Question

Which among the following expression after simplification would contain only $3$ terms?

• A. $21x+32y+2x-27y+5y$
• B. $5x+6y-4x-y$
• C. $3x+y+2x$
• D. $9x+7y+7x+9y+8$

Answer 1

The correct option is D $9x+7y+7x+9y+8$

In option (a.), we have $21x+32y+2x-27y+5y$.
Combining the like terms of this expression:

$=(21x+2x)+(32y-27y+5y)$
Simplifying the like terms:

$=\underset{–}{(21x+2x)}+\underset{–}{(32y-27y+5y)}$

$=23x+10y$

$\therefore$ The expression $21x+32y+2x-27y+5y$ on simplification gives $23x+10y$ which contains $2$ terms only.

In option (b.), we have $5x+6y-4x-y$. Combining the like terms of this expression:

$=(5x-4x)+(6y-y)$
Simplifying the like terms:

$=\underset{–}{(5x-4x)}+\underset{–}{(6y-y)}$

$=x+5y$

$\therefore$ The expression $5x+6y-4x-y$ on simplification gives $x+5y$ which contains $2$ terms only.

In option (c.), we have $3x+y+2x$. Combining the like terms of this expression:
$=(3x+2x)+y$
Simplifying the like terms:

$=\underset{–}{(3x+2x)}+y$

$=5x+y$

The expression $3x+y+2x$ on simplification would give $5x+y$ which contains $2$ terms only.

In option (d.), we have $9x+7y+7x+9y+8$. Combining the like terms of this expression:
$=(9x+7x)+(7y+9y)+8$
Simplifying the like terms:

$=\underset{–}{(9x+7x)}+\underset{–}{(7y+9y)}+8$

$=16x+16y+8$

$\therefore$ The expression $9x+7y+7x+9y+8$ on simplification would give $\underset{–}{16x+16y+8}$ which contains $2$ variable terms and $1$ constant term, i.e. in total $3$ terms.

Therefore, option (d.) is the correct answer.