Question

How do you find the sum or difference of, $\left( 4a-5{{b}^{2}}+3 \right)+\left( 6-2a+3{{b}^{2}} \right)$ ?

Answer 1

Hint: We have been asked to perform two of the four fundamental mathematical operations, that is, subtraction and addition. The meaning of two operations is that this particular operation demands both of them. Here, we have the two entities as variables and not certain numbers that have constant values, but still the basis of sum and difference remains the same.

Complete step-by-step solution:
Let us first assign some terms that we are going to use in our solution. Let us denote the two terms involved in subtraction be ‘P’ and ‘Q’, such that we have:
$\Rightarrow P=\left( 4a-5{{b}^{2}}+3 \right)$
$\Rightarrow Q=\left( 6-2a+3{{b}^{2}} \right)$
Then, according to the problem, we need to find the value of the expression: $\left( P+Q \right)\text{ and }\left( P-Q \right)$.
In this problem, we can see that both the terms ‘P’ and ‘Q’ contain constant terms and multiple variables. So, upon finding the sum and difference of these individual terms, we will keep the similar terms together so that they can be operated upon. This can be done as follows:
Our first required mathematical expression is:
$\Rightarrow P+Q=\left( 4a-5{{b}^{2}}+3 \right)+\left( 6-2a+3{{b}^{2}} \right)$
On separating like terms and then proceeding in our solution, we get:
$\begin{align}
  & \Rightarrow P+Q=\left( 4a-5{{b}^{2}}+3 \right)+\left( 6-2a+3{{b}^{2}} \right) \\
 & \Rightarrow P+Q=4a-5{{b}^{2}}+3+6-2a+3{{b}^{2}} \\
 & \Rightarrow P+Q=4a-2a-5{{b}^{2}}+3{{b}^{2}}+3+6 \\
 & \therefore P+Q=\left( 2a-2{{b}^{2}}+9 \right) \\
\end{align}$
Thus, we can see that for separate terms both the operations, that is sum and difference, have been used in our solution.
Hence, the sum of, $\left( 4a-5{{b}^{2}}+3 \right)\text{ and }\left( 6-2a+3{{b}^{2}} \right)$ comes out to be $\left( 2a-2{{b}^{2}}+9 \right)$.
Now, we need to find the difference between the two terms. In this case, our mathematical expression is equal to: $\left( P-Q \right)$ . This can be calculated as follows:
$\Rightarrow P-Q=\left( 4a-5{{b}^{2}}+3 \right)-\left( 6-2a+3{{b}^{2}} \right)$
On separating like terms and then proceeding in our solution, we get:
$\begin{align}
  & \Rightarrow P-Q=\left( 4a-5{{b}^{2}}+3 \right)-\left( 6-2a+3{{b}^{2}} \right) \\
 & \Rightarrow P-Q=4a-5{{b}^{2}}+3-6+2a-3{{b}^{2}} \\
 & \Rightarrow P-Q=4a+2a-5{{b}^{2}}-3{{b}^{2}}+3-6 \\
 & \therefore P-Q=\left( 6a-8{{b}^{2}}-3 \right) \\
\end{align}$
Hence, the difference of, $\left( 4a-5{{b}^{2}}+3 \right)\text{ and }\left( 6-2a+3{{b}^{2}} \right)$ comes out to be $\left( 6a-8{{b}^{2}}-3 \right)$.
Hence, the sum and the difference of the two terms has been calculated by collecting like terms and then performing the operators, that are, addition and subtraction on them.

Note: While calculating sum or difference of two variables or one constant and one variable, we should always combine like terms together at each step of the process. This is one of the best ways to proceed in our solution. Also, one should be very careful while performing these rather “easy” subtraction and addition operations and not concur any error just to be quick.