Notice that both equations coefficient of #x" is "+1/2#

This makes life easy as we can 'get rid' of one for the variables.

To solve an equation easily you need just one variable and a relationship to some constant.

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Given:

#1/2x=y-1#............................(1)

#1/2x+1/3y=5#......................(2)

It is common practice to convert them so that they are of form

#y=mx+c#. This not strictly necessary in this case as the process will sort this out any way.

Equation (2) has the lesser value of #y# so I am opting for:

#color(brown)("Subtract equation (2) from equation (1) giving:")#

#color(green)("Notice how I have used 0 as a place keeper")#

#" "1/2x+0ycolor(white)(.)=y-1#

#" "underline(1/2x+1/3y=0y+5 )->"subtract"#

#" "color(green)(0" " -1/3y=y-6)#

Add #color(magenta)(1/3y)# to both sides

#" "color(green)(-1/3ycolor(white)(.)color(magenta)(+1/3y)=ycolor(white)(.)color(magenta)(+1/3y)-6)#

#" "0=4/3y-6#

Add 6 to both sides giving:

#6=4/3y#

#color(blue)(y=(6xx3)/4 = 9/2)#

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I am choosing equation (1) for substitution of y

#1/2x=y-1" "->" "1/2x=9/2-1#

#color(blue)(x=7)#

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Check

#1/2x+1/3y=5 " becomes "1/2(7)+1/3(9/2) = 5# Confirmed!