How do you simplify #(6\times 7) + ( 4\times 5) - 12#?

Since this problem is an Order of Operation, I will be using PEMDAS:

P - Parentheses
E - Exponents
M - Multiplication
D - Division
A - Addition
S - Subtraction

Note: For multiplication and division, do whichever one that comes first in the equation and same for addition and subtraction. :)
Now, let's start working it out!

Let's solve inside the Parentheses first:

#(6times7)+(4times5)-12#

#(42)+(20)-12#

Note: You don't have to keep the parentheses now...

#42+20-12#

We will skip Multiplication and Division because there is none.
That will then leave us with Addition and Subtraction. Addition comes first in the problem, so we'll find the sum, now!

#42+20=62#

#62-12#

Now we have to do Subtraction!

#62-12=50#

#50# is now your answer. I hope you understand my explanation! If not, look at this Khan Academy's example on order of operations. My knowledge is my source for this answer.

Count the number of terms first. They are separated by the #+ and -# signs.

There are #3# terms.

Notice that the parentheses are not necessary at all .. Multiplication is a stronger operation than addition or subtraction, so would be done first anyway. I will leave them out to emphasise this concept

#color(blue)(6xx7)" "color(red)(+4xx5)" "color(green)(-12)#

#=color(blue)(42)" "color(red)(+20)" "color(green)(-12)#

#= 62color(green)(-12)#

#=50#

Note that you can add or subtract in any order as long as the signs stay with the correct number.

#=color(blue)(42)" "color(green)(-12)" "color(red)(+20)#

#=30" "color(red)(+20)#

#=50#

Or you could do:

#=color(blue)(42)" "color(red)(+20)" "color(green)(-12)#

#=color(blue)(42)" "+8#

#=50#