How do you solve the system of equations #x- 2y = 5# and #2/ 3x - 4/ 3y = 6#?

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Answer 1 · 2017-06-03T23:15:50

See a solution process below:

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

#x - 2y = 5#

#x - 2y + color(red)(2y) = 5 + color(red)(2y)#

#x - 0 = 5 + 2y#

#x = 5 + 2y#

Step 2) Substitute #(5 + 2y)# for #x# in the second equation and solve for #y#:

#2/3x - 4/3y = 6# becomes:

#2/3(5 + 2y) - 4/3y = 6#

#(2/3 * 5) + (2/3 * 2y) - 4/3y = 6#

#10/3 + 4/3y - 4/3y = 6#

#10/3 = 6#

Because #10/3 != 6# there is no solution to this problem. Or the solution is the null or empty set #{O/}#.

This indicates the two lines are parallel and are not the same line.