How do you simplify #3\times ( \frac { 5} { 18} + \frac { 1} { 6} ) + \frac { 4} { 5} #?

Let's look at the problem:
#color(red)(3*(5/18 + 1/6) + 4/5)#

According to PEMDAS or the Order of Operations, we need to solve the problem in the parentheses, #5/18 + 1/6.#

First, we need to make the denominators the same. Since 18 is a multiple of 6, we just need to change #color(green)(1/6)# to a denominator with 18. So we would do:

#3*(5/18+(1*3)/(6*3))+4/5#.

#darr#

#1/6 = 3/18#

Now, let's plug#3/18# into the problem:

#color(red)(3*( 5/18 + 3/18) + 4/5)#
Or,
#color(red)(3*((5+3)/18)+4/5)#
Thus, the problem is :

#color(red)(3*( 8/18) + 4/5)#
Here, we need to do #3 * 8/18#. Because 3 is equal to #3/1#, the problem becomes:

#color(red)[(3/1 * 8/18)]color(red)+color(red)(4/5)#

Now let's solve, #3*8=24, and 1*18=18#,

#color(red)[(24/18)]color(red)(+4/5)#
Since the top and bottom of #color(red)(24/18)# can both be divided by 2,
#darr#
#=color(red)([cancel24^12]/[cancel18^9])#
#darr#
#=color(red)(cancel12^4/cancel9^3)#

Now, the last step is to do #color(blue)(4/3 + 4/5)#. Just like before, we need to change the denominators to be the same. As 3 and 5's LCM is 15 , we'll need to change the two fractions denominator to 15. So the fractions become:

#color(blue)(4/3 = 20/15)#

#color(blue)(4/5=12/15)#

Finally, the last step is to add:

#color(red)(20/15 + 12/15 = 32/15)#

Then simplify:

#32/15 = 2 2/15 #

So #3*(5/18 + 1/6) + 4/5# is equal to #2 2/15.#

Given: #3xx(5/18+1/6)+4/5#

Multiply and divide have priority over add or subtract but remembering that brackets group things

Consider #(5/18+1/6)#

#color(green)(5/18+[1/6color(red)(xx1)])#

#color(green)(5/18+[1/6color(red)(xx3/3)])#

#color(green)(5/18+3/18=8/18#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all back together")#

#3xx8/18+4/5#

This this is the same as:

#3/1xx8/18+4/5#

#(3xx8)/(1xx18)+4/5#

#24/18+4/5#

5 will divide exactly into any number ending in 5 or 0. So we need to multiply the 18 until we have a 0 as the last digit.

Known that #5xx8=40# giving the 0 so #5xx 18= 90# again giving the 0

#color(green)([24/18color(red)(xx1)]color(white)("d")+color(white)("d")[4/5color(red)(xx1)]#

#color(green)([24/18color(red)(xx5/5)]+[4/5color(red)(xx18/18)]#

#color(green)(color(white)("dd")120/90color(white)("ddd")+color(white)("ddd")72/90 color(white)("ddd")=color(white)("d") 192/90color(white)("d")=color(white)("d")32/15)#