Subtract the sum of


Subtract the sum of 3l − 4m − 7n2 and 2l + 3m − 4n2 from the sum of 9l + 2m − 3n2 and − 3l + m + 4n2 .....


We have to subtract the sum of $\left(3 /-4 m-7 n^{2}\right)$ and $\left(2 l+3 m-4 n^{2}\right)$ from the sum of $\left(9 /+2 m-3 n^{2}\right)$ and $\left(-3 /+m+4 n^{2}\right)$

$\left\{\left(9 l+2 m-3 n^{2}\right)+\left(-3 l+m+4 n^{2}\right)\right\}-\left\{\left(3 l-4 m-7 n^{2}\right)+\left(2 l+3 m-4 n^{2}\right)\right\}$

$=\left(9 l-3 l+2 m+m-3 n^{2}+4 n^{2}\right)-\left(3 l+2 l-4 m+3 m-7 n^{2}-4 n^{2}\right)$

$=\left(6 l+3 m+n^{2}\right)-\left(5 l-m-11 n^{2}\right)$            (Combining like terms inside the parentheses)

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

$=6 l-5 l+3 m+m+n^{2}+11 n^{2}$            (Combining like terms )

$=l+4 m+12 n^{2}$                                   (Combining like terms )

Thus, the required solution is $l+4 m+12 n^{2}$.

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