How do you solve #\frac { - 2x } { 3} = \frac { - 5} { 2} + \frac { - x + 1} { 3}#?

3 Answers
Mar 3, 2017

I gor #x=13/2#

Explanation:

I would first take a common denominator such as #6# and change the numerator accordingly to get:
#(2*(-2x))/6=(3*(-5)+2(-x+1))/6#
get rid of the denominators:
#(2*(-2x))/cancel(6)=(3*(-5)+2(-x+1))/cancel(6)#
rearrange and solve for #x#:
#-4x=-15-2x+2#
#2x=13#
#x=13/2#

Mar 3, 2017

#x=6 1/2#

Explanation:

#(-2x)/3=-5/2+(-x+1)/3#

#:.(-4x=-15+2(-x+1))/6#

#:.-4x=-15-2x+2#

#:.-4x+2x=-15+2#

#:.-2x=-13#

#:.-x=-13/2#

multiply L.H.S and R.H.S. by#color(red)-color(red)1#

#:.color(red)xcolor(red)=color(red)13/color(red)2#

#:.color(red)xcolor(red)=color(red)6color(red) 1/color(red)2#

substitute #color(red)xcolor(red)=color(red)(13)/color(red)2#

#:.(-2(color(red)13/color(red)2))/3=-5/2+(-(color(red)13/color(red)2)+1)/3#

#:.(-26/2)/3=-5/2+((color(red)-color(red)13/color(red)2)+1)/3#

#:.-cancel26^13/cancel2^1 xx 1/3=-5/2+(-6 1/2+1)/3#

#:.-13/3=-5/2+(-5 1/2)/3#

#:.-13/3=-5/2+((-11/2)/3)#

#:.-13/3=- 5/2+(-11/2 xx 1/3)#

#:.-13/3=-5/2-11/6#

#:.-13/3=(-15-11)/6#

#:.- 13/3=-cancel26^13/cancel6^3#
#:.-13/3=-13/3#

Mar 3, 2017

#x = 13/2#

Explanation:

#(-2x)/3 = (-5)/2 +(-x+1)/3#

When you have an equation with fractions you can get rid of the denominators by multiplying each term by the LCM of the denominators. In this case it is #6#

#(-2x xxcolor(blue)(6))/3 = (-5xxcolor(blue)(6))/2 +(color(blue)(6xx)(-x+1))/3#

cancel the denominators

#(-2x xxcolor(blue)(cancel6^2))/cancel3 = (-5xxcolor(blue)(cancel6^3))/cancel2 +(color(blue)(cancel6^2)(-x+1))/cancel3#

#-4x=-15 +2(-x+1)#

#-4x=-15-2x+2#

#15-2 = 4x-2x#

#13=2x#

#x= 13/2#