Question

4th time posting this...........

Answer 1

Answer:

D is the correct representation.

$\frac{\frac{4x^2 + 2x}{x^{2}+x-2} }{ \frac{8x^2 +4x}{3x^2 +10x+8} }$ =  ${\frac{(3x+4)}{2(x-1)}$

Step-by-step explanation:

Here, the given equation is:

$\frac{\frac{4x^2 + 2x}{x^{2}+x-2} }{ \frac{8x^2 +4x}{3x^2 +10x+8} }$

here, numerator  = ${\frac{4x^2 + 2x}{x^{2}+x-2} }$

and denominator = $\frac{8x^2 +4x}{3x^2 +10x+8}$

Solving numerator and denominator separately, we get

NUMERATOR:

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

Denominator:

$\frac{8x^2 +4x}{3x^2 +10x+8} = \frac{4x(2x+1)}{3x^2 +6x+ 4x+8}\\ \implies \frac{4x(2x+1)}{3x(x+2)+ 4(x+2)} = \frac{4x(2x+1)}{(3x+4)(x+2)}$

Hence, the transformed  fraction is:

$\frac{\frac{2x(2x +1)}{(x+2)(x-1) }}{\frac{4x(2x+1)}{(3x+4)(x+2)}} = {\frac{2x(2x +1)}{(x+2)(x-1) }} \times{\frac{(3x+4)(x+2)}{4x(2x+1)}$

or, implied fraction is  ${\frac{(3x+4)}{2(x-1)}$

Hence, D is the correct representation.