Question

Find the points of intersection of the equation: xy=2 and x+y =4

Answer 1

Answer:

(2 - $\sqrt{2}$ , 2 + $\sqrt{2}$ ) and (2 + $\sqrt{2}$, 2 - $\sqrt{2}$ )

Step-by-step explanation:

xy = 2 → (1)

x + y = 4 ( subtract x from both sides )

y = 4 - x → (2)

substitute y = 4 - x into (1)

x(4 - x) = 2

4x - x² = 2 ( multiply through by - 1 )

x² - 4x = - 2

using the method of completing the square

add ( half the coefficient of the x- term)² to both sides

x² + 2(- 2)x + 4 = - 2 + 4

(x - 2)² = 2 ( take square root of both sides )

x - 2 = ± $\sqrt{2}$ ( add 2 to both sides )

x = 2 ± $\sqrt{2}$ , that is

x = 2 - $\sqrt{2}$ , x = 2 + $\sqrt{2}$

substitute these values of x into (2) for corresponding values of y

x = 2 - $\sqrt{2}$ , then

y = 4 - (2 - $\sqrt{2}$)

  = 4 - 2 + $\sqrt{2}$

  = 2 + $\sqrt{2}$ ⇒ (2 - $\sqrt{2}$ , 2 + $\sqrt{2}$ ) ← 1 point of intersection

x = 2 + $\sqrt{2}$ , then

y = 4 - (2 + $\sqrt{2}$ )

   = 4 - 2 - $\sqrt{2}$

   = 2 - $\sqrt{2}$ ⇒ (2 + $\sqrt{2}$ , 2 - $\sqrt{2}$ ) ← 2nd point of intersection

Answer 2

Answer:$\Large\boxed{(2+\sqrt{2},~2-\sqrt{x} )~~and~~ (2-\sqrt{2},~2+\sqrt{x} )}$

Step-by-step explanation:

Given the system of equations

$1)~xy=2$

$2)~x+y=4$

Divide x on both sides of the 1) equation

$xy=2$

$xy\div x=2\div x$

$y=\dfrac{2}{x}$

Current system

$1)~y=\dfrac{2}{x}$

$2)~x+y=4$

Substitute the 1) equation into the 2) equation

$x+(\dfrac{2}{x} )=4$

Multiply x on both sides

$x\times x+\dfrac{2}{x}\times x=4\times x$

$x^2+2=4x$

Subtract 4x on both sides

$x^2+2-4x=4x-4x$

$x^2-4x+2=0$

Use the quadratic formula to solve for the x value

$x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)} }{2(1)}$

$x=2\pm\sqrt{2}$

Substitute the x value into one of the equations to find the y value

$xy=2$

$(2+\sqrt{2} )y=2$

$y=2-\sqrt{2}$

$OR$

$xy=2$

$(2-\sqrt{2} )y=2$

$y=2+\sqrt{2}$

Therefore, the points of intersection are

$\Large\boxed{(2+\sqrt{2},~2-\sqrt{x} )~~and~~ (2-\sqrt{2},~2+\sqrt{x} )}$

Hope this helps!! :)

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