How do you solve #n/(n-3)=2/3#?

First, we can cross multiply to get #n# out of a fraction:
#3n=2(n-3)#

We can use the distributive property on the right-hand side:
#3n=2n-6#

By subtracting #2n# from both sides, we get #n# terms on one side:
#n=-6# which is our answer

First, multiply each side of the equation by #color(red)((n - 3))color(blue)(3)# to eliminate the fractions while keeping the equation balanced. #color(red)((n - 3))color(blue)(3)# is the Lowest Common Denominator of the two fractions:

#color(red)((n - 3))color(blue)(3) xx n/(n -3) = color(red)((n - 3))color(blue)(3) xx 2/3#

#cancel(color(red)((n - 3)))color(blue)(3) xx n/color(red)(cancel(color(black)(n - 3))) = color(red)((n - 3))cancel(color(blue)(3)) xx 2/color(blue)(cancel(color(black)(3)))#

#3n = 2(n - 3)#

Next, expand the terms in parenthesis on the right side of the equation:

#3n = (2 xx n) - (2 xx 3)#

#3n = 2n - 6#

Now, subtract #color(red)(2n)# from each side of the equation to solve for #n# while keeping the equation balanced:

#-color(red)(2n) + 3n = color(red)(2n) + 2n - 6#

#(-color(red)(2) + 3)n = 0 - 6#

#1n = -6#

#n = -6#