How do you solve the system of equations #-8x - 2y = - 1# and #3x - 12y = 4#?

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Answer 1 · 2018-07-07T23:02:05

#x=10/51# & #y=-29/102#

Given system of linear equations:

#-8x-2y=-1#

or #8x+2y=1\ ...... .......(1)#

#3x-12y=4\ ...... .......(2)#

Multiplying (1) by #6# & adding to (2) as follows

#3x-12y+6(8x+2y)=4+6\times 1#

#51x=10#

#x=10/51#

setting #x=10/51# in (1) we get

#y=\frac{1-8x}{2}=\frac{1-8\times 10/51}{2}=-29/102#

hence, the solution is #x=10/51# & #y=-29/102#

Answer 2 · 2018-07-07T23:11:48

#x=10/51#

#y=-29/102#

#color(blue)("Equation A: ") -8x-2y=-1#

#color(green)("Equation B: ") 3x-12y=4#

Multiply #color(blue)("Equation A")# by 6, then use elimination to first eliminate #y# from each equation.

#color(blue)("Equation A: ") 6*(-8x-2y) = 6*(-1)#

#color(blue)("Equation A: ") -48x-12y=-6#

#color(green)("Equation B: ") 3x-12y=4#

Subtract #color(green)("Equation B")# from #color(blue)("Equation A")#:

#-51x = -10#

#x = 10/51#

Substitute this #x# value into one of the original equations to find #y#:

#-8(10/51) - 2y =-1#

#-2y = -1 + 80/51#

#y = 1/2 - 40/51#

#y = frac{51-80}{102}#

#y = -29/102#