How do you solve #8/3 = 12/n#?

2 Answers
May 27, 2017

Answer:

See a solution process below:

Explanation:

First, multiply each side of the equation by #color(red)(3)/color(blue)(8)#:

#color(red)(3)/color(blue)(8) xx 8/3 = color(red)(3)/color(blue)(8) xx 12/n#

#cancel(color(red)(3))/cancel(color(blue)(8)) xx color(blue)(cancel(color(black)(8)))/color(red)(cancel(color(black)(3))) = 36/(8n)#

#1 = 36/(8n)#

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

#color(red)(n) xx 1 = color(red)(n) xx 36/(8n)#

#n = cancel(color(red)(n)) xx 36/(8color(red)(cancel(color(black)(n))))#

#n = 36/8#

#n = (4 xx 9)/(4 xx 2)#

#n = (color(red)(cancel(color(black)(4))) xx 9)/(color(red)(cancel(color(black)(4))) xx 2)#

#n = 9/2#

May 27, 2017

Answer:

#n=9/2" "->" " 4 1/2#

Explanation:

Given: #8/3=12/n#

We need to find #n# so lets make it the top value (numerator)

Turn everything upside down (yes you can do this!)#

#3/8=n/12#

We now need to 'get' #n# on its own so we need to 'get rid' of the 12. We do this by turning it into 1 as 1 times anything does not change the value.

Multiply both sides by 12

#color(green)(3/8color(red)(xx12)" "=" "n/12color(red)(xx12))#

#color(green)(" "36/8" "=" "ncolor(red)(xx12/(color(green)(12)))#

But #12/12=1" "# and #" "nxx1=n#

#9/2=n#

#n=9/2" "->" " 4 1/2#