An algebra problem by Abhijith Asokan

Yes, since the roots are $a$ , $b$ , and $c$ , this means that:

$(x-a)(x-b)(x-c) = x^3 - \frac{3}{2}x^2 + \frac{5}{2}x - \frac{7}{2}$

$\Rightarrow x^3 -(a+b+c)x^2 + (ab+bc+ca)x - abc = x^3 - \frac{3}{2}x^2 + \frac{5}{2} x- \frac{7}{2}$

$\Rightarrow ab+bc+ca = \frac{5}{2} \quad abc = \frac{7}{2}$

Since $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} = \dfrac{ab + bc +ca}{abc} = \dfrac {\frac{5}{3}}{\frac{7}{2}} = \dfrac{5}{7}= \dfrac{x}{y}$

Therefore, $x +y = 5 +7 = \boxed{12}$

$\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } =\frac { bc+ac+ab }{ abc } =\frac { C/A }{ -D/A } =\frac { 5/2 }{ -(-7)/2 }$ n=5,m=7 n+m=12

For any equation, (quad, tri., biquad,..) for an eq. with roots a1,b1,c1,d1......, the eq with roots 1/a1, 1/a2, 1/a3.... can be obtained by just reversing the coefficients..

In this case coeff are 2, -3 ,+5, -7. .. reversing coeffs, we have the new eq: -7(x^3) +5(x^2) -3(x)=2=0 this equations sum= 1/a +1/b + 1/c = -(-5/7)=5/7= x+y= 5+7==12..