Question

$$\frac{ 2t }{ t-5 } + \frac{ 12 }{ { t }^{ 2 } -25 }$$

Answer

(2*(t*(t+5)+6))/((t-5)*(t+5))

Solution


Rewrite \({t}^{2}-25\) in the form \({a}^{2}-{b}^{2}\), where \(a=t\) and \(b=5\).
\[\frac{2t}{t-5}+\frac{12}{{t}^{2}-{5}^{2}}\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[\frac{2t}{t-5}+\frac{12}{(t+5)(t-5)}\]
Rewrite the expression with a common denominator.
\[\frac{2t(t+5)+12}{(t-5)(t+5)}\]
Factor out the common term \(2\).
\[\frac{2(t(t+5)+6)}{(t-5)(t+5)}\]