X-y=3=>x=3+y
X+2y=-6
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3+y+2y=-6
3+3y=-6
3y=-6-3
3y=-9
y=-3
X+2y=-6
X+2(-3)=-6
X-6=-6
X=0
The answer to this question is 1686.
Answer:
b.
Step-by-step explanation:
b and c have the same amount of sqaures but b is facing the same way.
The answer is A.
If a redundant conclusion is reached in basic algebra this states that the variable holds all possible real values.
If you algebraically solve Kendra's you do achieve the true statement 5 = 5 (leaving out D). And if you test any value of x for the equation it does hold true (getting rid of B).
Hopefully this makes sense.
