Answer:
The solution for multiplication of
![\bold{(4)/(3) * (6)/(7) \text { is } (8)/(7)}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lmqp00dhwvx9m7amy3y083x5e0ykel3cvg.png)
Solution:
From question, given that
![= (4)/(3) * (6)/(7)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/4ilz4zbx57yhq61nqu04cuij34e47602nh.png)
Rewrite the above term, the expression becomes,
![=(4 * 6)/(3 * 7)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/q17y8n0s88171ersfwik0pgmb2a6cvnquc.png)
Factored form of above term,
![=(4 * 2 * 3)/(3 * 7)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/o802ibcb8m7vdcrex7sm3ar4o3hz16hc44.png)
Canceling 3 in numerator and denominator. Since 3 is common term in numerator and denominator both. So the above expression becomes,
![=(4 * 2)/(7)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/h86a7p8gg2invse6d4ynepfoqw9mpj3py8.png)
On simplifying above term by multiplying 4 and 2, the above expression becomes,
![=(8)/(7)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/grmvuvhgqpqi3lhy8mr4h1rletzsqp38v4.png)
Hence the answer for the multiplication of
![(4)/(3) * (6)/(7) \text { is } (8)/(7)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/g9m6k66hubgav52jd8j52p6da7fhc0v3du.png)