Question

From the sum of $ 2{y^2} + 3yz, - {y^2} - yz - {z^2} $ and $ yz + 2{z^2} $ subtract the sum of $ 3{y^2} - {z^2} $ and $ - {y^2} + yz + {z^2} $

Answer 1

Hint: Terms with the same variables and powers can be added or subtracted which changes the coefficient of that particular term. Add the given terms separately and then subtract one from another. If terms have two or more variables then terms with the same variables can be added or subtracted.

Complete step-by-step answer:
Given to us ,two sets of expressions. The first set contains $ 2{y^2} + 3yz, - {y^2} - yz - {z^2} $ and $ yz + 2{z^2} $ and these are to be added. The second set contains $ 3{y^2} - {z^2} $ and $ - {y^2} + yz + {z^2} $ to be added.
Firstly, let us add the expressions in the first set. This can be written as follows.
 $\Rightarrow 2{y^2} + 3yz + ( - {y^2} - yz - {z^2}) + yz + 2{z^2} $
By removing the brackets, we get
$ 2{y^2} + 3yz - {y^2} - yz - {z^2} + yz + 2{z^2} $
Now we know that terms with the same variables and powers can be added or subtracted so let us segregate all the y terms, z terms and yz terms together.
This expression now becomes
$ 2{y^2} - {y^2} + 3yz - yz + yz - {z^2} + 2{z^2} $
Now, we can take common terms out and write the coefficients together.
$\Rightarrow {y^2}\left( {2 - 1} \right) + yz\left( {3 - 1 + 1} \right) + {z^2}\left( { - 1 + 2} \right) $
By solving this, we get $ \left( 1 \right){y^2} + \left( 3 \right)yz + \left( 1 \right){z^2} $
So now this expression becomes $ {y^2} + 3yz + {z^2} $
Similarly, we can add the expressions of the other set $ 3{y^2} - {z^2} $ and $ - {y^2} + yz + {z^2} $
 $\Rightarrow 3{y^2} - {z^2} - {y^2} + yz + {z^2} $
Let us segregate similar terms together to get $ 3{y^2} - {y^2} + yz - {z^2} + {z^2} $
Taking common variables out and writing coefficients together
$\Rightarrow \left( {3 - 1} \right){y^2} + yz + \left( { - 1 + 1} \right){z^2} $
By solving we get $ 2{y^2} + yz $
Now, we have to subtract the second expression from the first.
We can write this as
$\Rightarrow {y^2} + 3yz + {z^2} - \left( {2{y^2} + yz} \right) $
By removing the bracket we get $ {y^2} + 3yz + {z^2} - 2{y^2} - yz $
We follow the same steps and write this as
$\Rightarrow \left( {1 - 2} \right){y^2} + \left( {3 - 1} \right)yz + {z^2} $
On solving, we get $ - {y^2} + 2yz + {z^2} $
Therefore the answer is $ 2yz + {z^2} - {y^2} $
So, the correct answer is “ $ 2yz + {z^2} - {y^2} $ ”.

Note: One should always check whether the variables have the same power before adding or subtracting their coefficients. It should also be noted that when one expression is subtracted from another, the minus sign should be multiplied by the respective signs of every term after it.