How do you solve the following system of equations?: #x+y= -2 , 4x+3y=8 #?

2 Answers

Answer:

the answer: #x=14# and #y = -16#

Explanation:

In the two equations, get #y# alone. So you have:

#y= -x-2# and

#y= -4/3 x +8/3#.

Then you set the two equations equal to each other:

#-x-2= -4/3 x +8/3#.

Then you solve for #x#, and you get #x=14#. You then substitute #14# into both equations to get #y#, and then you have your coordinates.

Feb 2, 2016

Solve by substitution and elimination:

#x+y=-2#

#4x+3y=8#

We can eliminate #4x# from the second equation by #x# from the first equation if we multiply it by #-4# to get #-4x#:

#rarr-4(x+y=-2)#

#rarr-4x-4y=8#

Now add both of the equations:

#rarr(-4x-4y=8)+(4x+3y=8)#

#rarr-1y=16#

#rarr-y=16#

So,if #-y =16 # then ,#y =-16#

Now substitute the value of #y # to the first equation:

#x+(-16)=-2#

#x-16=-2#

#x=-2+16=14#

So,#(x,y)=(14,-16)#

You can check the answer and it will be correct! :)