How do you solve the following linear system: # x=1-y , x-y=5 #?

1 Answer
Nov 13, 2015

Answer:

#color(blue)(x=3)#
A large amount of detail given for the method on purpose!
Using my methods you should be able to find the value of y#-> color(brown)("Practice makes it much faster")#

Explanation:

#color(brown)("You will get used to this and be able to do simple ones in your head")#

Given:

#x=1-y...........................................(1)#
#x-y=5..........................................(2)#

In each case make y the dependant variable (answer) then equate to each other through y to find x

#color(blue)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#
#color(red)("For (1)")#

Add y to both sides giving:

#x+y=1-y+ycolor(white)(xxx) "Note that" -y +y=0 " so disappears!" #
#x+y=1#

Subtract #x# from both sides
#x-x+y=1-x#
#y=1-x..................................(1_a)#

#color(blue)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#
#color(red)("For (2)")#

Add y to both sides giving:
#x-y+y=5+y#
#x=5+y#

Subtract 5 from bot sides giving:
#x-5 = 5-5+y#
#x-5=y#

Reverse so that y is on the left
#y=x-5.......................................(2_a)#

#color(blue)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#
#"Equation "(1_a) = color(blue)(y) = "Equation "(2_a)#

#color(brown)(1-x)=color(blue)(y)=color(brown)(x-5)#
#color(brown)(1-x)=color(brown)(x-5)#

Add #x# to both sides

#1-x+x=x+x-5#
#1=2x-5#

Add 5 to both sides

#1+5=2x-5+5#

#2x=6#

#color(brown)(underline("DIVIDE"))# BOTH SIDES BY 2

#2/2x=6/2#

#color(blue)(x=3)#

#color(blue)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#

Using the methods I have shown you first substitute #x=3# in equations #(1_a) "or " (2_a) then manipulate to find y