Question

13x²y²{2z-12} by 39xy{z-6}

Divide using factorization and verify the answer​

Answer 1

Answer:

$\frac{13 {x}^{2} {y}^{2} (2z - 12) }{39xy(z - 6)} \\ \\ \frac{xy(2z - 12}{3(z - 6)} \\ \\ \frac{2xy(z - 6)}{3(z - 6)} \\ \\ \frac{2xy}{3}$

Answer 2

Given equations are

$13x^{2} y^{2} (2z-12)$-----------(1)

and $39xy(z-6)$----------(2)

Now divide the equation (1) by (2)

$=\frac{13x^{2} y^{2} (2z-12)}{39xy(z-6)}$

$=\frac{x^{2} y^{2}2(z-6) }{3xy(z-6)}$

$=\frac{2xy}{3}$

The value after division of equation (1) and (2) we got the value $\frac{2xy}{3}$

Verification: If we multiply the equation(2) and the final we will get equation (1)

$=\frac{2xy}{3} \times 39xy(z-6)$

$=13xy\times xy\times2(z-6)$

$=13x^{2} y^{2} (2z-12)$

This is equal to equation (1)

Hence, the final answer is verified.