Question

The denominator of a fraction is one less than three times its numerator. On adding one to the fraction becomes 1/3. Find the fraction.

Answer 1

Answer:

Let numerator be x

denominator be y

y=3x-1 - - - - - - - - - - - - - equation 1

x/y + 1 =1/3 - - - - - - - - - - - equation 2

Substitute y in equation 2

x/3x-1 +1 =1/3

x+3x-1 =3x-1/3

3x+9x-3 =3x-1

12x-3x=3-1

9x =2

x=2/9

y=3*2/9 - 1

y=-1/3

the fraction =x/y

=2/9*-3/1

=-2/3

Please mark it as a brainliest answer

Answer 2

The given question is The denominator of a fraction is one less than three times its numerator. On adding one to the fraction becomes 1/3.

we have to find the fraction.

In the fraction let the numerator be x and the denominator be y.

The general expression (fraction)for the given question is

$\frac{x}{y}$

If the denominator is one less than 3 times it's numerator means 3x-1.which is y

y= 3x-1.

On adding 1 to the fraction becomes

$\frac{x}{y} + 1 = \frac{1}{3}$

On substituting the value of y in the above equation we get the value as

$\frac{x}{3x - 1} + 1 = \frac{1}{3} \\ \frac{x + 3x - 1}{3x - 1} = \frac{1}{3} \\ 3(x + 3x - 1) = 1(3x - 1) \\ 3x + 9x - 3 = 3x - 1 \\ 3x - 3x + 9x - 3 + 1 = 0 \\ 9x - 2 = 0 \\ 9x = 2 \\ x = \frac{2}{9}$

The value of x is

$\frac{2}{9}$

substitute the value of x in y=3x-1.

$y = 3( \frac{2}{9} ) - 1 \\ y = \frac{6}{9} - 1 \\ y = \frac{6 - 9}{9} \\ y = \frac{ - 3}{9} \\ y = \frac{ - 1}{3}$

Therefore, the fraction

$\frac{x}{y} = \frac{ \frac{2}{9} }{ \frac{ - 1}{3} } \\ \frac{2}{9} \times \frac{ - 3}{1} \\ = \frac{ - 2}{3}$

Therefore, the final answer is

$\frac{x}{y} = \frac{ - 2}{3}$

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