How do you solve the system #3x-y=4# and #2x-3y=-9# using substitution?

2 Answers
Feb 13, 2017

See the entire solution process below:

Explanation:

Step 1) Solve the first equation for #y#:

#3x - y = 4#

#-color(red)(3x) + 3x - y = -color(red)(3x) + 4#

#0 - y = -3x + 4#

#-y = -3x + 4#

#-1 xx -y = -1(-3x + 4)#

#y = 3x - 4#

Step 2) Substitute #3x - 4# for #y# in the second equation and solve for #x#:

#2x - 3y = -9# becomes:

#2x - 3(3x - 4) = -9#

#2x - 9x + 12 = -9#

#-7x + 12 - color(red)(12) = -9 - color(red)(12)#

#-7x + 0 = -21#

#-7x = -21#

#(-7x)/color(red)(-7) = -21/color(red)(-7)#

#(color(red)(cancel(color(black)(-7)))x)/cancel(color(red)(-7)) = 3#

#x = 3#

Step 3) Substitute #3# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = 3x - 4# becomes:

#y = (3 xx 3) - 4#

#y = 9 - 4#

#y = 5#

The solution is: #x = 3# and #y = 5# or #(3, 5)#

Feb 13, 2017

#color(red)("Extreme detail")# given for determining #x# using first principles.

The shared point of these two equations is#" "(x,y)->(3,5)#

Explanation:

Given:
#3x-y=4" "..............Equation(1)#
#2x-3y=-9" ".........Equation(2)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the value of x")#

Consider #Equation(1)#

Add #color(red)(y)# to both sides

#" "color(green)(3x-ycolor(red)(+y)" "=" "4 color(red)(+y))#

But #-y+y=0# giving:

#" "3x+0=4+y#

#" "3x=4+y#

Subtract #color(red)(4)# from both sides

#" "color(green)(3xcolor(red)(-4)" "=" "4color(red)(-4)+y)#

#" "3x-4=y#

#" "y=3x-4" "......Equation(1_a)#

As we have just used equation(1) we now need to use equation(2)

Using #Equation(1_a)# substitute for #y# in #Equation(2)#

#color(green)(2x-3color(red)(y)=-9" "->" "2x-3(color(red)(3x-4))=-9)#

#" "2x-9x+12=-9#

#" "-7x+12=-9#

Subtract 12 from both sides

#" "color(green)(-7x+12color(red)(-12)" "=" "-9color(red)(-12))#

#" "-7x+0=-21#

Divide both sides by -7

#" "color(green)((-7)/(color(red)(-7))color(white)(.) x=(-21)/(color(red)(-7)))#

#" "+1xx x=+3#

#" "x=3#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the value of y")#

I chose equation 1 as it is the most strait forward one to use.

Substitute for x in equation 1 giving:

#3x-y=4" "->" "3(3)-y=4#

#y=5#
Tony B