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)​

$\left\{\left(9l+2m-3n^2\right)+\left(-3l+m+4n^2\right)\right\}-\left\{\left(3l-4m-7n^2\right)+\left(2l+3m-4n^2\right)\right\}$
$=\left(9l-3l+2m+m-3n^2+4n^2\right)-\left(3l+2l-4m+3m-7n^2-4n^2\right)$
$=\left(6l+3m+n^2\right)-\left(5l-m-11n^2\right)$ (Combining like terms inside the parentheses)
$=6l+3m+n^2-5l+m+11n^2$
$=6l-5l+3m+m+n^2+11n^2$ (Collecting like terms)
$=l+4m+12n^2$ (Combining like terms)

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