Respuesta :
A) Only the first is true: we can solve the equation for y:
[tex]y = 7-2x[/tex]
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
[tex]x^2-2xy+y^2=(x-y)^2[/tex]
So, whatever value we fix for x, we can choose y=x and the equation becomes
[tex](x-x)^2=0[/tex]
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: [tex]x^2[/tex] and [tex]y^2[/tex] are positive numbers, because their are squares, and their sum can't be -1.