二元一次方程式混函數及比例式問題

有兩題數學問題要發問求解析
1.設2x-y和3y+1成正比
當x=-1，y=3則所形成的函數為
y=ax+b，則求a+2b?
答案-6求詳細解析
(2x-y)=k(3y+1)
(-2-3)=k(9+1)
-5=10k
k=-1/2

(2x-y)=-1/2(3y+1)
4x-2y=-3y-1
y=-4x-1
a=-4, b=-1

a+2b
=-4+2*(-1)
=-4-2
=-6


2.有壹個二元一次聯立方程式
{199(x+y)+200(x-y)=1
{199^2(x+y)+200^2(x-y)=399
答案為x=0，y=-1

設x+y=A, x-y=B

199A+200B=1(1)
199^2A+200^2B=399(2)
(1)*200-(2)
199A=-199
A=-1
-199+200B=1
200B=200
B=1

x+y=-1
x-y=1
2x=0
x=0
y=-1

不懂請問!

1.2x-y和3y+1成正比==>設2x-y=k(3y+1)==>2x=(3k+1)y+k...(1)
又當x=-1，y=3則所形成的函數為y=ax+b，則(-1,3)代入(1)
==>2(-1)=(3k+1)*3+k==>-2=9k+3+k==>10k=-5==>k=-0.5代入(1)
==>2x=(-1.5+1)y+(-0.5)==>2x+0.5y=-0.5==>4x+y=-1==>y=-4x-1=ax+b
==>a=-4,b=-1===>a+2b=(-4)+2(-1)=-4-2=-6...ans
2.199(x+y)+200(x-y)=1...(1)
1992(x+y)+2002(x-y)=399...(2)
(1)+(2)==>[199(x+y)+1992(x+y)]+[200(x-y)+2002(x-y)]=400
==>199(x+y)(1+199)+200(x-y)(1+200)=400
==>199*(x+y)*200+200(x-y)*201=400
===>199*(x+y)+201(x-y)=2.........(等號左右同除以200)
===>400x-2y=2....(乘開合併)==>200x-y=1==>y=200x-1...(3)
199(x+y)+200(x-y)=1...(1)
1992(x+y)+2002(x-y)=399...(2)
(2)-(1)===>1992(x+y)-199(x+y)+2002(x-y)-200(x-y)=399-1
===>199(x+y)(199-1)+200(x-y)(200-1)=398
===>199(x+y)*198+200(x-y)*199=398=2*199
===>198(x+y)+200(x-y)=2....(等號兩邊同除以199)
===>398x-2y=2...(合併整理)
===>y=199x-1...(4)
(3)=(4)===>y=200x-1=199x-1==>x=0,y=-1...ans