Answer:
Step-by-step explanation:
Let consecutive odd integers be (2x + 1) and (2x + 3)
Their reciprocals are [tex]\frac{1}{2x + 1 }[/tex] and [tex]\frac{1}{2x + 3 }[/tex]
[tex]\frac{1}{2x+1}+\frac{1}{2x+3}=\frac{44}{483}\\\\\frac{1*(2x+3)}{(2x+1)(2x+3)}+\frac{1*(2x+1)}{(2x+3)(2x+1)}=\frac{44}{483}\\\\\frac{2x + 3 + 2x +1}{(2x + 1)(2x+3)}=\frac{44}{483}\\\\\frac{4x + 4}{4x^{2}+8x+3}=\frac{44}{483}\\[/tex]
483*(4x+4) = 44*(4x² + 8x + 3)
483*4x + 483*4 = 44*4x² + 44*8x + 44*3
1932x + 1932 = 176x² + 352x + 132
176x² + 352x + 132 -1932x - 1932 = 0
Combine like terms
176x² + 352x - 1932x - 1932 + 132 = 0
176x² - 1580x - 1800 = 0
Divide the equation by 4
44x² - 395x - 450 = 0
