How do you solve this system of equations: #-x - 2y = - 4 and 2x - y = 3#?

Question

4296 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-14T02:21:08

#x = 2, y=1#

Algebra

#2x-y=3#

#4x-2y=6# -- (i)

#-x-2y=-4# -- (ii)

equation i minus equation ii,

#(4x-2y) - (-x-2y) = 6-(-4)#

#4x-2y +x+2y= 10#

#5x = 10#

#x = 2#

Putting it in equation (i),

#4(2)-2y=6#

#4-y=3#

#y=1#

So, the answer is #x = 2, y=1#.

#2x-y=3 => y = 2x-3#

#-x-2y=-4 => y = -x/2+2#

Plotting these lines,

graph{(2x-y-3)(x+2y-4)=0}

These cross each other at #(2,1)# so the solution to these equations will be #x=2, y=1#.