+0  
 
0
236
1
avatar

-12((3x-4)/(3)-(3-4x)/(4)-(6-2x)/(6))

Please solve with all steps included

Guest Oct 27, 2017

Best Answer 

 #1
avatar+7336 
+1

I answered a question nearly identical to this yesterday, but maybe my answer was unclear.

Here is another way of looking at it.

 

-12((3x-4)/(3)-(3-4x)/(4)-(6-2x)/(6))        In fraction form, this is

 

\(=\,-12(\frac{3x-4}{3}-\frac{3-4x}{4}-\frac{6-2x}{6})\)

 

\(=\,-12[\frac13(3x-4)-\frac14(3-4x)-\frac16(6-2x)]\)        We can distribute the  -12  like this...

 

\(=\,(-12)(\frac13)(3x-4)-(-12)(\frac14)(3-4x)-(-12)(\frac16)(6-2x) \\~\\ =\,(\frac{-12}3)(3x-4)-(\frac{-12}4)(3-4x)-(\frac{-12}6)(6-2x) \\~\\ =\,(-4)(3x-4)-(-3)(3-4x)-(-2)(6-2x) \\~\\ =\,(-4)(3x-4)+(3)(3-4x)+(2)(6-2x) \\~\\ =\,(-4)(3x)+(-4)(-4)+(3)(3)+(3)(-4x)+(2)(6)+(2)(-2x) \\~\\ =\,-12x+16+9-12x+12-4x \\~\\ =\,-28x+37\)

hectictar  Oct 27, 2017
 #1
avatar+7336 
+1
Best Answer

I answered a question nearly identical to this yesterday, but maybe my answer was unclear.

Here is another way of looking at it.

 

-12((3x-4)/(3)-(3-4x)/(4)-(6-2x)/(6))        In fraction form, this is

 

\(=\,-12(\frac{3x-4}{3}-\frac{3-4x}{4}-\frac{6-2x}{6})\)

 

\(=\,-12[\frac13(3x-4)-\frac14(3-4x)-\frac16(6-2x)]\)        We can distribute the  -12  like this...

 

\(=\,(-12)(\frac13)(3x-4)-(-12)(\frac14)(3-4x)-(-12)(\frac16)(6-2x) \\~\\ =\,(\frac{-12}3)(3x-4)-(\frac{-12}4)(3-4x)-(\frac{-12}6)(6-2x) \\~\\ =\,(-4)(3x-4)-(-3)(3-4x)-(-2)(6-2x) \\~\\ =\,(-4)(3x-4)+(3)(3-4x)+(2)(6-2x) \\~\\ =\,(-4)(3x)+(-4)(-4)+(3)(3)+(3)(-4x)+(2)(6)+(2)(-2x) \\~\\ =\,-12x+16+9-12x+12-4x \\~\\ =\,-28x+37\)

hectictar  Oct 27, 2017

8 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.