Step-by-step explanation:
Taking the provided equation ,
[tex]\implies \dfrac{3x+4}{5}-\dfrac{2}{x+3} =\dfrac{8}{5} [/tex]
1) Here denominator of two fractions are 5 and x +3 . So their LCM will be 5(x+3)
[tex]\implies \dfrac{(x+3)(3x+4)-(2)(5)}{5(x+3)}=\dfrac{8}{5} [/tex]
2) Transposing 5(x+3) to Right Hand Side . And 5 to Left Hand Side .
[tex]\implies 5(3x^2+4x+9x+12 -10) = 40(x+3)[/tex]
3) Multiplying the expressions.
[tex] \implies 5(3x^2+13x+2) = 40x + 120 [/tex]
4) Opening the brackets .
[tex]\implies 15x^2+ 65x + 10 = 40x + 120 [/tex]
5) Transposing all terms to Left Hand Side .
[tex]\implies 15x^2 + 65x - 40x + 10 - 120 = 0 \\\\\implies 15x^2 +25x - 110 = 0[/tex]
6) Solving the quadratic equation .
[tex]\implies 5(3x^2+5x -22) = 0 \\\\\implies 3x^2+13x-22 = 0 \\\\ \implies x = \dfrac{-b\pm \sqrt{b^2-4ac}}{4ac} \\\\\implies x = \dfrac{-5\pm \sqrt{5^2-4(-22)(3)}}{2(3)} \\\\\implies x = \dfrac{-5\pm \sqrt{289}}{6}\\\\\implies x =\dfrac{-5\pm 17}{6} \\\\\implies x = \dfrac{17-5}{6},\dfrac{-17-5}{6}\\\\\implies x = \dfrac{12}{6},\dfrac{-22}{6} \\\\\underline{\boxed{\red{\bf\implies x = 2 , \dfrac{-11}{3}}}}[/tex]
