Question

1

Answer 1

same reason you need them for numbers

Answer 2

suppose you have \(\frac{1}{x}+\frac{7}{x+3}\) since \(x\) is a variable, your addition must work for any real value of \(x\) so for example if you had \(x=5\)
you would be adding \(\frac{1}{5}+\frac{7}{8}\)
you need a common denominator here, namely \(5\times 8=40\) so you need a common denominator for \(\frac{1}{x}+\frac{7}{x+3}\) , namely \(x(x+3)\)