Subtract the sum of $3l- 4m - 7n^2$ and $2l + 3m - 4n^2$ from the sum of $9l + 2m - 3n^2$ and $-3l + m + 4n^2$.

To do:

We have to subtract the sum of $3l- 4m - 7n^2$ and $2l + 3m - 4n^2$ from the sum of $9l + 2m - 3n^2$ and $-3l + m + 4n^2$.

Solution:

Sum of $9l + 2m - 3n^2$ and $-3l + m + 4n^2$ is,

$9l + 2m - 3n^2 + (-3l + m + 4n^2)= 9l + 2m - 3n^2 - 3l + m + 4n^2$

$= 9l- 3l+ 2m + m - 3 n^2 + 4n^2$

$= 6l + 3m + n^2$

Sum of $3l - 4m - 7n^2$ and $2l +3m- 4n^2$ is,

$3l- 4m - 7n^2 +(2l+ 3m- 4n^2)= 3l + 2l - 4m + 3m- 7n^2 - 4n^2$

$= 5l -m- 11n^2$

Subtracting $5l -m- 11n^2$ from $6l + 3m + n^2$

$(6l + 3m + n^2) - (5l - m - 11n^2) = 6l + 3m + n^2 - 5l + m + 11n^2$

$= 6l - 5l + 3m + m + n^2 + 11n^2$

$= l + 4m+ 12n^2$

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Updated on: 10-Oct-2022

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