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

2 Answers
mizoo
Mar 25, 2018

#(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)

Jim G.
Mar 25, 2018

(-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]}

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