Question

What should be taken away from $3x^2-4y^2+5xy+20$ to obtain $-x^2-y^2+6xy+20$?

• A. $3x^2-3y^2-xy$
• B. $4x^2+3y^2-xy$
• C. $4x^2-3y^2-xy$
• D. $4x^2-3y^2+xy$

Answer 1

The correct option is C $4x^2-3y^2-xy$

'Take away' represents subtraction.
We must subtract a certain quantity from $3x^2-4y^2+5xy+20$ to obtain $-x^2-y^2+6xy+20$.
i.e., $(3x^2-4y^2+5xy+20)-\left(\underset{–}{}\underset{–}{}\underset{–}{}\underset{–}{}\right)=-x^2-y^2+6xy+20$

We see that $3x^2$ in the LHS becomes $-x^2$ in the RHS.
So, the term that has to be subtracted from $3x^2$ in order to get $-x^2$ is $4x^2$.

Similarly, we see that $-4y^2$ in the LHS becomes $-y^2$ in the RHS.
So, the term that has to be subtracted from $-4y^2$ in order to get $-y^2$ is $-3y^2$.

Similarly, by comparing the terms with variable 'xy', we see that the term that has to be subtracted from $5xy$ in order to get $6xy$ is $-xy$.

Also we see that 20 in the LHS remains as it is. So, we need not add or subtract anything in this case.

Therefore, the expression that has to be subtracted from $3x^2-4y^2+5xy+20$ to obtain $-x^2-y^2+6xy+20$ is $4x^2-3y^2-xy$.

Verification:
$(3x^2-4y^2+5xy+20)-(4x^2-3y^2-xy)=3x^2-4y^2+5xy+20-4x^2+3y^2+xy=3x^2-4x^2-4y^2+3y^2+5xy+xy+20=-x^2-y^2+6xy+20$