# X variable on both side- need help pls!

• Nov 19th 2006, 08:12 AM
architech
X variable on both side- need help pls!
OK...
New to this forum....
I'm a CAD geek by nature & would think Algebra would be a snap... :confused:
But it ain't as I remember it...

So here's my problem...
Please provide the step by step solutions... so I can learn this... please... ;)

Problem:
(x-5)/3 -3/2=(4x+7)/6

Soultion:
?????
• Nov 19th 2006, 08:18 AM
CaptainBlack
Quote:

Originally Posted by architech
OK...
New to this forum....
I'm a CAD geek by nature & would think Algebra would be a snap... :confused:
But it ain't as I remember it...

So here's my problem...
Please provide the step by step solutions... so I can learn this... please... ;)

Problem:
(x-5)/3 -3/2=(4x+7)/6

Soultion:
?????

$\displaystyle (x-5)/3 -3/2=(4x+7)/6$

Expand all the brackets:

$\displaystyle x/3 -5/3-3/2=x/3-19/6=2x/3 + 7/6$

Subtract $\displaystyle x/3$ from both sides:

$\displaystyle x/3 - 19/6 -x/3=-19/6 = 2x/3 +7/6 -x/3=x/3+7/6$

or:

$\displaystyle -19/6=x/3+7/6$

subtract $\displaystyle 7/6$ form both sides:

$\displaystyle x/3=-13/3$

Multiply by $\displaystyle 3$:

$\displaystyle x=-13$

RonL
• Nov 19th 2006, 08:53 AM
Soroban
Hello, architech!

Welcome aboard!

Quote:

$\displaystyle \frac{x-5}{3} -\frac{3}{2}\;=\;\frac{4x+7}{6}$

With an equation, we can eliminate the fractions immediately . . .

Multiply both sides by the LCD, 6:

. . . . . . .$\displaystyle 6\left(\frac{x-5}{3} - \frac{3}{2}\right) \;= \;6\left(\frac{4x+7}{6}\right)$

.$\displaystyle \not{6}^{^2}\left(\frac{x-5}{\not{3}}\right) - \not{6}^{^3}\left(\frac{3}{\not{2}}\right) \;= \;\not{6}\left(\frac{4x+7}{\not{6}}\right)$

. . . . . . .$\displaystyle 2(x - 5) - 3(3) \;= \;4x+7$

. . . . . . . . .$\displaystyle 2x - 10 - 9 \;=\;4x+7$

. . . . . . . . . . . . . $\displaystyle -2x \;= \;26$

. . . . . . . . . . . . . . . $\displaystyle \boxed{x \;= \;-13}$

• Nov 19th 2006, 09:17 AM
architech
YOU GUYS ROCK!!!! :p :D :)
THANKS!!!! ;)

CRYSTAL CLEAR!!!!