For each of the following equations, determine which of the following statements are true:
(1) for all real numbers x, there exists a real number y such that the equation is true.
(2) there exists a realnumber x, such that for allreal numbers y, the equation is true.
note that it is possible for both statements to be true or for both to be false.
a. 2x +y =7
b. y+x = x +y
c. x2 −2xy+y2 =0
d. (x −5)(y−1) =0
e. x2 +y2 =− 1

A) Only the first is true: we can solve the equation for y:

So, if you fix any value for x, you can choose y=7-2x and the equation will be true. The other claim is not true: the value for x that solves the equation depends on the value of y, so we can't hope to find a "universal" x-value such that the equation is true for all y-values.

B) Both are trivially true, because the equation is an identity: addition is commutative.

C) Only the first is true: we have

So, whatever value we fix for x, we can choose y=x and the equation becomes

which is clearly true. The other claim is not true: the value for x that solves the equation depends on the value of y, so we can't hope to find a "universal" x-value such that the equation is true for all y-values.

D) Both claims are false: claim number 1 is false because the only solution to the equation is x=5, so you can't choose any value for x and expect to find a solution by choosing an appropriate value for y. Claim number 2 is false because, even if we choose x=5, the equation is true for all the possible values of y, except y=1, which is an excluded value for this expression

E) Both claims are false: and are positive numbers, because their are squares, and their sum can't be -1.

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3

step-by-step explanation:

what was your answer?