How do you solve the following system: #3x+y=8, 2x +5y = 5 #?

1 Answer
Jul 26, 2018

Answer:

The more formal layout approach.

This #ul("clearly")# shows the thinking processes and is easy to follow for teacher marking.

I have determined the value of #x# to start you off.

Explanation:

Given:
#3x+y=8" "......................Equation(1)#
#2x+5y=5" "...................Equation(2)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("This bit is not part of the formal approach.")#

We need one equation with just one unknown.

I am using method by example: known that #color(green)(2=2) color(red)(larr" True"#

Multiply both sides by #color(purple)(3)#

#color(green)(ubrace([2color(purple)(xx3)]) = ubrace([2color(purple)(xx3)])#
#color(green)( color(white)(".d") darr color(white)("dddd.d")darr )#
#color(green)(color(white)("dd") 6color(white)("dd") =color(white)("dd") 6 color(red)(larr" Still true"))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Back to the solution.")#

Multiply both sides of #Eqn(1)# by 5 #color(white)("d") # ( 'to get rid' of the #y#)

#15x+5y=40" "........Equation(1_a)#

#Eqn(1_a) - Eqn(2)#

#15x+5y=40#
#ul(color(white)(1)2x+5y=color(white)(4)5larr" Subtract")#
#13x+0color(white)("d") =35#

Divide both sides by 13

#(13x)/13=35/13#

#x=35/13#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Your bit")#

Substitute this into either #Eqn(1) or Eqn(2)# to determine the value of #y#

I will let you do that