How do you solve the following linear system: # -2x+5y=4, 3x+8y=1 #?

1 Answer
Alan P.
May 20, 2016

#(x,y)=color(blue)(""(-27/31,14/31))#

Explanation:

Given
[1]#color(white)("XXX")-2x+5y=4#
[2]#color(white)("XXX")3x+8y=1#

To eliminate the #x# components we can convert the given equations
into ones with identical coefficients for #x#
by multiplying [1] by #3# and [2] by #2#
[3]#color(white)("XXX")-6x+15y=12#
[4]#color(white)("XXX")6x+16y=2#

Adding [3] and [4]
[5]#color(white)("XXX")31y=14color(white)("XX")rarrcolor(white)("XX")y=14/31#

We cold plug this value back into one of the original equations and solve for #x#
or (the method I find simpler in this case) repeat the above process to eliminate #y# from the original given equations.
[6]#color(white)("XXX")-16x+40y=32#
[7]#color(white)("XXX")15x+40y=5#

Subtracting [7] from [6]
[8]#color(white)("XXX")-31x = 27color(white)("XX")rarrcolor(white)("XX")x=-27/31#

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