Use your definitions of logs to change the equations:
first one:
log(x+y) = 0 or
10^0 = x+y
x+y = 1 ------> y = 1-x
second
2logx = log(y-1)
log x^2 = log(y-1)
x^2 = y-1 -----> y = x^2+1
so x^2+1 = 1-x
x^2 + x = 0
x(x+1) = 0
x=0 or x=-1
then y = 1 or y = 0
now logx is only defined for positve values of x, so after all this we conclude that there is no solution, since neither solution pair works in the second equation.
Solve the pair of simultaneous equations
log(x+y)=0,
2logx=log(y-1)
4 answers
Log(x+y) = 10^0 x+y =1 2logx= Log(y+1) Logx^2 =log (y+1) x^2 =y+1
When X=2,y=-2
X=1,y=-1
X=1,y=-1
Not satisfy with the working out