`4a)` $\begin{cases} x-3|y|=-2 (1)\\2(x+1)-5(y-1)=6 (2) \end{cases}$

Có `(2) <=> 2x+2-5y+5=6`

`<=> x=(5y-1)/(2)` thay vào `(1)` được:

`(5y-1)/(2)-3|y|=-2`

`<=> 5y-6|y|=-3`

`_` `TH1: y=>0 => |y|=y`

`=> 5y-6y=3`

`<=> y=3 (tm)`

`=> x=7`

`_` `TH2: y<0 => |y|=-y`

`=> 5y-6.(-y)=3`

`<=> y=-(3)/(11)`

`=> x=-(13)/(11)`

Vậy `(x; y) in {(-(13)/(11); -(3)/(11)), (7; 3)}`

`5b)` $\begin{cases} 4x=|y+1|-4 (1)\\(x+1)(2-y)=5-xy (2) \end{cases}$

`(2) <=> 2x-xy+2-y=5-xy`

`<=> y=2x-3` thay vào `(1)` được: 

`4x=|2x-3+1|-3`

`<=> |2x-2|=4x+3 (x>=-3/4)`

`_` `TH1: 2x-2=4x+3`

`<=> x=-5/2 (ktm)`

`_` `TH2: 2-2x=4x+3`

`<=> x=-1/6 (tm)`

`=> y=-(10)/(3)`

Vậy `(x; y)=(-1/6; -(10)/(3))`

`6b)` $\begin{cases} 2|x+2|-|y-3|=0 (1)\\(x+3)-2(y+1)=-6 (2)\end{cases}$

`(2) <=> x+3-2y-2=-6`

`<=> x=2y-7` thay vào `(1)` được:

`2|2y-7+2|-|y-3|=0`

`<=> |4y-10|=|y-3|`

`_` `TH1: 4y-10=y-3`

`<=> y=7/3`

`=> x=-7/3`

`_` `TH2: 4y-10=3-y`

`<=> y=(13)/(5)`

`=> x=-9/5`

Vậy `(x; y) in {(-7/3; 7/3), (-9/5; (13)/(5))}`