To work this problem we first need to convert the amount of dark chocolate Holly has from a mixed number to an improper fraction:
#2 7/8 => 2 + 7/8 => (8/8 xx 2) + 7/8 => 16/8 + 7/8 => (16 + 7)/8 => 23/8#
Next, let's call a full batch of cookies: #b#
We can write and solve this equation to find out how many full batches of cookies Holly can make:
#3/4b = 23/8#
Multiply each side of the equation by #color(red)(4)/color(blue)(3)# to solve for #b# while keeping the equation balanced:
#color(red)(4)/color(blue)(3) xx 3/4b = color(red)(4)/color(blue)(3) xx 23/8#
#cancel(color(red)(4))/cancel(color(blue)(3)) xx color(blue)(cancel(color(black)(3)))/color(red)(cancel(color(black)(4)))b = cancel(color(red)(4))/color(blue)(3) xx 23/(color(red)(cancel(color(black)(8)))2)#
#b = 23/6#
Now, convert this improper fraction to a mixed number.
#b => 23/6 => (18 + 5)/6 => 18/6 + 5/6 => 3 + 5/6 => 3 5/6#
Holly can make 3 full batches of cookies
Because Holly has enough chocolate for #5/6# of a batch of cookies we can multiply #5/6# by the amount of chocolate that goes into a batch of cookies to find out how much chocolate she has left over:
#5/6 xx 3/4"lb" =>#
#(5 xx 3)/(6 xx 4)"lb" =>#
#(5 xx color(red)(cancel(color(black)(3))))/(color(red)(cancel(color(black)(6)))2 xx 4)"lb" =>#
#5/8"lb"#
Holly would have #5/8# of a pound of dark chocolate left over after making 3 batches of cookies.
