How do you solve #6/10=(p-7)/(p+10)#?

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Answer 1 · 2018-03-25T06:50:44

The solution is #p=65/2#.

First, reduce the fraction:

#6/10=(p-7)/(p+10)#

#color(Red)cancelcolor(Black)6^3/color(red)cancelcolor(black)10^5=(p-7)/(p+10)#

#3/5=(p-7)/(p+10)#

#3(p+10)=5(p-7)#

#3*p+3*10=5*p-5*7#

#3p+30=5p-35#

#3p+30color(blue)+color(blue)35=5p-35color(blue)+color(blue)35#

#3p+30color(blue)+color(blue)35=5pcolor(red)cancelcolor(black)(color(black)-35color(blue)+color(blue)35)#

#3p+30color(blue)+color(blue)35=5p#

#3p+65=5p#

#3p+65color(blue)-color(blue)(3p)=5pcolor(blue)-color(blue)(3p)#

#color(red)cancelcolor(black)(3p)+65color(red)cancelcolor(blue)(color(blue)-color(blue)(3p))=5pcolor(blue)-color(blue)(3p)#

#65=5pcolor(blue)-color(blue)(3p)#

#65=2p#

#color(blue)(color(black)65/2)=color(blue)(color(black)(2p)/2)#

#color(blue)(color(black)65/2)=color(blue)(color(black)(color(red)cancelcolor(black)2p)/color(red)cancelcolor(blue)2)#

#color(blue)(color(black)65/2)=p#

This is the solution. Hope this helped!