Carcass wrote:
If \(\frac{a}{b}=\frac{3}{5}\) , then what is the value \(\frac{b(a+b)(\frac{b}{a}-1)}{a(a-b)(\frac{a}{b}-1)}\)
Divide the numerator and denominator by ab
\(\frac{\frac{b(a+b)}{ab}(\frac{b}{a}-1)}{\frac{a(a-b)}{ab}(\frac{a}{b}-1)}\)
\(\frac{\frac{(a+b)}{a}(\frac{b}{a}-1)}{\frac{(a-b)}{b}(\frac{a}{b}-1)}\)
\(\frac{(1+\frac{b}{a})(\frac{b}{a}-1)}{(\frac{a}{b}-1)(\frac{a}{b}-1)}\)
\(\frac{(\frac{b}{a}+1)(\frac{b}{a}-1)}{(\frac{a}{b}-1)(\frac{a}{b}-1)}\)
\(\frac{(\frac{b}{a})^2-1}{(\frac{a}{b}-1)^2}\)
Now substitute the value a/b=2/5 and b/a = 5/3
\(\frac{(\frac{5}{3})^2-1}{(\frac{3}{5}-1)^2}\)
\(\frac{\frac{25}{9}-1}{(\frac{-2}{5})^2}\)
\(\frac{\frac{16}{9}}{\frac{4}{25}}\)
\((\frac{16}{9})(\frac{25}{4})\)
\(\frac{100}{9}\)