How do you solve #9x+6y=3# and #-3x+y=14# using substitution?

1 Answer
Jun 1, 2018

Answer:

First, you take 1 equation and you make it so that the #y# is the main thing like #y =#something... Then you take that and you substitute it into your other equation to get #x = #something...You then take that and you substitute it into your first equation and done.

Explanation:

NOTE:
color #color(gold)(gold# and the bolded text is the main stuff
Other writing is explanations

So, to start you got these 2 equations...
#color(gold)(9x + 6y = 3#
#color(gold)(-3x + y = 14#

Let's call equation 1 A and equation 2 B. So...
#9x + 6y = 3#: #color(red)(A#
#-3x + y = 14#: #color(blue)(B#

So to start substituting, you first take #color(blue)(B# because it has the lowest number of '#y#'s...
#-3x +y = 14# #color(blue)(B#
#9x + 6y = 3# #color(red)(A#
Equation #color(blue)(B# has the lowest number of #y#.

Now, you take equation #color(blue)(B# and you twist and turn it to make the #y# in the equation the main letter. Like so...

#color(gold)(-3x + y = 14# | This is your starting equation
#y = 14 + 3x# | you move over the '#-3x#' to the other side of the '#=#' sign. Since it was a 'negative' number before, going over to the other side makes '#-3x#' a positive number, '#3x#'.
#color(gold)(y = 14 + 3x# | now '#y#' is the main letter.

Now you have this, you get your other equation (equation #color(red)(A#) and you 'combine' it with your new equation #color(blue)(B#. Like so...

#9x + 6y = 3#: #color(red)(A#
#y = 14 + 3x#: NEW #color(blue)(B#

You substitute your NEW #color(blue)(B# into #color(red)(A#...
#9x+ 6color(gold)((14 + 3x)) = 3# | In the NEW #color(blue)(B#, it says the #y =# 'that'. Therefore, you can make the #y# in #color(red)(A# (#6y#) also equal 'that'.

Now you just have to do some switch-a-roos and stuff to get your answers: x = something and y = something, like this...

#color(gold)(9x + 6(14 + 3x) = 3# | what you got now
#9x + 84 + 18x = 3# | expanded the brackets
#27x + 84 = 3# | added the similar numbers
#27x = 3 - 84# | moved #84# to the other side
#27x = -81#
#x = -81/27# | moved #27# to the other side.
#color(gold)(x = -3# | your first answer.

Now...
#color(gold)(-3x + y = 14#: #color(blue)(B# | The old equation #color(blue)(B#
Since #x = -3# so...
#-3(-3) + y = 14# | substituting # x = -3#
#9 + y = 14#
#y = 14 - 9# | moving 9 to the other side
#color(gold)(y = 5# | your second answer.

That's how you do it.