Question

The sum of the numerator and the denominator of a fraction is 8 if 3 is added to both the numerator and the denominator the fraction becom
es 3 upon 4 find the fraction

Answer 1

Step-by-step explanation:

let the numerator be x

and the denominator be y.

Case 1.

x+y = 8 ........(1)

case 2.

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

cross multiplying,

4(x+3)=3 (y+3)

4x +12=3y+9

4x -3y = -3.......... (2)

subtracting both the eq.

4x-3y = -3......(2)

x + y =8 .... (1)

multiply the eq. 1 with 4 and subtract

4x + 4y = 32

4x - 3y = -3

(-) (+)___(+)______

7y = 35

y= 5

now,

x+y =8

x + 5 =8

x = 3

x = 3

y=5

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \mathfrak{ \: \: Given, \: \: }}\\ \\ \text{The sum of the numerator and denominator = 8} \\ \\ \text{If 3 is added to both numerator and denominator, } \\ \mathtt{then \: \: the \: \: fraction \: \: becomes \: \: \: \frac{3}{4} } \\ \\ \underline{ \mathfrak{ \: \: To \: \: find : \mapsto \: }} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text{The \: fraction = ? }$

$\underline{ \mathfrak{ \: \: Suppose, \: \: }} \\ \\ \: \: \: \: \: \: \: \text{The numerator = x } \\ \text{And, } \\ \: \: \: \: \: \: \: \text{The denominator = y} \: \\ \\ \mathtt{ \therefore \: \: The \: \: \: fraction \: \: \: is = \frac{x}{y} }$

$\underline{ \bold{ \: \: A.T.Q., \: \: }} \\ \\ \: \: \: \: \: \: \mathtt{x + y = 8 \: \: \: - - - - - - > (1)} \\ \\ \mathtt{And,} \\ \\ \: \: \: \: \: \: \mathtt{ \frac{x + 3}{y + 3} = \frac{3}{4} } \\ \\ \mathtt{\Longrightarrow \:4( x + 3) = 3(y + 3) } \\ \\ \mathtt{\Longrightarrow \:4x + 12 = 3y + 9 } \\ \\ \mathtt{\Longrightarrow \: 4x - 3y = 9 - 12} \\ \\ \mathtt{\Longrightarrow \:4x - 3y = - 3 \: \: \: - - - - - - > (2) } \\ \\ \mathfrak{Again,} \\ \\ \mathtt{(1) \times 3\Longrightarrow \:3x + 3y = 24 \: \: \: \: - - - - - > (3)}$

$\mathfrak{Now, } \\ \\ \mathtt{(2) + (3) \Rightarrow 7x = 21 } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{\Longrightarrow x = \frac{21}{7} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{\therefore x = 3} </p><p>$

$\underline{ \text{ Putting the value of x in eq. (1) , we get}} \\ \\ \mathtt{(1) \Rightarrow 3 + y = 8} \\ \\ \: \: \: \: \mathtt{\Longrightarrow \: y = 8 - 3} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{ \therefore\: \: y = 5}$

$\therefore \: \text{Numerator, x = 3} \\ \: \: \: \text{ Denominator, y = 5}$

$\mathtt{ \therefore \: \: The \: \: \: fraction \: \: \: is = \frac{x}{y} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\mathtt{ = \frac{3}{5} }}$

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: VERICATION \: \: } \mid}}}}}$

$\underline{\text{ \: Numerator = 3 \: \: and \: \: Denominator = 5 \: } } \\ \\ : \rightsquigarrow \: \text{Sum of the numerator and denominator = 3 + 5 = 8 } \\ \\ \mathfrak{Again,} \\ \\ : \rightsquigarrow \: \text{When 3 is added to both the numerator and denominator,} \\ \mathtt{ then \: \: the \: \: fraction \: \: becomes = \frac{3 + 3}{5 + 3 } = \frac{6}{8} = \frac{3}{4} }$