How do you solve the following linear system: # 3y+8x=7 , 8x=y+1 #?

1 Answer
MG
Apr 15, 2018

#x=(5/16)# and #y=(3/2)#

Explanation:

Firstly, re-arrange the equations so that the #x# and #y# and their respective coefficients are on the lhs of the equal sign, and the constants on the rhs. Number the equations accordingly:

#8x+3y=7# #(1)#
#8x-y=1# #(2)#

subtract #(2)# from #(1)#:

#8x-8x=0# immediately allows us to solve for #y#:

#3y-(-y)=4y# and #7-1=6#

#4y=6#
#y=(6/4)=(3/2)#
Next, substitute #y=(3/2)# back into equation #(2)#, giving:
#8x-(3/2)=1#
#8x=1+(3/2)=(5/2)#
#x=(5/2)/8#
#x=(5/16)#
The solution is: #x=(5/16)# and #y=(3/2)#

CHECK SOLUTION (IMPORTANT): Substitute these values back into equations #(1)# and #(2)# to make sure you arrive at #7# for equation #(1)# and #1# for equation #(2)#.

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