Given:
#color(brown)(x+2y=5)#................................(1)
#color(brown)(2x-3y=-4)#........................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using short cut methods: (jumping steps)
#color(blue)("Making y the dependant variable:")#
#y=-x+5/2....................(1_a)#
#y=2/3x+4/3......................(2_a)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To determine the value of "x)#
By substitution we can equate equation #(1_a)# to equation #(2_a)# through #y# giving:
#-x+5/2=y=2/3x+4/3#
Collecting like terms
#2/3x+x=5/2-4/3#
#5/3x= (15-8)/6#
#color(blue)(x=7/6xx3/5=7/10)#...........................(3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To determine the value of "y)#
Substitute equation (3) into equation #(1_a)#. The easier of the two!
#y=-7/10+5/2 #
#y= (25-7)/10#
#y=18/10 = 9/5#
#color(blue)(y=9/5)#
Note: #y=9/5 = 1.8" and "x=7/10=0.7#
#=> P_("crossing point") = (x,y)->(0.7,1.8) #
