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

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**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.