Solve by substitution of simultaneous equations?

Question

x = 2y

y = x^2 - 4x + 4

Please help........

Answer 1

since x=2y, substitute it.

y = x^2 - 4x + 4

y=(2y)^2-4(2y)+4

y=2y^2-8y+4

2y^2-8y-y+4=0

2y^2-9y+4

(2y-1)(y-4)

y=1/2 and 4

to get x

substute the values of y

when y is 1/2, x=1

when y is 4, x=8

hope this helps

Answer 2

y=x/2

so plug x/2 in y.

x/2=x^2-4x+4

x^2-7x/2+4=0

Then use quadratic formula to solve for x.

x=7/4 plus or minus the square root of 17. that is the value of x.

after you find the value x. plug to x. to find y

Answer 3

x = 2y

y = x^2 - 4x + 4

y = (2y)^2 - 4 ( 2y) +4

x = 2y

y = 4y^2 -8y+4

4y^2 - 8y+4 -y = 0

4y^2 -9y+4 = 0.

delta = (-9)^ 2 - 4*4*4 = 81 - 64= 17.

y1 = [9- sqrt(17) ] / 8

y2 = [9+sqrt(17) ] /8.

x1 = [9- sqrt(17) ] / 4

x2 = [9+ sqrt(17) ] / 4.

Answer 4

y=x/2 ... (1)

x/2=x^2-4x+4

x=2x^2-8x+8

2x^2-9x+8=0

Use the quadratic formula to solve for 2 values of x then substitute back into (1) to get the y.

Answer 5

y = 4y² - 8y + 4

4y² - 9y + 4 = 0

y = [ 9 ± √ (81 - 64 ) ] / 8

y = [ 9 ± √ (17) ] / 8

x = 2 [ 9 ± √ (17) ] / 8

x =  [ 9 ± √ (17) ] / 4