Question

3y-3/y ÷ 7y-7/3(y to the power of 2)​

Answer 1

Answer:

(hope it helps you )it is a easy method

Answer 2

On simplification we get

$\frac{9y^{3}-9y}{y^{2} -7}$ =$\frac{3y-3}{y}$ ÷  $\frac{7y-7}{3y^{2} }$

Step-by-step explanation:

Given:

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

To find:

Simplify the term

Solution:

We need to simplify the fractional form into simple fractional terms.

$\frac{3y-3}{y}$ gets divided by $\frac{7y-7}{3y^{2} }$ , On simplifying the form we get the higher-order power terms.

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

 $\frac{3y-3}{y}$

           

  $\frac{7y-7}{3y^{2} }$

On splitting the numerator and thus taking lcm 3y² in the denominator we have,

 $\frac{3y}{y}$ - $\frac{3}{y}$

               

 $\frac{7y}{3y^{2} }$ - $\frac{7}{3y^{2} }$

Canceling the like term, therefore we get,

3 - $\frac{3}{y}$

           

$\frac{7}{3y}$ - $\frac{7}{3y^{2} }$

$\frac{3y-3}{y}$

           

$\frac{7y}{3y^{2} }$ -  $\frac{7}{3y^{2} }$

$\frac{3y-3}{y}$

         

$\frac{7y-7}{3y^{2} }$

Thus the fractional denominator term changed as an inverse term of the fraction and we get,

$\frac{3y-3}{y}$ × $\frac{3y^{2} }{7y-7}$

Canceling y term we get

$\frac{3y-3}{7y-7}$ ×$3y}$

We take $\frac{7y+7}{7y+7}$ the term as the inverse of the denominator for simplification,

$\frac{9y^{2} -9y}{7y-7}$ ×$\frac{7y+7}{7y+7}$

Applying formula (a+b)(a-b)=a²-b² we get,

$\frac{63y^{3}-63y^{2} +63y^{2}-63y }{7y^{2}-7^{2} }$

(-63y²+63y²=0) canceling the term we get,

$\frac{63y^{3} -63y}{7y^{2}-49 }$

Taking 7 as common we get,

$\frac{9y^{3}-9y}{y^{2} -7}$

Hence,

By simplification we get $\frac{9y^{3}-9y}{y^{2} -7}$ .