Do the graphs of 4y=(x-3)^2 and 3x-y=14 intersect at (5,1)

it will happen if both y and x components be equal

so

y= (x-3)²/4 , y = 3x- 14

3x - 14 = (x-3)²/4

12x - 56 = x²-6x + 9

x² - 18x + 65 = 0

delta = 18²-4*65 = 64

x1 = (18 + (64)^(1/2))/2 = 13

x2 = (18 - (64)^(1/2))/2 = 5

one point is x = 5 but we should check if on that point y component is 1 or not

y = (5-3)²/4 = 1

so the functions will intersect at (5,1)

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