If x=6 and the equation is z+3y=4x+y then what is y

y = 9

Step-by-step explanation:

First, simplify the equation by collecting like terms (with the same variable).

I will move all the "y" to the left and the "x" to the right.

x + 3y = 4x + y

x - x + 3y = 4x - x + y               Subtract x on both sides

3y = 3x + y                         "x" is cancelled out on the left

3y - y = 3x + y - y        Subtract y on both sides to cancel out on the right.

2y = 3x                      Simplified equation

Now, substitute any "x" with 6.

2y = 3(6)                  Simplify the multiplication

2y = 18

2y/2 = 18/2                Divide both sides by 2 to isolate "y"

y = 9              Solved for "y"

Check your

Substitute "x" with 6 and "y" with 9.

x + 3y = 4x + y

6 + 3(9) = 4(6) + 9                  Simplify multiplication first

6 + 27 = 24 + 9                 Simplify each side by adding

33 = 33               Same answers

LS = RS               Left side equal right side

Therefore the answer is correct.

∠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.