How do you solve the following system?: #3x - 2y = 1 , 6x+5y=18 #

1 Answer
Mar 30, 2018

Answer:

See a solution process below: #(41/27, 16/9)#

Explanation:

Step 1) Solve each equation for #6x#:

  • Equation 1:

#3x - 2y = 1#

#3x - 2y + color(red)(2y) = 1 + color(red)(2y)#

#3x - 0 = 1 + 2y#

#3x = 1 + 2y#

#color(red)(2) xx 3x = color(red)(2)(1 + 2y)#

#6x = (color(red)(2) xx 1) + (color(red)(2) xx 2y)#

#6x = 2 + 4y#

  • Equation 2:

#6x + 5y = 18#

#6x + 5y - color(red)(5y) = 18 - color(red)(5y)#

#6x + 0 = 18 - 5y#

#6x = 18 - 5y#

Step 2) Because the left side of both equations are equal we can equate the right side of each equation and solve for #y#:

#2 + 4y = 18 - 5y#

#2 - color(red)(2) + 4y + color(blue)(5y) = 18 - color(red)(2) - 5y + color(blue)(5y)#

#0 + (4 + color(blue)(5))y = 16 - 0#

#9y = 16#

#(9y)/color(red)(9) = 16/color(red)(9)#

#(color(red)(cancel(color(black)(9)))y)/cancel(color(red)(9)) = 16/9#

#y = 16/9#

Step 3) Substitute #16/9# for #y# in either of the equations in Step 1 and solve for #x#:

#6x = 2 + 4y# becomes:

#6x = 2 + 4(16/9)#

#6x = (9/9 xx 2) + 64/9#

#6x = 18/9 + 64/9#

#6x = 82/9#

#6x = 82/9#

#1/6 xx 6x = 1/6 xx 82/9#

#6/6x = 82/54#

#x = 41/27#

The Solution Is:

#x = 41/27# and #y = 16/9#

Or

#(41/27, 16/9)#