How do solve the following linear system?: # x+2y=1 , 6 x+3y=11 #?

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Answer 1 · 2018-05-18T12:14:16

See a solution process below:

Step 1) Solve the first equation for #x#:

#x + 2y = 1#

#x + 2y - color(red)(2y) = 1 - color(red)(2y)#

#x + 0 = 1 - 2y#

#x = 1 - 2y#

Step 2) Substitute #(1 - 2y)# for #x# in the second equation and solve for #y#:

#6x + 3y = 11# becomes:

#6(1 - 2y) + 3y = 11#

#(6 * 1) - (6 * 2y) + 3y = 11#

#6 - 12y + 3y = 11#

#6 + (-12 + 3)y = 11#

#6 + (-9)y = 11#

#6 - 9y = 11#

#6 - color(red)(6) - 9y = 11 - color(red)(6)#

#0 - 9y = 5#

#-9y = 5#

#(-9y)/color(red)(-9) = 5/color(red)(-9)#

#(color(red)(cancel(color(black)(-9)))y)/cancel(color(red)(-9)) = -5/9#

#y = -5/9#

Step 3) Substitute #-5/9# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = 1 - 2y# becomes:

#x = 1 - (2 * -5/9)#

#x = 1 - (-10/9)#

#x = 1 + 10/9#

#x = 9/9 + 10/9#

#x = 19/9#

The Solution Is:

#x = 19/9# and #y = -5/9#

Or

#(19/9, -5/9)#