How do you solve #(p-2)/5=p/8#?

3 Answers
Apr 17, 2017

#p=16/3#

Explanation:

#"if "a/b=c/d" than "a*d=b*c #

#(p-2)/5=p/8#

#8*(p-2)=5*p#

#8*p-8*2=5p#

#8p-16=5p#

#"add -5p to both sides of equation "#

#8p-5p-16=cancel(5p)-cancel(5p)#

#3p-16=0#

#"add +16 to both sides of equation."#

#3p-cancel(16)+cancel(1)6=16#

#3p=16#

#"divide both sides of equation by 3"#

#(cancel(3)p)/cancel(3)=16/3#

#p=16/3#

Apr 17, 2017

See the entire solution process below:

Explanation:

First, multiply each side of the equation by the lowest common denominator of the two fractions which is #color(red)(40)# to eliminate the fractions while keeping the equation balanced:

#color(red)(40) xx (p - 2)/5 = color(red)(40) xx p/8#

#cancel(color(red)(40))8 xx (p - 2)/color(red)(cancel(color(black)(5))) = cancel(color(red)(40))5 xx p/color(red)(cancel(color(black)(8)))#

#8(p - 2) = 5p#

Next, expand the terms in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

#color(red)(8)(p - 2) = 5p#

#(color(red)(8) xx p) - (color(red)(8) xx 2) = 5p#

#8p - 16 = 5p#

Then, add #color(red)(16)# and subtract #color(blue)(5p)# from each side of the equation to isolate the #p# term while keeping the equation balanced:

#-color(blue)(5p) + 8p - 16 + color(red)(16) = -color(blue)(5p) + 5p + color(red)(16)#

#(-color(blue)(5) + 8)p - 0 = 0 + 16#

#3p = 16#

Now, divide each side of the equation by #color(red)(3)# to solve for #p# while keeping the equation balanced:

#(3p)/color(red)(3) = 16/color(red)(3)#

#(color(red)(cancel(color(black)(3)))p)/cancel(color(red)(3)) = 16/3#

#p = 16/3#

Apr 17, 2017

#p=16/3#

Explanation:

Cross multiply

#(color(green)(p-2))/color(red)5=color(red)p/color(green)8#

#color(blue)8(p-2)=5p#

Distribute the #8# into #p-2#

#color(blue)8p-2(color(blue)8)=5p#

#8p-16=5p#

Subtract #5p# from both sides

#8pcolor(blue)(-5p)-16=cancel(5pcolor(blue)(-5p))#

#3p-16=0#

Add #16# to both sides

#3pcancel(-16color(blue)(+16))=0color(blue)(+16)#

#3p=16#

Divide both sides by 3

#(cancel3p)/cancelcolor(blue)3=16/color(blue)3#

#p=16/3#

This is the final answer (it cannot be simplified furthermore)