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

2 Answers
Mar 25, 2018

Answer:

#(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)#

Mar 25, 2018

Answer:

#(-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]}