Вопрос в картинках...

Решите задачу:

$\mathtt{ \dfrac{4}{x+3}+ \dfrac{5}{2x+3}=2}\\\mathtt{ \dfrac{4(2x+3)+5(x+3)-2(x+3)(2x+3)}{(x+3)(2x+3)}=0 }\\\mathtt{ \dfrac{8x+12+5x+15-2(2x^2+3x+6x+9)}{(x+3)(2x+3)} =0}\\\mathtt{ \dfrac{8x+12+5x+15-4x^2-18x-18}{(x+3)(2x+3)}=0 }\\\mathtt{ \dfrac{-4x^2-5x+9}{(x+3)(2x+3)}=0}\\\boxed{\mathtt{ \left \{ {{-4x^2-5x+9=0} \atop {(x+3)(2x+3)\neq0}} \right. }}\\\mathtt{1)(x+3)(2x+3)=0}\\\mathtt{x+3\neq0\ our\ 2x+3\neq0}\\\mathtt{x\neq-3\ \ \ \ \ \ \ \ \ \ x\neq-1 \dfrac{1}{2}}\\\mathtt{2)-4x^2-5x+9=0}$
$\mathtt{4x^2+5x-9=0}\\\mathtt{D=5^2-4*4*(-9)=25+144=169=(\pm13)^2}\\\mathtt{x_1= \dfrac{-5+13}{2*4};\ x_1=1}\\\mathtt{x_2= \dfrac{-5-13}{2*4}=-2 \dfrac{1}{4}}\\\mathbf{OTVET}\mathtt{x_1=1;\ x_2=-2 \dfrac{1}{4}. }$