Question

plz answer my question
find 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 1

Answer:

here is your answer

hope it helps you

Step-by-step explanation:

A number that can be expressed in the form p/q, where p and q are integers and q ≠ 0, is called a rational number. All integers and fractions are rational numbers.

Examples: -2/7, 3/8, etc.

Numerator and denominator

In p/q, the integer p is the numerator, and the integer q (≠ 0) is the denominator.

Equivalent rational numbers: By multiplying the numerator and denominator of a rational number by the same non zero integer, we obtain another rational number equivalent to the given rational number.  

Positive rational numbers: When the numerator and denominator both are positive

negative rational numbers: When the numerator is positive and denominator is negative.

The number 0 is neither a positive nor a negative rational number.

Subsequently, the concept of rational numbers on a number line is also explained.

After this, in the first half of the chapter, some more topics like rational numbers in standard form, comparison of rational numbers and rational numbers between two rational numbers are explained.

The standard form of rational numbers must be studied which states that a rational number is in the standard form if its denominator is a positive integer and the numerator and denominator have no common factor other than 1.  

There are an unlimited number of rational numbers between two rational numbers.

In the other half of the chapter, Operations on rational numbers are discussed. OPERATIONS discussed in the chapter are:

1. Addition

2. Additive inverse

3. Subtraction

4. Multiplication

5. Division

have a nice day

Answer 2

★ 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.

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

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,

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

$\:$

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