How do you solve #5y – x – 6 = 0# and #-3x + 6y = -9#?

1 Answer
May 19, 2016

Answer:

#x=9, y=3#

Explanation:

First Method, using substitution

By rearranging the first equation we have,

#x = 5y -6#

substitute value of #x# in second equation

#-3*(5y -6) + 6y = -9#

#-15y +18 + 6y = -9#

#-9y +18 = -9#

#rArry-2 = 1#
#or y = 3#

by substituting value of #y# in either of the equations
we have

#x = 9#

Second Method,
Multiply first equation with 3,

#3*(5y-x-6)=0#

#15y-3x-18=0#

Now subtract LHS of the second equation from the LHS above equation and do the same with the RHS,

#(15y-3x-18)-(-3x+6y) = 0-(-9)#

#15y -3x -18 +3x-6y=9#

#9y -18 =9#
#rArr y =3#

and by substituting #y# in either of the equations we have

#x=9#