Question

For $x=\frac{4}{7}$ and $y=\frac{3}{13}$, verify that: $(x\times y{)}^{-1}=x^{-1}\times y^{-1}$ [4 MARKS]

Answer 1

Concept : 1 Mark
Simplification : 2 Marks
Verification : 1 Mark

We have, $x=\frac{4}{7},y=\frac{3}{13}$

Now, $(x\times y{)}^{-1}={(\frac{4}{7}\times \frac{3}{13})}^{-1}$

$={\left(\frac{4\times 3}{7\times 13}\right)}^{-1}$

$={\left(\frac{12}{91}\right)}^{-1}$

$=\frac{91}{12}$

Again, $x^{-1}\times y^{-1}={\left(\frac{4}{7}\right)}^{-1}\times {\left(\frac{3}{13}\right)}^{-1}$

$=\frac{7}{4}\times \frac{13}{3}$

$=\frac{7\times 13}{4\times 3}$

$=\frac{91}{12}$

$\therefore (x\times y{)}^{-1}=x^{-1}\times y^{-1}$

hence, we have verified the result.