a.x+y=3
⇔y=3-x
x+3y=5
⇔x+3(3-x)=5
⇔x+9-3x=5
⇔x-3x=5-9
⇔-2x=-4
⇔x=-4:(-2)
⇔x=2.
Ta có:x+y=3
⇔2+y=3
⇔y=1.
Vậy x=2;y=1.
b.$x^{2}$ -5x+4=0
⇔$x^{2}$ -x-4x+4=0
⇔($x^{2}$ -x)-(4x-4)=0
⇔x(x-1)-4(x-1)=0
⇔(x-1)(x-4)=0
⇔\(\left[ \begin{array}{l}x-1=0\\x-4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0+1\\x=0+4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=4\end{array} \right.\)
Vậy x∈(1;4).