Question

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

Answer 1

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)​

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

\[= \left( 9l - 3l + 2m + m - 3 n^2 + 4 n^2 \right) - \left( 3l + 2l - 4m + 3m - 7 n^2 - 4 n^2 \right)\]

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

\[= 6l + 3m + n^2 - 5l + m + 11 n^2\]

\[= 6l - 5l + 3m + m + n^2 + 11 n^2\]               (Collecting like terms)

\[= l + 4m + 12 n^2\]                              (Combining like terms)

Thus, the required solution is \[l + 4m + 12 n^2\].