Given:
#-x+3y=-9.........................................(1)#
#5x-2y=35..............................................(2)#
#color(red)("~~~~~~~~~~~~~~~~~~ All calculation shown ~~~~~~~~~~~~~~~")#
For equation (1): making #y# the dependant variable (y=..)
Add #color(blue)(x)# to both sides so that it is removed from the left.
#(3y-x) color(blue)(+x)=(-9)color(blue)(+x)#
#color(brown)("The brackets serve no purpose other than to show what")#
#color(brown)("is being altered or to group things so that they are obvious.")#
#3y=x-9#
Divide both sides by 3
#3/(color(blue)(3)) times y= x/(color(blue)(3)) -9/(color(blue)(3)) #
#color(green)(y= 1/3 x-3......................................(1_a))#
#color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#
For equation (2): making #y# the dependant variable (y=..)
By sight!
#color(green)(y= 5/2x-35/2...................................................(2_a))#
#color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#
Both equation #(1_a)# and #(2_a)# have a common value in y so
adopting #"Equation "(1_a) =y = "Equation "(2_a)# to solve for #x#
#color(blue)(1/2x -3 )color(brown)(= y =) color(blue)(5/2x-35/2)#
Giving:
#color(blue)(1/2x -3 )color(brown)(=)color(blue)(5/2x-35/2)#
Collecting like terms:
#5/2x -1/2x =35/2-3#
#2x=29/2#
Divide both sides by 2 giving
#(2x) divide 2 =(29/2) divide 2#
#2/2x = 29/2 times 1/2#
#color(green)(x= 29/4............................................(3))#
#color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#
Now substitute for #x# using equation (3) into either equation #(1_a) " or " (2_a)# to determine the value of y.
I will let you do that!
