Subtract the number \[4a-7ab+3b+12\] from \[12a-9ab+5b-3\]

Hint: We solve this problem by using the simple subtraction of numbers.
During the subtraction, we need to keep in mind that only the terms of the same variable are allowed to subtract or add which is simply adding or subtracting the coefficients of the same variable.
We use the condition that subtraction of \[x\] from \[y\] is given as \[y-x\]

Complete step by step answer:
We are given that the two numbers to be subtracted as \[4a-7ab+3b+12\] and\[12a-9ab+5b-3\]
Let us assume that the first number as
\[\Rightarrow P=4a-7ab+3b+12\]
Similarly, let us assume that the second number as
\[\Rightarrow Q=12a-9ab+5b-3\]
We are asked to subtract the number \[4a-7ab+3b+12\] from \[12a-9ab+5b-3\]
The above statement can be modified as to subtract \[P\] from \[Q\]
Let us assume that the result in the above subtraction as \[R\]
We know that the condition that subtraction of \[x\] from \[y\] is given as \[y-x\]
By using the above result to given numbers we get
\[\Rightarrow R=Q-P\]
By substituting the required values in above equation we get
\[\Rightarrow R=\left( 12a-9ab+5b-3 \right)-\left( 4a-7ab+3b+12 \right)\]
We know that we need to add or subtract the terms with the same variable.
So, let us rearrange the terms in the above equation such that we get the terms of same variable as a group then we get
\[\begin{align}
  & \Rightarrow R=\left( 12a-4a \right)+\left( -9ab+7ab \right)+\left( 5b-3b \right)+\left( -3-12 \right) \\
 & \Rightarrow R=8a-2ab+2b-15 \\
\end{align}\]
Therefore we can conclude that the result after subtracting \[4a-7ab+3b+12\] from \[12a-9ab+5b-3\] is given as \[8a-2ab+2b-15\]

Note:
Students may make mistakes in the subtraction of numbers.
We have the condition that subtraction of \[x\] from \[y\] is given as \[y-x\]
Here, we can see that the number \[y\] is taken as the first term in the subtraction \[y-x\] because it gives that the subtraction from \[y\]
In other words we may also consider that \[y>x\] so that the subtraction \[y-x\] is correct.