How do you solve the system of equations #4x ^ { 2} - 3y ^ { 2} = - 11# and #5x ^ { 2} + 2y ^ { 2} = 38# by the addition method?

1 Answer
Apr 29, 2018

The solutions are #{(x=+-2),(y=+-3):}#

Explanation:

The equations are

#{(4x^2-3y^2=-11),(5x^2+2y^2=38):}#

#<=>#, #L2larr(4L2-5L1)#, #{(4x^2-3y^2=-11),(0+23y^2=207):}#

#<=>#, #L2larr((L2)/23)#, #{(4x^2-3y^2=-11),(y^2=9):}#

#<=>#, #L1larr(L1+3L2)#, #{(4x^2-27=-11),(y^2=9):}#

#<=>#, #{(4x^2=27-11=16),(y^2=9):}#

#<=>#, #{(x^2=4),(y^2=9):}#

#<=>#, #{(x=+-2),(y=+-3):}#