How do you solve #2x+4y=-1# and #4x-3y=-2#?

2 Answers
Don't Memorise
Apr 22, 2016

The solution for the system of equations is:
#color(green)(x = - 1/2#
#color(green)(y =0 #

Explanation:

#2x + 4y = -1#, multiplying by #2#
#color(blue)(4x) + 8y = -2#................equation #(1)#

#color(blue)(4x) - 3y = -2#.................equation #(2)#

Solving by elimination:

Subtracting equation #2# from #1# :
#cancelcolor(blue)(4x) + 8y = -2#

#-cancelcolor(blue)(4x) + 3y = +2#

#11 y = 0#

#y = 0/11#

#color(green)(y =0 #

Finding #x# from the first equation.

#2x + 4y = -1#

#2x = -1 - 4y #

#2x = -1 - 4 * 0 #

#2x = -1 #

#color(green)(x = - 1/2#

Don't Memorise
Apr 22, 2016

The solution for the system of equations is:
#color(green)(x = - 1/2#
#color(green)(y =0 #

Explanation:

#2x + 4y = -1#, multiplying by #2#
#color(blue)(4x) + 8y = -2#................equation #(1)#

#color(blue)(4x) - 3y = -2#.................equation #(2)#

Solving by elimination:

Subtracting equation #2# from #1# :
#cancelcolor(blue)(4x) + 8y = -2#

#-cancelcolor(blue)(4x) + 3y = +2#

#11 y = 0#

#y = 0/11#

#color(green)(y =0 #

Finding #x# from the first equation.

#2x + 4y = -1#

#2x = -1 - 4y #

#2x = -1 - 4 * 0 #

#2x = -1 #

#color(green)(x = - 1/2#

Related questions
See all questions in Linear Systems with Multiplication
Impact of this question
1880 views around the world
You can reuse this answer
Creative Commons License