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)