Question

Write any three fractions equivalent to the following fraction \[\dfrac{7}{9}\].

Answer 1

Hint:Fraction is a number which is of the form \[\dfrac{p}{q}\] where \[q\] is not equal to zero. That is, \[\dfrac{p}{q},(q \ne 0)\]
An Equivalent fraction is a fraction which has different numerators and denominators that represents the same value or proportion of whole. Here we make many ways to find the equivalent fraction of the given fraction.

Complete step-by-step answer:
It is given that the fraction \[\dfrac{7}{9}\]
We can write it as,
     \[\dfrac{7}{9} = 0.777....\]
That is approximate to \[\dfrac{7}{9} = 0.\overline 7 \]
We can solve many way in this problem,
Here we can solve one way,
That is, \[7\]multiply by \[a\]and \[9\] multiply by \[b\]where \[a \ne 0,b \ne 0\] we can
We take \[a\] and \[b\]has variables
\[ \Rightarrow \dfrac{{7 \times a}}{{9 \times b}} = \dfrac{{7a}}{{9b}}.... \to \left( 1 \right)\]
Now we have some cases,
If \[(i)\] $\text{a = b}$
    \[(ii)\] $\text{a} \ne\text{b}$

Case \[(i)\] $\text{a = b}$
We can take a number from 1 to 9, and put into a and b
Suppose \[a = b = 3\]
\[ \Rightarrow \dfrac{7}{9} \times \dfrac{3}{3} = \dfrac{{21}}{{27}}\]
\[ \Rightarrow \dfrac{{21}}{{27}} = 0.777..\]
\[ \Rightarrow \dfrac{{21}}{{27}} = 0.\overline 7 \]
Suppose we take \[a = b = 4\] substitute in equation \[\left( 1 \right)\] and we get \[\dfrac{{28}}{{36}}\]
Now we find,
     \[\dfrac{{28}}{{36}} = 0.77..\]
      \[\dfrac{{28}}{{36}} = 0.\overline 7 \]
Similarly, for any number \[a( = b) = \{ 1,2,3,4,5,6,7,8,9\} \]
It gives the same value, \[0.\overline 7 \]

Now we take second case
Case \[(ii)\] $\text{a} \ne\text{b}$
Put \[a = 6,b = 5\]
\[\dfrac{{7 \times 6}}{{9 \times 5}} = \dfrac{{42}}{{45}}\]
\[\dfrac{{42}}{{45}} \ne 0.\overline 7 \]
Here we can take any number for \[a\] and any number for \[b\](which is not the same number) is not equal to the given fraction value.
It is satisfied only when \[a\] and \[b\] are equal.
Then we take any arbitrary values for a and b (is equal)
In the sense, take \[a( = b) = \{ 2,5,9\} \]
We get,
\[ \Rightarrow \dfrac{{7a}}{{9b}} = \dfrac{{14}}{{18}} = 0.\overline 7 \]
\[ \Rightarrow \dfrac{{7a}}{{9b}} = \dfrac{{35}}{{45}} = 0.\overline 7 \]
\[ \Rightarrow \dfrac{{7a}}{{9b}} = \dfrac{{63}}{{81}} = 0.\overline 7 \]
$\therefore $The three fractions are \[\dfrac{{14}}{{18}},\dfrac{{35}}{{45}},\dfrac{{63}}{{81}}\] equivalent fraction to the fraction\[\dfrac{7}{9}\]

Note:If we notice, we use only a single digit in both cases and all through the sum, if you want a double digit or more than it will also be in the way of our solving.
If we take \[a = 7,b = 9\] in the given fraction that same as a numerator and denominator value for \[a\] and \[b\] we get the same result,
That is
\[ \Rightarrow \dfrac{{7a}}{{9b}} = \dfrac{{7 \times 7}}{{9 \times 9}} = \dfrac{{49}}{{81}}\]
\[ \Rightarrow \dfrac{{49}}{{81}} = 0.\overline 7 \]
This is the way to approach another method.