Question

(b) If L= (3y + 1/7) , M = (3y - 1/8 ) ,N = (3y -1/7)

(i) Find the sum of L , M and N

(ii) Find the product of L and M​

Answer 1

$\large\sf{\boxed{\underline{Answer :-}}}$

Given :-

L = 3y + $\dfrac{1}{7}$

M = 3y - $\dfrac{1}{8}$

N = 3y - $\dfrac{1}{7}$

To Find :-

1. L + M + N

2. L × M

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1. L + M + N

$\implies$ 3y + $\dfrac{1}{7}$ + 3y - $\dfrac{1}{8}$ + 3y - $\dfrac{1}{7}$

$\implies$ 3y + 3y + 3y + $\dfrac{1}{7}$ - $\dfrac{1}{7}$ - $\dfrac{1}{8}$

$\implies$ 9y + $\cancel{\dfrac{1}{7}}$ - $\cancel{\dfrac{1}{7}}$ - $\dfrac{1}{8}$

$\implies$ 9y - $\dfrac{1}{8}$

$\therefore$ L + M + N = 9y - $\dfrac{1}{8}$

2. L × M

$\implies$ ( 3y + $\dfrac{1}{7}$ ) × ( 3y - $\dfrac{1}{8}$ )

$\implies$ 3y × 3y - 3y × $\dfrac{1}{8}$ + $\dfrac{1}{7}$ × 3y - $\dfrac{1}{7}$ × $\dfrac{1}{8}$

$\implies$ 9y² - $\dfrac{3y}{8}$ + $\dfrac{3y}{7}$ - $\dfrac{1}{56}$

$\implies$ 9y² - $\dfrac{21}{56}$ + $\dfrac{24}{56}$ - $\dfrac{1}{56}$

$\implies$ 9y² + $\dfrac{3y}{56}$ - $\dfrac{1}{56}$

$\therefore$ L× M = 9y² + $\dfrac{3y}{56}$ - $\dfrac{1}{56}$

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Answer completed :)

Answer 2

To Find :-

1. L + M + N

2. L × M

___________________________

1. L + M + N

=> 3y+1/7 + 3y-1/8 + 3y-1/7

=> 3y+ 3y + 3y + 1/7- 1/7- 1/8

=> 9y - 1/8 Ans.

2. L × M

=> (3y + 1/7) + (3y-1/8)

=> 3y × 3y - 3y × 1/8 + 1/7 × 3y - 1/7 × 1/8

=> 9y² - 3y/8 + 3y/7 - 1/56

=> 9y²- 21/56 + 24/56 - 1/56

=> 9y² + 3y/56 - 1/56 Ans.