How do you solve the system of equations #x+ 4y = - 8# and #x + y = - 5#?

2 Answers
Aritra G.
Mar 27, 2017

We would use elimination process here as I'd outline below.

#x = -4# and #y = -1#

Explanation:

We have #x + 4y = -8# be equation #(1)#
And #x + y = -5# be #(2)#

Then subtracting #(2)# from #(1)#,

#x + 4y - (x + y) = - 8 + 5#
#implies 3y = -3#

Which gives #y = -1#

Substituting the value of #y# in any one equations, say #(1)#

#x - 4 = -8#
Gives, #x = -4#

seol
Mar 27, 2017

#x=28/5,y=-3/5#

Please note that this method is not the only way to solve a system of equations! The other answer is just as effective.

Explanation:

Given information
#x+4y=-8#
#x+y=5#

Being a system of equations problem, you can apply a substitution, where you find the "value" of one variable by calculating a new equation for it.

Let's find #x# from the second equation.

Working it out
#x+y=5#
#x=5-y# or #x=-y+5#

Now that we have the "value" for #x#, plug it into the first equation
#\color(red)(x)+4y=-8#

  • #\therefore(\color(red)(-y+5))+4y=8#
  • #-y+5-4y=8#
  • #5-y-4y=8#
  • #5-5y=8#
  • #-5y=(8-5)#
  • #-5y=3#
  • #y=3/(-5)#

Based on the resulting value for #y# plugin to the equation found for "value" of #x#
#x=-\color(green)(y)+5#

  • #\thereforex=-(\color(green)(3/(-5)))+5#
  • #x=3/5+5#
  • #x=3/5+25/5#
  • #x=28/5#
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