+0

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

0
32
1

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

Please solve with all steps included

Guest Oct 27, 2017

#1
+5198
+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
Sort:

#1
+5198
+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

### 30 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details