How do you solve the following system?: # 4x +5y =-4 , -3x -8y = 4#

2 Answers
Jun 10, 2018

Answer:

#x=-12/17,y=-4/17#

Explanation:

Multiplying the first equation by #3# and the second one by #4# and adding both we get
#-17y=4#
so

#y=-4/17#
now we can calculate #x#
#4x+5(-4/17)=-4#
so

#x=1/4*(-4+20/17)#

so

#x=-12/17#

Jun 10, 2018

Answer:

#x=-12/17color(white)("dddd")y=-4/17#

Explanation:

Given:

#4x+5y=-4" "...........................Equation(1)#
#-3x-8y=4" "............................Equation(2)#

2 equations and 2 unknowns. Thus solvable.

We need to end up with one equation and 1 unknown.

#color(blue)("'Getting rid' of the "x" term")#

#[color(white)("d")3xxEqn(1)color(white)(2/2)]+[color(white)("d")4xxEqn(2)color(white)(2/2)]#

#color(white)("d.")12x+15y=-12" "........................Equation(1_a)#
#ul(-12x-32y=color(white)("d..")16)" "........................Equation(2_a)#
#color(white)("dd")0color(white)(".d")-17y=+4#

Divide both sides by #-17#

#color(white)("dddddddd")y=-4/17#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("'Getting rid' of the "y" term")#

Substitute #-4/17# for #y#

I choose #Eqn(1)# as all the left side of = is positive.
It will still work if you choose #Eqn(2)#

#color(green)( 4x+5color(red)(y)=-4 color(white)("dddd")->color(white)("dddd")4x+5(color(red)(-4/17))=-4 )#

#color(white)("dddddddddddddddd")->color(white)("dddd")4xcolor(white)("ddd")-20/17color(white)("dd")=-4#

Add #20/17# to both sides

#color(white)("dddddddddddddddd")->color(white)("dddd")4xcolor(white)("dd")=-4+20/17#

#color(white)("dddddddddddddddd")->color(white)("dddd")4x=-48/17#

Divide both sides by 4

#color(white)("ddddddddddddddd")->color(white)("dddd")x=-12/17#

Tony B