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$(a)$ From the sum of $3x-y+11$ and $-y-11$ subtract $3x-y-11$.
$(b)$ From the sum of $4+3x$ and $5-4x+2x^2$ subtract the sum of $3x^2-5x$ and $-x^2+2x+5$.
To do: $(a)$. To subtract $3x-y-11$ from the sum of $3x-y+11$ and $-y-11$.
$(b)$. To subtract the sum of $3x^2-5x$ and $-x^2+2x+5$ from the sum of $4+3x$ and $5-4x+2x^2$.
Solution:
a). The sum of $3x-y+11$ and $-y-11$.
$=(3x-y+11)+(-y-11)$
$=3x-y+11-y-11$
$=3x-2y$
Now, we subtract $3x-y-11$ from $3x-2y$
$(3x-2y)-(3x-y-11)$
$=3x-2y-3x+y+11$
$=-y+11$
b). The sum of $4+3x$ and $5-4x+2x^2$.
$=(4+3x)+(5-4x+2x^2)$
$=4+3x+5-4x+2x^2$
$=-x+2x^2+9$
Sum of $3x^2-5x$ and $-x^2+2x+5$
$=(3x^2-5x)+(-x^2+2x+5)$
$=3x^2-5x-x^2+2x+5$
$=2x^2-3x+5$
Now, we subtract $-x+2x^2+9$ from $2x^2-3x+5$
$=2x^2-x+9-2x^2+3x-5$
$=2x+4$
Updated on: 10-Oct-2022
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