How do you solve the following system?: #-8x -7y =7, 5x -3y = -4#

1 Answer
Mar 26, 2016

#(x,y)=(-49/59,-3/59)#

Explanation:

Using matrices and Cramer's Rule
(with a bit of practice this can be done mentally, but I will lay out the steps in detail)

Converting
#color(white)("XXX")-8x-7y=7#
and
#color(white)("XXX")5x-3y=-4#
into matrix form:
#color(white)("XXXXX")xcolor(white)("XX")ycolor(white)("XX)c#
#color(white)("XXX")((-8,-7,7),(5,-3,-4))#
Using the determinants
#color(white)("XXX")D_c# (the original matrix without the #c# column)
#color(white)("XXXXX")=|(-8,-7),(5,-3)|=(-8) * (-3)-(-7) * (5) = 59#
Similarly
#color(white)("XXX")D_x=|(7,-7),(-4,-3)|=7 * (-3) - (-7) * (-4)=-49#
and
#color(white)("XXX")D_y=|(-8,7),(5,-4)|= (-8) * (-4) -5 * 7 = -3#

By Cramer's Rule
#color(white)("XXX")x=(D_x)/(D_c) = -49/59#
and
#color(white)("XXX")y=(D_y)/(D_c) = -3/59#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Alternately: By Elimination
[1]#color(white)("XXX")-8x-7y=7#
[2]#color(white)("XXX")5x-3y=-4#

[3]=[1]#xx 5##color(white)("XXX")-40x-35y=35#
[4]=[2]#xx8##color(white)("XXX")40x-24y=-32#

[5]=[3]+[4]#color(white)("XXX")-59y=3#
[6]#color(white)("XXXXXXXXX")y=-3/59#

...and similarly for #x#