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

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Solution

We have to subtract the sum of (3l $-$ 4m $-$ 7n²) and (2l + 3m $-$ 4n²) from the sum of (9l + 2m $-$ 3n²) and ( $-$ 3l + m + 4n²)

$\{(9 l + 2 m - 3 n^{2}) + (- 3 l + m + 4 n^{2})\} - \{(3 l - 4 m - 7 n^{2}) + (2 l + 3 m - 4 n^{2})\}$
$= (9 l - 3 l + 2 m + m - 3 n^{2} + 4 n^{2}) - (3 l + 2 l - 4 m + 3 m - 7 n^{2} - 4 n^{2})$
$= (6 l + 3 m + n^{2}) - (5 l - m - 11 n^{2})$ (Combining like terms inside the parentheses)
$= 6 l + 3 m + n^{2} - 5 l + m + 11 n^{2}$
$= 6 l - 5 l + 3 m + m + n^{2} + 11 n^{2}$ (Collecting 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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