Plz answer my questionfind the rational number in each pair and examin

1.	Plz answer my questionfind the rational number in each pair and examine if they are equal or not :(3/7+7/9)-2/7 , 3/7-(-7/9-2/7)
Answer» ★ How to do :- Here, we are GIVEN with some of the rational numbers that should be solved separately. On the other side, we are given with other three rational numbers in which they should be solved separately. The FIRST PART can be considered as LHS and the second part can be considered as RHS. We can solve both of them separately and then, we can compare the results of both pairs. This concept can do be known as associative property in which on LHS, we group first two numbers and on LHS, we group the last two numbers and vise versa. So, let's solve!! $\:$ ➤ Solution :- ${\tt \leadsto \bigg( \dfrac{3}{7} + \dfrac{7}{9} \bigg) - \dfrac{2}{7} \neq \dfrac{3}{7} - \bigg( \dfrac{(-7)}{9} - \dfrac{2}{7} \bigg)}$ LHS :- ${\tt \leadsto \bigg( \dfrac{3}{7} + \dfrac{7}{9} \bigg) - \dfrac{2}{7}}$ First solve the numbers in bracket. LCM of 7 and 9 is 63. ${\tt \leadsto \bigg( \dfrac{3 \times 9}{7 \times 9} + \dfrac{7 \times 7}{9 \times 7} \bigg) - \dfrac{2}{7}}$ Multiply the numerators and denominators of both fractions. ${\tt \leadsto \bigg( \dfrac{27}{<klux>63</klux>} + \dfrac{49}{63} \bigg) - \dfrac{2}{7}}$ Write both numerators with a common denominator. ${\tt \leadsto \bigg( \dfrac{27 + 49}{63} \bigg) - \dfrac{2}{7}}$ Add them now. ${\tt \leadsto \dfrac{76}{63} - \dfrac{2}{7}}$ LCM of 63 and 7 is 63. ${\tt \leadsto \dfrac{76}{63} - \dfrac{2 \times 9}{7 \times 9}}$ Multiply the numerator and denominator of second fraction. ${\tt \leadsto \dfrac{76}{63} - \dfrac{18}{63}}$ Subtract the fractions now. ${\tt \leadsto \dfrac{76 - 18}{63} = \dfrac{58}{63}}$ $\:$ RHS :- ${\tt \leadsto \dfrac{3}{7} - \bigg( \dfrac{(-7)}{9} - \dfrac{2}{7} \bigg)}$ First solve the numbers in bracket. LCM of 9 and 7 is 63. ${\tt \leadsto \dfrac{3}{7} - \bigg( \dfrac{(-7) \times 7}{9 \times 7} - \dfrac{2 \times 9}{7 \times 9} \bigg)}$ Multiply the numerators and denominators of both fractions. ${\tt \leadsto \dfrac{3}{7} - \bigg( \dfrac{(-49)}{63} - \dfrac{18}{63} \bigg)}$ Write both numerators with a common denominator. ${\tt \leadsto \dfrac{3}{7} - \bigg( \dfrac{(-49) - 18}{63} \bigg)}$ Subtract them now. ${\tt \leadsto \dfrac{3}{7} - \dfrac{(-67)}{63}}$ LCM of 7 and 63 is 63. ${\tt \leadsto \dfrac{3 \times 9}{7 \times 9} - \dfrac{(-67)}{63}}$ Multiply the numerators and denominators of first fraction. ${\tt \leadsto \dfrac{27}{63} - \dfrac{(-67)}{63}}$ Write the second NEVER in numerator with one sign. ${\tt \leadsto \dfrac{27 - (-67)}{63} = \dfrac{27 + 67}{63}}$ Add the numerators now. ${\tt \leadsto \dfrac{94}{63}}$ $\:$ Now, let's compare the results of LHS and RHS. ${\tt \leadsto \dfrac{58}{63} \: \: and \: \: \dfrac{94}{63}}$ We can see that they are like fractions, by which we can compare them easily. ${\tt \leadsto \dfrac{58}{63} \neq \dfrac{94}{63}}$ So, ${\sf \leadsto LHS \neq RHS}$ $\:$ ${\red{\underline{\boxed{\bf So, \: they \: aren't \: equal}}}}$

Plz answer my questionfind the rational number in each pair and examine if they are equal or not :(3/7+7/9)-2/7 , 3/7-(-7/9-2/7)

Answer»

★ How to do :-

Here, we are GIVEN with some of the rational numbers that should be solved separately. On the other side, we are given with other three rational numbers in which they should be solved separately. The FIRST PART can be considered as LHS and the second part can be considered as RHS. We can solve both of them separately and then, we can compare the results of both pairs. This concept can do be known as associative property in which on LHS, we group first two numbers and on LHS, we group the last two numbers and vise versa. So, let's solve!!

