How do you solve #(x-1) /3 - (x-1) /2 = 6#?

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Answer 1 · 2016-04-02T08:47:18

#x=-35#

#color(blue)((x-1)/(3)-(x-1)/2=6#

First do fractional part. For that find the LCM of #3# and #6# which is #6#

So,

#rarr(2(x-1))/(2(3))-(3(x-1))/(3(2))=6#

#rarr(2(x-1))/(6)-(3(x-1))/(6)=6#

Use distributive property #color(brown)(a(b+c)=ab+ac#

#rarr(2x-2)/6-(3x-3)/6=6#

#rarr((2x-2)-(3x-3))/6=6#

Remember to change the signs

#rarr(2x-2-3x+3)/6=6#

Collect like terms

#rarr(-x+1)/6=6#

#rarr-x+1=6*6#

#rarr-x+1=36#

#rarr-x=36-1#

#rarr-x=35#

#color(green)(rArrx=-35#