$\left \{ {{\dfrac{a}{2} - \dfrac{b}{2} = 1} \atop {\dfrac{4}{3}a + \dfrac{4}{3}.b=1}} \right.$
$⇒$ $\left \{ {{\dfrac{a}{2} - \dfrac{b}{2} = \dfrac{2}{2}} \atop {\dfrac{4a}{3} + \dfrac{4b}{3}=\dfrac{3}{3}}} \right.$
$⇔$ $\left \{ {{a-b=2} \atop {4a+4b=3}} \right.$
$⇔$ $\left \{ {{4a-4b=8} \atop {4a+4b=3}} \right.$
$⇔$ $\left \{ {{4a-4b=8} \atop {4a-4b-4a-4b=8-3}} \right.$
$⇔$ $\left \{ {{4a-4b=8} \atop {-8b=5}} \right.$
$⇔$ $\left \{ {{4a-4b=8} \atop {b=-\dfrac{5}{8}}} \right.$
$⇔$ $\left \{ {{4a-4b=8} \atop {b=-\dfrac{5}{8}}} \right.$
$⇔$ $\left \{ {{a=\dfrac{11}{8}} \atop {b=-\dfrac{5}{8}}} \right.$
Vậy `(a;b)=(\frac{11}{8};-\frac{5}{8})`.
