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

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

Please solve with all steps included

Guest Oct 27, 2017

#1
+7266
+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
+7266
+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