Question

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

Answer 1

(2 -
`√(2)` , 2 +
`√(2)` ) and (2 +
`√(2)`, 2 -
`√(2)` )

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 = ±
`√(2)` ( add 2 to both sides )

x = 2 ±
`√(2)` , that is

x = 2 -
`√(2)` , x = 2 +
`√(2)`

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

x = 2 -
`√(2)` , then

y = 4 - (2 -
`√(2)`)

= 4 - 2 +
`√(2)`

= 2 +
`√(2)` ⇒ (2 -
`√(2)` , 2 +
`√(2)` ) ← 1 point of intersection

x = 2 +
`√(2)` , then

y = 4 - (2 +
`√(2)` )

= 4 - 2 -
`√(2)`

= 2 -
`√(2)` ⇒ (2 +
`√(2)` , 2 -
`√(2)` ) ← 2nd point of intersection

Answer 2

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

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/ x=2/ x`

`y=(2)/(x)`

Current system

`1)~y=(2)/(x)`

`2)~x+y=4`

Substitute the 1) equation into the 2) equation

`x+((2)/(x) )=4`

Multiply x on both sides

`x* x+(2)/(x)* x=4* 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=(-(-4)\pm√((-4)^2-4(1)(2)) )/(2(1))`

`x=2\pm√(2)`

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

`xy=2`

`(2+√(2) )y=2`

`y=2-√(2)`

`OR`

`xy=2`

`(2-√(2) )y=2`

`y=2+√(2)`

Therefore, the points of intersection are

`\Large\boxed{(2+√(2),~2-√(x) )~~and~~ (2-√(2),~2+√(x) )}`

Hope this helps!! :)

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