$\:$

➤ Solution :-

${\tt \leadsto \bigg( \dfrac{3}{7} + \dfrac{7}{9} \bigg) - \dfrac{2}{7} \neq \dfrac{3}{7} - \bigg( \dfrac{(-7)}{9} - \dfrac{2}{7} \bigg)}$

LHS :-

${\tt \leadsto \bigg( \dfrac{3}{7} + \dfrac{7}{9} \bigg) - \dfrac{2}{7}}$

First solve the numbers in bracket.

LCM of 7 and 9 is 63.

${\tt \leadsto \bigg( \dfrac{3 \times 9}{7 \times 9} + \dfrac{7 \times 7}{9 \times 7} \bigg) - \dfrac{2}{7}}$

Multiply the numerators and denominators of both fractions.

${\tt \leadsto \bigg( \dfrac{27}{<klux>63</klux>} + \dfrac{49}{63} \bigg) - \dfrac{2}{7}}$

Write both numerators with a common denominator.

${\tt \leadsto \bigg( \dfrac{27 + 49}{63} \bigg) - \dfrac{2}{7}}$

Add them now.

${\tt \leadsto \dfrac{76}{63} - \dfrac{2}{7}}$

LCM of 63 and 7 is 63.

${\tt \leadsto \dfrac{76}{63} - \dfrac{2 \times 9}{7 \times 9}}$

Multiply the numerator and denominator of second fraction.

${\tt \leadsto \dfrac{76}{63} - \dfrac{18}{63}}$

Subtract the fractions now.

${\tt \leadsto \dfrac{76 - 18}{63} = \dfrac{58}{63}}$

$\:$

RHS :-

${\tt \leadsto \dfrac{3}{7} - \bigg( \dfrac{(-7)}{9} - \dfrac{2}{7} \bigg)}$

First solve the numbers in bracket.

LCM of 9 and 7 is 63.

${\tt \leadsto \dfrac{3}{7} - \bigg( \dfrac{(-7) \times 7}{9 \times 7} - \dfrac{2 \times 9}{7 \times 9} \bigg)}$

Multiply the numerators and denominators of both fractions.

${\tt \leadsto \dfrac{3}{7} - \bigg( \dfrac{(-49)}{63} - \dfrac{18}{63} \bigg)}$

Write both numerators with a common denominator.

${\tt \leadsto \dfrac{3}{7} - \bigg( \dfrac{(-49) - 18}{63} \bigg)}$

Subtract them now.

${\tt \leadsto \dfrac{3}{7} - \dfrac{(-67)}{63}}$

LCM of 7 and 63 is 63.

${\tt \leadsto \dfrac{3 \times 9}{7 \times 9} - \dfrac{(-67)}{63}}$

Multiply the numerators and denominators of first fraction.

${\tt \leadsto \dfrac{27}{63} - \dfrac{(-67)}{63}}$

Write the second NEVER in numerator with one sign.

${\tt \leadsto \dfrac{27 - (-67)}{63} = \dfrac{27 + 67}{63}}$

Add the numerators now.

${\tt \leadsto \dfrac{94}{63}}$

$\:$

Now, let's compare the results of LHS and RHS.

${\tt \leadsto \dfrac{58}{63} \: \: and \: \: \dfrac{94}{63}}$

We can see that they are like fractions, by which we can compare them easily.

${\tt \leadsto \dfrac{58}{63} \neq \dfrac{94}{63}}$

So,

${\sf \leadsto LHS \neq RHS}$

$\:$

${\red{\underline{\boxed{\bf So, \: they \: aren't \: equal}}}}$