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\(\cfrac{12x+17}{3x+1}-\cfrac{2x+15}{x+7}=\cfrac{31}{5} \)
বা, \(\cfrac{(12x+17)(x+7)-(2x+15)(3x+1)}{(3x+1)(x+7)} =\cfrac{16}{5}\)
বা, \(\cfrac{(12x^2+84x+17x+119)-(6x^2+2x+45x+15)}{3x^2+21x+x+7}=\cfrac{16}{5}\)
বা, \(\cfrac{12x^2+101x+119-6x^2-47x-15}{3x^2+22x+7}=\cfrac{16}{5} \)
বা, \(\cfrac{6x^2+54x+104}{3x^2+22x+7}=\cfrac{16}{5} \)
বা, \(16(3x^2+22x+7)=5(6x^2+54x+104) \)
বা, \(48x^2+352x+112=30x^2+270x+520 \)
বা, \(48x^2+352x+112-30x^2-270x-520=0 \)
বা, \(18x^2+82x-408=0\)
বা, \(9x^2+41x-204=0 \)
বা, \(9x^2+(68-27)x-204=0 \)
বা, \(9x^2+68x-27x-204=0\)
বা, \(x(9x+68)-3(9x+68)=0 \)
বা, \((9x+68)(x-3)=0 \)

অর্থাৎ,হয় \((9x+68)=0∴x=-\cfrac{68}{9} \)
নয়, \((x-3)=0∴x=3 \)

\(∴x=-\cfrac{68}{9}\) ও \(x=3\) হল \(\cfrac{12x+17}{3x+1}-\cfrac{2x+15}{x+7}=\cfrac{31}{5}\) দ্বিঘাত সমীকরনের সমাধান ।