How do you solve the system of linear equations #5x - 4y = 9# and #x - 2y = -3#?

Question

4765 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-09T02:18:34

See a solution process below:

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

#z - 2y + color(red)(2y) = -3 + color(red)(2y)#

#z - 0 = -3 + 2y#

#z = -3 + 2y#

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

#5x - 4y = 9# becomes:

#5(-3 + 2y) - 4y = 9#

#(5 xx -3) + (5 xx 2y) - 4y = 9#

#-15 + 10y - 4y = 9#

#-15 + (10 - 4)y = 9#

#-15 + 6y = 9#

#color(red)(15) - 15 + 6y = color(red)(15) + 9#

#0 + 6y = 24#

#6y = 24#

#(6y)/color(red)(6) = 24/color(red)(6)#

#(color(red)(cancel(color(black)(6)))y)/cancel(color(red)(6)) = 4#

#y = 4#

Step 3) Substitute #4# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = -3 + 2y# becomes:

#x = -3 + (2 xx 4)#

#x = -3 + 8#

#x = 5#

The solution is: #x = 5# and #y = 4# or #(5, 4)#