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

2 Answers
May 12, 2018

Answer:

#x=1#
#y=-4#

Explanation:

There are 3 ways to solve this. Here is one way:

Elimination:

Line them up:

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

Add all that goes together:

#2x+x=3x#
#-y+y=0#
#6-3=3#

Put it back into an equation:

#3x=3#

#x=1#

Plug what x equals (1) into one of the previous equations:

#(2•1)-y=6#
#(-2)-y=6-2#
#-y=4 or y=-4#

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

May 12, 2018

Answer:

#x=1#

#y=-4#

Explanation:

These are called simultaneous equations. Multiply both equations so they both have the same leading coefficient. Use the coefficient of #x# in one equation to multiply the entire equation of the other one. Do this for both.

#[2x-y=6]" " xx1#
#[x+y=-3]" " xx2#

Equals

#2x-y=6#
#2x+2y=-6#

Then subtract the two equations

#2x-2x=0#
#[-y]-[2y]=-3y#
#6-[-6]=12#

Result:

#-3y=12#

Simplify:

#-y=4#

so

#y=-4#

Replace the #y# in one of the equations with #-4# to solve for #x#.

#2x-y=6#

#2x- (-4)=6#

#2x+4=6#

#2x=6-4#

#2x=2#

#x=1#