Question

What are possible value(s) of x and y if `y^2=x^2-64` and `3y=x+8?`?

Answer 1

#(x, y) = (-8, 0), (10, 6)

Explanation:

`3y = x + 8 => x = 3y - 8`

`y^2 = x^2 - 64`
`y^2 = (3y - 8)^2 - 64`
`y^2 = 9y^2 - 48y + 64 - 64`
`8y^2 - 48y = 0`
`8y(y - 6) = 0`
`y = 0, 6`

`x = 3y - 8 and y = 0`:
`x = 0 - 8`
`= -8`

`x = 3y - 8 and y = 6`:
`x = 3 xx 6 - 8`
`x =10`

`(x, y) = (-8, 0), (10, 6)`

Answer 2

`(-8,0),(10,6)`

Explanation:

`y^2=x^2-48to(1)`

`3y=x+8to(2)`

`"from equation "(2)" we can express x in terms of y"`

`rArrx=3y-8to(3)`

`"substitute "x=3y-8" in equation "(1)`

`rArry^2=(3y-8)^2-64larrcolor(blue)"expand "(3y-8)^2`

`rArry^2=9y^2-48ycancel(+64)cancel(-64)`

`rArr8y^2-48y=0larrcolor(blue)"factorise"`

`8y(y-6)=0`

`"equate each factor to zero and solve for y"`

`8y=0rArry=0`

`y-6=0rArry=6`

`"substitute these values into equation "(3)`

`y=0rArrx=-8rArr(-8,0)`

`y=6rArrx=18-8=10rArr(10,6)`
graph{(y^2-x^2+64)(y-1/3x-8/3)((x+8)^2+(y-0)^2-0.04)((x-10)^2+(y-6)^2-0.04)=0 [-20, 20, -10, 10]}