می‌دانیم جواب معادله در خود معادله صدق می‌کند، لذا با جایگذاری $x = 5$ در معادله مقدار $a$ را می‌یابیم:

$\frac{4}{{x - a}} + \frac{{2x - 2}}{{x + a}} = 3 \to x = 5 \to \frac{4}{{5 - a}} + \frac{{2 \times 5 - 2}}{{5 + a}} = 3$

$\frac{4}{{5 - a}} + \frac{8}{{5 + a}} = 3 \Rightarrow \frac{{4(5 + a)}}{{(5 - a)(5 + a)}} + \frac{{8(5 - a)}}{{(5 - a)(5 + a)}} = 3$

$ \Rightarrow \frac{{20 + 4a + 40 - 8a}}{{25 - {a^2}}} = 3 \Rightarrow \frac{{60 - 4a}}{{25 - {a^2}}} = 3$

$ \Rightarrow 60 - 4a = 75 - 3{a^2} \Rightarrow 3{a^2} - 4a - 15 = 0 \Rightarrow (3a + 5)(a - 3) = 0$

$\eqalign{
  &  \Rightarrow 3a + 5 = 0 \Rightarrow a =  - \frac{5}{3}  \cr 
  & a - 3 = 0 \Rightarrow a = 3 \cr} $

حال با جایگذاری $a = 3$ ریشهٔ دیگر را می‌یابیم:

$\frac{4}{{x - 3}} + \frac{{2x - 2}}{{x + 3}} = 3 \Rightarrow \frac{{2x - 2}}{{x + 3}} = 3 - \frac{4}{{x - 3}} \Rightarrow \frac{{2x - 2}}{{x + 3}} = \frac{{3x - 9}}{{x - 3}} - \frac{4}{{x - 3}}$

$\frac{{2x - 2}}{{x + 3}} = \frac{{3x - 13}}{{x - 3}} \Rightarrow (2x - 2)(x - 3) = (x + 3)(3x - 13)$

$ \Rightarrow 2{x^2} - 6x - 2x + 6 = 3{x^2} - 13x + 9x - 39 \Rightarrow {x^2} + 4x - 45 = 0$

$(x + 9)(x - 5) = 0 \Rightarrow $

$\eqalign{
  & x + 9 = 0 \Rightarrow x =  - 9  \cr 
  & x - 5 = 0 \Rightarrow x = 5 \cr} $

بار دیگر با جایگذاری $a =  - \frac{5}{3}$ معادله را حل می‌کنیم:

$\frac{4}{{x + \frac{5}{3}}} + \frac{{2x - 2}}{{x - \frac{5}{3}}} = 3$

$ \Rightarrow \frac{4}{{3x + 5}} + \frac{{2x - 2}}{{3x - 5}} = 1 \Rightarrow \frac{{2x - 2}}{{3x - 5}} = 1 - \frac{4}{{3x + 5}} \Rightarrow \frac{{2x - 2}}{{3x - 5}} = \frac{{3x + 1}}{{3x + 5}}$

$ \Rightarrow (2x - 2)(3x + 5) = (3x - 5)(3x + 1)$

$ \Rightarrow 6{x^2} + 10x - 6x - 10 = 9{x^2} - 12x - 5$

$ \Rightarrow 3{x^2} - 16x + 5 = 0 \to a{x^2} + bx + c = 0 \to $

$\eqalign{
  & a = 3  \cr 
  & b =  - 16  \cr 
  & c = 5 \cr} $

$\Delta  = {b^2} - 4ac \Rightarrow \Delta  = {( - 16)^2} - 4 \times (3) \times (5) = 256 - 60 = 196$

${x_1} = \frac{{ - b + \sqrt \Delta  }}{{2a}} \Rightarrow {x_1} = \frac{{ - ( - 16) + \sqrt {196} }}{{2 \times 3}} = \frac{{16 + 14}}{6} = \frac{{30}}{6} = 5$

${x_2} = \frac{{ - b - \sqrt \Delta  }}{{2a}} \Rightarrow {x_2} = \frac{{ - ( - 16) - \sqrt {196} }}{{2 \times 3}} = \frac{{16 - 14}}{6} = \frac{2}{6} = \frac{1}{3}$