Question

${\displaystyle \frac{3y-3}{y}÷\frac{7y-7}{3y^2}}$ is

Options

• ${\displaystyle \frac{9y}{7}}$

• ${\displaystyle \frac{9y^3}{(21y-21)}}$

• ${\displaystyle \frac{21y^2-42y+21}{3y^3}}$

• ${\displaystyle \frac{7(y^2-2y+1)}{y^2}}$

Answer 1

${\displaystyle \frac{9y}{7}}$

Explanation;

Hint: ${\displaystyle \frac{3y-3}{y}÷\frac{7y-7}{3y^2}}$

= ${\displaystyle \frac{3(y-1)}{y}÷\frac{7(y-1)}{3y^2}}$

= ${\displaystyle \frac{3(y-1)}{y}\times \frac{3y^2}{7(y-1)}}$

= ${\displaystyle \frac{9y}{7}}$