How do you solve the system of equations #x^2+6y^2=40# and #y^2-3x^2=-44# by the elimination method?

1 Answer
Jim G.
May 27, 2018

#(-4,+-2),(4,+-2)#

Explanation:

#x^2+6y^2=40to(1)#

#y^2-3x^2=-44to(2)#

#"rearranging equation "(1)" gives"#

#x^2=40-6y^2to(3)#

#color(blue)"substitute "x^2=40-6y^2" into "(2)#

#y^2-3(40-6y^2)=-44#

#y^2-120+18y^2=-44#

#19y^2-120=-44#

#"add 120 to both sides"#

#19y^2=-44+120=76#

#"divide both sides by 19"#

#y^2=76/19=4#

#color(blue)"take the square root of both sides"#

#y=+-sqrt4=+-2#

#"substitute "y=+-2" into equation "(3)#

#x^2=40-24=16#

#x=+-4#

#"solutions are "(-4,+-2),(4,+-2)#
graph{(x^2+y^2-40)(y^2-3x^2+44)=0 [-20, 20, -10, 10]}

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