Find the point of intersection

Question

of y=2x-4 and y=2cos2π/3x-4

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Answer 1

Both expressions are equal to y, so equate the two to each other and solve for x.

2x-4 = 2cos2π/3x-4

cos2π = 1

2x-4 = 2/(3x-4)  multiply both sides by (3x-4)

(2x-4)(3x-4) =  2

6x²-20x+16 =2

6x²-20x+14 = 0

Simplify

3x²-10x+7 = 0

Use AC factoring to solve for x.  AC = 21, factors of -21 are -7 and -3. Split the middle term (-10x=-7x-3x).

3x²-7x-3x+7 = 0  Group the terms like this:

(3x²-7x)+(-3x+7) = 0  (This helps avoid making a sign error!)

x(3x-7)-1(3x-7) = 0

(x-1)(3x-7) = 0

x = 1 or 7/3

 

There are two intersections.

For x=1

y = 2(1)-4 = -2

(x,y) = (1,-2)

 

For x=7/3

y=2(7/3)-4 = 14/3 -  12/3 = 2/3

(x,y) = (7/3,2/3)