Question #2d385

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Answer 1 · 2016-06-23T16:01:42

First part in a lot of detail to demonstrate algebraic manipulation.
Second part not so.

Point common to both equations is:
#(x,y)->(3,2)#

Given:#" "7x+y=23# ........................Equation (1)
#" "3x=4y+1#.........................Equation (2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider equation(1)

Subtract #7x# from both sides

#y=-7x+23" ............................Equation" (1_a)#

#color(green)("Substitute for "color(blue)(y)" in equation (2) using "(1_a))#

#color(brown)(3x=4color(blue)(y)+1" "->" "3x=4color(blue)((-7x+23))+1)#

#color(white)()#

#color(brown)(3x=-28x+92+1)" "color(green)(larr" multiplied out the bracket")#
#color(white)()#

#color(brown)(3xcolor(blue)(+28x)=-28xcolor(blue)(+28x)+93)color(green)(larr" add "color(blue)(28x)" to both sides")#
#color(white)()#

#color(brown)(31x=0+93)#
#color(white)()#

#color(brown)(31/(color(blue)(31)) x=93/(color(blue)(31)) )color(green)(larr" divide both sides by "color(blue)(31))#

#color(white)()#

#color(brown)(x=3)" "color(green)(larr31/31=1" and "93/31 = 3)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute for #x# in (1)

#7x+y=23" " =>" " y=2#