I need help, -3x+9+4x=10 just wondering if the answer is X=1

Step-by-step explanation:

 Fortunately, you don’t have to prove that x or y must be 1 or -1 in order for x*y to not be an element of G.

You only need to prove that x*y is an element of G for any x,y in G.  (That’s what it means for * to be an operation on G.)

I haven’t thought yet about how to prove that, but let me give you some thoughts off the top of my head.

We know -1 < x, y < 1.  We must show -1 < (x + y) / (xy + 1) and (x + y) / (xy + 1) < 1.

First, let’s show that -1 < (x + y) / (xy + 1).

I want to multiply xy + 1 to both sides, but we need to know whether it’s positive or not, so we know whether to reverse the inequality.

Well, -1 < x, y < 1 implies that xy > -1, which implies that xy +1 > 0.

So we must show that -(xy + 1) < x + y.

That is, let’s show that 0 < x + y + xy + 1.  (I don’t know if this is going anywhere useful; I’m just playing around with algebra.)

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.