How do you solve #4/x+7=6/x#?

2 Answers
Apr 6, 2018

Answer:

The solution is #x=2/7#.

Explanation:

Multiply everything by #x# to get rid of the #x# in the denominator:

#4/x+7=6/x#

#4/xcolor(blue)(*x)+7color(blue)(*x)=6/xcolor(blue)(*x)#

#4/color(red)cancelcolor(black)xcolor(blue)(*color(red)cancelcolor(blue)x)+7color(blue)(*x)=6/color(red)cancelcolor(black)xcolor(blue)(*color(red)cancelcolor(blue)x)#

#4+7color(blue)(*x)=6#

#4+7x=6#

#4+7xcolor(blue)-color(blue)4=6color(blue)-color(blue)4#

#color(red)cancelcolor(black)4+7xcolor(red)cancelcolor(black)(color(blue)-color(blue)4)=6color(blue)-color(blue)4#

#7x=6color(blue)-color(blue)4#

#7x=2#

#color(blue)(color(black)(7x)/7)=color(blue)(color(black)2/7)#

#color(blue)(color(black)(color(red)cancelcolor(black)7x)/color(red)cancelcolor(blue)7)=color(blue)(color(black)2/7)#

#x=color(blue)(color(black)2/7)#

That's the solution. Hope this helped!

Apr 6, 2018

Answer:

#2/7#

Explanation:

#4/x +7=6/x#

#rArr 4/x - 6/x=-7#

#rArr (4-6)/x = -7#

#rArr -2/x =-7#

#rArr x=2/7#

Hope this helps :)