Question

a proper fraction whose denominator is greater than its numerator by 5 and when 3 is added to both numerator and denominator it becomes 3 by 4 let's write the equation
and solve it​
please itd urgent

Answer 1

Answer:-

Let the numerator be "x" and denominator be "y".

The fraction => x/y.

Given:

Denominator is greater than the numerator by 5.

=> y = x + 5 -- equation (1)

And,

If 3 is added to numerator and denominator the fraction becomes 3/4.

According to the question,

$\frac{x + 3}{y + 3} = \frac{3}{4}$

Substitute "y" value here,

$\frac{x + 3}{x + 5 + 3} = \frac{3}{4} \\ \\ \frac{x + 3}{x + 8} = \frac{3}{4}$

After cross multiplication we get,

$4(x + 3) = 3(x + 8) \\ \\ 4x + 12 = 3x + 24 \\ \\ 4x - 3x = 24 - 12 \\ \\ x = 12$

Substitute "x" value in equation (1).

y = x + 5

y = 12 + 5

y = 17

Therefore, the fraction x/y = 12/17.

Answer 2

Given,

• A fraction whose denominator is greater than its numerator by 5.
• When 3 is added to both numerator and denominator then it becomes 3/4.

To Find,

• The Fraction

Solution,

⇒Suppose the numerator be a

And Suppose the denominator be b

Therefore ,

$\boxed{\bold{\red{The \ Fraction = \frac{a}{b}}}}$

According to the First Condition :-

• A fraction whose denominator is greater than its numerator by 5.

$\implies \boxed{\sf{b = a + 5}}$

According to the Second Condition :-

• When 3 is added to both numerator and denominator then it becomes 3/4.

$\implies \sf{\dfrac{a + 3}{b + 3} = \dfrac{3}{4}}$

$\implies \sf{4(a + 3) = 3(b + 3)}$

$\implies \sf{4a + 12 = 3b + 9}$

$\implies \sf{4a - 3b = 9 - 12}$

$\implies \sf{4a - 3b = -3}$

$\implies \sf{4a - 3(a + 5) = -3}$

(Put the value of b From the First Condition)

$\implies \sf{4a -3a - 15 = -3}$

$\implies \sf{a = -3 + 15}$

$\implies \boxed{\sf{a = 12}}$

Now Put the Value of a in First Condition :-

$\implies \sf{b = a + 5}$

$\implies \sf{b = 12 + 5}$

$\implies \boxed{\sf{b = 17}}$

Therefore,

$\boxed{\sf{\purple{The \ Fraction = \dfrac{a}{b} = \dfrac{12}{17}}}}$

$\rule{200}2$