How do solve the following linear system?: # 4x+3y=1 , -6x-2y=-8 #?

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Answer 1 · 2018-06-06T17:22:13

the solution is:

#(x,y)-=(2.2,-2.6)#

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

#-6x-2y=-8#------(2)

If
#a11x+a12y=b11#
#a21x+a22y=b21#

Then,

#x=(b11a22-b21a12)/(a11a22-a21a12)#

#y=(a11b21-a21b11)/(a11a22-a21a12)#

Here,

#a11=4; a12=3; b11=1#

#a21=-6; a22=-2; b21=-8#

Substituting

#x=(1xx(-2)-(-8)xx3)/(4xx(-2)-(-6)xx3)=(-2+24)/(-8+18)=22/10=2.2#

#y=(4xx(-8)-(-6)xx1)/(4xx(-2)-(-6)xx3)=(-32+6)/(-8+18)=-26/10=-2.6#

Check:

#lhs=4x+3y=4xx2.2+3xx(-2.6)=8.8-7.8=1=rhs#

#lhs=-6x-2y=-6xx2.2-2xx(-2.6)=-13.2+5.2=-8=rhs#

Hence,
the solution is:

#(x,y)-=(2.2,-2.6)#