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Given, $\frac{x^2 - x - 2}{x^2 - 4}$

Step 1:

Factorized polynomials:

x^{2} - x - 2 = x^{2} - 2x + x - 2

= x(x - 2) + (x - 2)

= (x + 1)(x - 2)

and x^{2} - 4 = x^{2} - 2^{2}

= (x - 2)(x + 2)

Step 2:

$\frac{x^2 - x - 2}{x^2 - 4}$ = $\frac{(x + 1)(x - 2)}{(x - 2)(x + 2)}$

= $\frac{x + 1}{x + 2}$

=> $\frac{x^2 - x - 2}{x^2 - 4}$ = $\frac{x + 1}{x + 2}$

Step 1:

Factorized polynomials:

x

= x(x - 2) + (x - 2)

= (x + 1)(x - 2)

and x

= (x - 2)(x + 2)

Step 2:

$\frac{x^2 - x - 2}{x^2 - 4}$ = $\frac{(x + 1)(x - 2)}{(x - 2)(x + 2)}$

= $\frac{x + 1}{x + 2}$

=> $\frac{x^2 - x - 2}{x^2 - 4}$ = $\frac{x + 1}{x + 2}$

Let present age of Sasra's son = x

Then age of Sasra = 3x

5 years later :

Age of Sasra's son = x + 5

and age of Sasra = 3x + 5

Since she will be two and a half times as old as her son:

=> 3x + 5 = $2\frac{1}{2}$(x + 5)

=> 3x + 5 = $\frac{5}{2}$(x + 5)

=> 2(3x + 5) = 5(x + 5)

=> 6x + 10 = 5x + 25

=> 6x - 5x = 25 - 10

=> x = 15, age of Sasra's son

So age of Sasra = 3x = 3 * 15 = 45

Answer:She is 45 years old and her son is 15 years old.

Then age of Sasra = 3x

5 years later :

Age of Sasra's son = x + 5

and age of Sasra = 3x + 5

Since she will be two and a half times as old as her son:

=> 3x + 5 = $2\frac{1}{2}$(x + 5)

=> 3x + 5 = $\frac{5}{2}$(x + 5)

=> 2(3x + 5) = 5(x + 5)

=> 6x + 10 = 5x + 25

=> 6x - 5x = 25 - 10

=> x = 15, age of Sasra's son

So age of Sasra = 3x = 3 * 15 = 45

Answer:

Step 1:

Let the fraction will be $\frac{x}{y}$

Case 1:

Increased numerator by 2 and denominator by 1, it becomes $\frac{5}{8}$

=> $\frac{x + 2}{y + 1}$ = $\frac{5}{8}$ .......................(1)

Case 2:

Increased numerator and denominator by 1, the fraction becomes equal to half

=> $\frac{x + 1}{y + 1}$ = $\frac{1}{2}$ .........................(2)

Step 2:

(1) => 8(x + 2) = 5(y + 1)

=> 8x + 16 = 5y + 5

=> 8x - 5y + 11 = 0 .............................(3)

(2) => 2(x + 1) = y + 1

=> 2x + 2 = y + 1

=> 2x - y + 1 = 0

=> y = 2x + 1 ...............................(4)

Step 3:

Put y = 2x + 1 in equation (3)

=> 8x - 5(2x + 1) + 11 = 0

=> 8x - 10x - 5 + 11 = 0

=> -2x + 6 = 0

=> 2x = 6

=> x = 3

Step 4:

Put x = 3 in equation (4)

=> y = 2 * 3 + 1

=> y = 7

Hence the fraction is $\frac{3}{7}$.

Let the fraction will be $\frac{x}{y}$

Case 1:

Increased numerator by 2 and denominator by 1, it becomes $\frac{5}{8}$

=> $\frac{x + 2}{y + 1}$ = $\frac{5}{8}$ .......................(1)

Case 2:

Increased numerator and denominator by 1, the fraction becomes equal to half

=> $\frac{x + 1}{y + 1}$ = $\frac{1}{2}$ .........................(2)

Step 2:

(1) => 8(x + 2) = 5(y + 1)

=> 8x + 16 = 5y + 5

=> 8x - 5y + 11 = 0 .............................(3)

(2) => 2(x + 1) = y + 1

=> 2x + 2 = y + 1

=> 2x - y + 1 = 0

=> y = 2x + 1 ...............................(4)

Step 3:

Put y = 2x + 1 in equation (3)

=> 8x - 5(2x + 1) + 11 = 0

=> 8x - 10x - 5 + 11 = 0

=> -2x + 6 = 0

=> 2x = 6

=> x = 3

Step 4:

Put x = 3 in equation (4)

=> y = 2 * 3 + 1

=> y = 7

Hence the fraction is $\frac{3}{7}$.