SOLUTION: determine whether the given ordered pair is a solution of the system. #1 (6,-1); x-y=3 2x+5y=6 #2 (2,-3); 2x+y=1 x-y=5 #3 (-2,1); x+3y=1

x=3+y
2x+5y=6
2(3+y)+5y=6
6+2y+5y=6
7y=0
y=0
x=3+y=3
solution pair=(3,0)

Substituting given pair (6,-1) in eqn1 we get
x-y=3
6-(-1)=3
7=3 which is false
hence given values are wrong,correct answer=(3,0)

#2 (2,-3)
2x+y=1
x-y=5

Substituting in eqn1 we get
2(2)+(-3)=1
4-3=1
which is correct
in eqn2,x-y=5
2-(-3)=5
5=5
which is also correct
Hence given pair (2,-3) is a valid solution

#3
(-2,1)
x+3y=1
2x-y=-5
eqn1: (-2)+3(1)=-2+3=1=RHS
eqn2: 2(-2)-(1)=-4-1=-5=RHS
hence given pair (-2,1)is a valid solution

#4
(-4,-1)
x-y=3
x+y=-5

eqn1: -4-(-1)=-4+1=-3 not equal to RHS
eqn2: -4+(-1)=-4-1=-5=RHS
as the given pair only satisfies one eqn,
it is not a valid solution

x-y=3
x+y=-5
2x=3-5
2x=-2
x=-1
y=-5-x=-5-(-1)=-5+1=-4
Solution=(-1,-4)