For solving using substitution method you have to separate out the value of any one variable (x or y)(xory) in any of the equation and put it in the other equation.
Assuming 6x + 5y = -16x+5y=−1 to be equation 1 1 and -5x - 9y = 5−5x−9y=5 be equation 22
Seperating out the value of xx
x = (-1 - 5y)/6x=−1−5y6 and now putting it in the second equation
-5((-1-5y)/6) - 9y = 5−5(−1−5y6)−9y=5
5((1+5y)/6) - 9y = 55(1+5y6)−9y=5
((5 + 25y)/6) - 9y = 5(5+25y6)−9y=5
Now on taking LCMLCM on LHS
((5 + 25y - 54y)/6) = 5(5+25y−54y6)=5
(5 + 25y - 54y) = 30(5+25y−54y)=30
5 - 29y = 305−29y=30
29y = -2529y=−25
y = -25/29y=−2529
Now put the value of yy in any of the two equations
(6x + 5(-25/29)) = -1(6x+5(−2529))=−1
(6x -125/29) = -1(6x−12529)=−1
6x = 125/29 - 16x=12529−1
6x = (125-29)/296x=125−2929
6x = 96/296x=9629
x = 16/29x=1629
