Question

ax+by+c=0:
bx+ay+c=0
solve using substitution method

Answer 1

ax+by+c=0
ax=-by-c
x=(-by-c)/a

bx+ay+c=0
b[(-by-c)/a]+ay+c=0
(-b^2y-bc)/a+ay+c=0
(-b^2y-bc+a^2y+ac)/a=0
a^2y-b^2y-bc+ac=0
y(a^2-b^2)+c(a-b)=0
y(a+b)(a-b)+c(a-b)=0
(a-b)[y(a+b)+c]=0
y(a+b)+c=0
y=-c/a+b

x=(-by-c)/a
x=[-b(-c/a+b)-c]/a
x=[(bc/a+b)-c]/a
x=[(bc-c(a+b))/a+b]/a
x=[c(b-a+b)]/a(a+b)
x=[c(2b-a)]/a(a+b)