How do you solve #x + 6y = 28# and #2x - 3y = -19#?

1 Answer
Jun 4, 2018

Answer:

Double the second equation, and then add the two equations together linearly, solving for #x#, and then solve for #y# in either. You will find that #x=-2# and #y=5#

Explanation:

First, line up the equations with one above the other:

#x+6y=28#
#2x-3y=-19#

next, multiply the ENTIRE bottom equation by 2:

#x+6y=28#
#color(red)(4)x-color(red)(6)y=color(red)(-38)#

Now, add the equations together:

#(x+6y=28)#
#ul(+(4x-6y=-38)#
#color(red)((4+1))x+color(red)((6-6))y=color(red)((28-38))#

#color(red)(5)x+color(red)(0)y=color(red)(-10)#

#5x=-10#

Now, solve for #x#:

#(cancel(5)x)/color(red)(cancel(5))=(-10)/color(red)(5)#

#color(blue)(x=-2)#

Since we now have a solution for #x#, we can plug it back into either equation to solve for #y#. Doing it in both proves that our solution for #x# is valid:

Equation 1:

#color(blue)(x)+6y=28#

#color(blue)(-2)+6y=28#

#color(blue)(cancel(-2))+6ycolor(red)(cancel(+2))=28color(red)(+2)#

#6y=30#

#(cancel(6)y)/color(red)(cancel(6))=30/color(red)(6)#

#color(green)(y=5)#

Equation 2:

#2color(blue)(x)-3y=-19#

#2color(blue)((-2))-3y=-19#

#-4-3y=-19#

#cancel(-4)-3ycolor(red)(cancel(+4))=-19color(red)(+4)#

#-3y=-15#

#(cancel(-3)y)/color(red)(cancel(-3))=(-15)/color(red)(-3)#

#color(green)(y=5)#

Both equations support the solution, we're done!