Is it possible to draw a quadrilateral with 4 congruent sides that is not a parallelogram?
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Answer:
see below
Step-by-step explanation:
y = 8x − 9
y = 4x − 1
Since both equations are equal to y, we can set them equal to each other
8x-9 = 4x-1
Subtract 4x from each side
8x-4x-9 = 4x-4x-1
4x-9 = -1
Add 9 to each side
4x-9+9 = -1+9
4x = 8
Divide each side by 4
4x/4 = 8/4
x=2
Now find y
y = 4x-1
y = 4(2)-1
y = 8-1
y = 7
(2,7)
The point of intersection when the 2 lines are graphed is (2,7)
