ to remove quotients} & x - 3(x + 2) &= 36 \\
&\text{Expand the brackets} & x - 3x - 6 &= 36 \\
&\text{Collect similar terms} & -2x &= 42 \\
&\text{Divide by $-2$} & \boxed{x = -21}
\end{align*}
 
Jun 2019
23
0
New York
Excellent algebra skills

Explanation was great and made sense.

A couple of links that were a big help and refresher for me.
[youtube]OX6fV75mdPI[/youtube]
[youtube]9Ze-XQdClso[/youtube]
 

skipjack

Forum Staff
Dec 2006
21,478
2,470
Or turn on your calculator, then enter 24 × 21 =

After solving the algebra problem, he insists on checking the answer, but he doesn't mention checking your answer in the first video after doing the multiplication problems.

This teacher seems well-intentioned, but he hasn't thought things through and hasn't linked his videos into an overall course. Also, saying "blow your mind" several times (or just once for that matter) is inappropriate for young children.