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