Let z be any number (like x would be)
if we set z equal to 2x-6y, then we get z = 2x-6y
The system
2x-6y = 8
2x-6 = 3
turns into this new system
z = 8
z = 3
but z is a single number. It can't be both 8 and 3 at the same time. So there are no solutions
Answer: 44
EC is basically EF times 2
2x + 6y = 20
3x - 2y = 8
I'll just substitute x and y with its corresponding values per option.
A) x = 4 ; y = 2
2(4) + 6(2) = 20 3(4) - 2(2) = 8
8 + 12 = 20 12 - 4 = 8
20 = 20 8 = 8
B) x = 4 ; y = -2
2(4) + 6(-2) = 20
8 - 12 = 20
-4 ≠ 20
C) x = -2 ; y = 4
2(-2) + 6(4) = 20 3(-2) - 2(4) = 8
-4 + 24 = 20 -6 - 8 = 8
20 = 20 - 14 ≠ 8
D) x = 2 ; y = 4
2(2) + 6(4) = 20
4 + 24 = 20
28 ≠ 20
Choice A. x = 4 ; y = 2 is the correct answer.
Based on the picture above,
The theorem or postulate that justifies that Angle HEF ~ angle HGE is :
A. AA similarity postulate
(s is used for equal side, that is used for congruent)
hope this helps
So my take on this
15 + 3 -2(2 - 1)
15 + 3 -2(1)
15 +3 -2
18 - 2 = 16
