How do you solve #4m+n=2# and #m-4n=9# using substitution?

1 Answer
Apr 27, 2016

#m=1" ; "n=-2#

Explanation:

Given:
#4m+n=2# -------------------------------Equation (1)
#m-4n=9#..........................................Equation (2)

#color(blue)("Determine the value of n")#

#color(brown)("Showing every step in the first part to demonstrate method")#

#color(green)("Consider Equation (2)")#

#color(green)("Add 4n to both sides")#

#" "m-4n+4n=9+4n#

#color(green)("But "4n-4n=0)#

#" "m=9+4n# .....................................(3)

#color(green)("Substitute for m in (1) using (3)")#

#" "4m+n=2" "->" "4(9+4n)+n=2#

#color(green)("Multiply out the bracket")#
#" "=>36+16n+n=2#

#" "36+17n=2#

#color(green)("Subtract 36 from both sides")#
#" "17n=-34#

#color(green)("Divide both sides by 17")#
#" "color(blue)(n=-34/17=-2)#
'.........................................................
#color(blue)("Determine the value of m")#

Substitute #n=-2# into (2)

#m-4n=9" "->" "m-4(-2)=9#

#m+8=9#

#m=9-8#

#m=1#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check
#4m+n=2" "->" "4(1)+(-2)=2#
#m-4n=9" "->" "1-4(-2)=9#