How do solve the following linear system?: # 8x + 3y= -9 , x - 3y = 1 #?

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

#x - 3y = 1#

#x - 3y + color(red)(3y) = 1 + color(red)(3y)#

#x - 0 = 1 + 3y#

#x = 1 + 3y#

Step 2) Substitute #1 + 3y# for #x# in the first equation and solve for #y#:

#8x + 3y = -9# becomes:

#8(1 + 3y) + 3y = -9#

#(8 * 1) + (8 * 3y) + 3y = -9#

#8 + 24y + 3y = -9#

#8 + 27y = -9#

#-color(red)(8) + 8 + 27y = -color(red)(8) - 9#

#0 + 27y = -17#

#27y = -17#

#(27y)/color(red)(27) = -17/color(red)(27)#

#(color(red)(cancel(color(black)(27)))y)/cancel(color(red)(27)) = -17/27#

#y = -17/27#

Step 3) Substitute #-17/27# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = 1 + 3y# becomes:

#x = 1 + (3 xx -17/27)#

#x = 1 - 17/9#

#x = (9/9 xx 1) - 17/9#

#x = 9/9 - 17/9#

#x = -8/9#

The solution is: #x = -8/9# and #y = -17/27# or #(-8/9, -17/27)#

#8x + 3y = -9" "# Equation 1

#x - 3y = 1" "# Equation 2

Eliminate #y# by adding the two equations. The result becomes

#9x = -8#

Hence,

#x = -8/9#

Find y by substituting the already known #x# to any of the original equation (or both just to double-check)

Using Equation 1,

#8(-8/9) + 3y = -9#

#-64 + 27y = -81#

#27y = -17#

Hence,

#y = -17/27#

To double-check, use also Equation 2 to get #y#

#-8/9 - 3y = 1#

#-8 -27y = 9#

#-17 = 27y#

Hence,

#y = -17/27#