How do you solve the following system?: #5x +y =2, 13x -5y = -2#

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Answer 1 · 2017-03-29T02:48:08

See the entire solution process below:

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

#5x + y = 2#

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

#0 + y = -5x + 2#

#y = -5x + 2#

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

#13x - 5y = -2# becomes:

#13x - 5(-5x + 2) = -2#

#13x - (5 xx -5x) - (5 xx 2) = -2#

#13x + 25x - 10 = -2#

#38x - 10 = -2#

#38x - 10 + color(red)(10) = -2 + color(red)(10)#

#38x - 0 = 8#

#38x = 8#

#(38x)/color(red)(38) = 8/color(red)(38)#

#(color(red)(cancel(color(black)(38)))x)/cancel(color(red)(38)) = (2 xx 4)/color(red)(2 xx 19)#

#x = (color(red)(cancel(color(black)(2))) xx 4)/color(red)(cancel(2) xx 19)#

#x = 4/19#

Step 3) Substitute #4/19# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = -5x + 2# becomes:

#y = (-5 xx 4/19) + 2#

#y = -20/19 + 2#

#y = -20/19 + (19/19 xx 2)#

#y = -20/19 + 38/19#

#y = 18/19#

The solution is: #x = 4/19# and #y = 18/19# or #(4/19, 18/19)#