The first thing to notice is that the coefficients of the #x# terms are the same. (Both are #1x#).
There are two approaches:
Subtract the two equations:#" "A-B#
#" "x" "+" "y=+11" " ..................A#
#" "ul(x" "-" "9y=-19" "..................B)#
Remember to change the signs:
#" "x" "+" "y=+11" " ..................A#
#ul(color(red)(-)x" "color(red)(+)" "9y=color(red)(+)19" "..................B)#
#color(white)(xxxxxxxx)10y =30#
#color(white)(xxxxxxxxxx)y =3#
Substitute #3# for #x# in one of the equations (#A#)
#x+3=11#
#x" "=8#
Check in #B" "8-9(3) =8-27 =-19#
Make #x# the subject and then equate the expressions:
#x= 11-y" "and" "x = 9y-19#
We know that
#" "x " "=" "x#, therefore:
#9y-19 = 11-y" "larr# solve for #y#
#9y+y = 11+19#
#" "10y = 30#
#" "y=3#
Substitute #3# for #y# in each equation and check if you get the same answer for #x#.
#x=11-(3)" "rarr x = 8#
#x = 9(3)-19" "rarrx = 27-19 =8#
The given equations are equations of straight lines. Solving them gives the point of intersection of the lines. #(8,3)#
graph{(y-1/9x-19/9)(y+x-11)=0 [-14.24, 14.23, -7.12, 7.12]}
