Question

16a^2/4a+11b minus 121b^2/4a+11b

Please help

Answer 1

I think this is your problem:\[\frac{ 16a^2 }{ 4a+11b }-\frac{ 121b^2 }{ 4a+11b }\]so luckily, because the denominators are the same, it can be written as one fraction:\[\frac{ 16a^2-121b^2 }{ 4a+11b }\]Are you familiar with the factorisation of the diference of two squares?
I mean:\[a^2-b^2=(a+b)(a-b)\]You now should realise that 16a² and 121b² are also squares, of 4a and 11b.
Now factor the numerator and then simplify!

Answer 2

Thank you for your help ZeHanz