Question

The sum of numerator and denominator of a fraction is 4 more than twice the numerator if the numerator and denominator are increased by 3 they are in ratio 2:3 find the fraction using elimination method

Answer 1

Answer:

Plz mark it as branliest

Answer 2

Answer:

The given fraction will be 5/9.

Solution:

Let n be the numerator and d be the denominator.

Given,  

The sum of the given n and d is equal to twice the numerator plus 4, i.e.

$\begin{array} { c } { n + d = 2 n + 4 } \\\\ { n + d - 2 n - 4 = 0 } \\\\ { - n + d = 4 } \\\\ { n - d = - 4 \ldots ( i ) } \end{array}$

From given, the numerator n and the denominator d must be increased by 3 to get the ratio 2:3

$\begin{array} { c } { \frac { n + 3 } { d + 3 } = \frac { 2 } { 3 } } \\\\ { 3 n + 9 = 2 d + 6 } \\\\ { 3 n - 2 d = 6 - 9 } \\\\ { 3 n - 2 d = - 3 \quad \ldots . ( i i ) } \end{array}$

From equations (i) and (ii), we get,

$\begin{aligned} ( i ) \times 3 \Rightarrow & 3 n - 3 d = - 12 \\\\ (ii) & \Rightarrow 3 n - 2 d = - 3 \end{aligned}$

Solving (i) and (ii) we get,

$-d=-9\\\\d=9$

Substituting d=9 in equation (i), we get,

$n-d=-4\\\\n-9=-4\\\\n=-4+9\\\\n=5$

∴ The given fraction is  

$\frac {n}{d}=\frac{5}{9}$