Question

Subtract $4x-2y+z$ from the sum of $3x+4y+z$ and $-6x+7y-8z$

Answer 1

Lets evaluate the sum of the given expressions $3x+4y+z$ and $-6x+7y-8z$,

$(3x+4y+z)+(-6x+7y-8z)=3x+4y+z-6x+7y-8z$

$=(3x-6x)+(4y+7y)+(z-8z)$

$=-3x+11y-7z$ $\dots \dots$(1)

Now, subtract $4x-2y+z$ from the sum obtained in the above step,

Sum of $\left[\right(3x+4y+z)+(-6x+7y-8z\left)\right]-(4x-2y+z)$

$=(-3x+11y-7z)-(4x-2y+z)$ [ $\text{From (1)}$ ]

$=-3x+11y-7z-4x+2y-z$

$=(-3x-4x)+(11y+2y)+(-7z-z)$

$=-7x+13y-8z$

Therefore, the required difference is $-7x+13y-8z$ .