How do you solve this system of equations #x+ 4y = 5;3x - 2y = - 13#?

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Answer 1 · 2017-08-13T11:38:20

See a solution process below:

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

#x + 4y = 5#

#x + 4y - color(red)(4y) = 5 - color(red)(4y)#

#x + 0 = 5 - 4y#

#x = 5 - 4y#

Step 2) Substitute #(5 - 4y)# for #x# in the second equation and solve for #y#:

#3x - 2y = -13# becomes:

#3(5 - 4y) - 2y = -13#

#(3 * 5) - (3 * 4y) - 2y = -13#

#15 - 12y - 2y = -13#

#15 + (-12 - 2)y = -13#

#15 + (-14)y = -13#

#15 - 14y = -13#

#-color(red)(15) + 15 - 14y = -color(red)(15) - 13#

#0 - 14y = -28#

#-14y = -28#

#(-14y)/color(red)(-14) = (-28)/color(red)(-14)#

#(color(red)(cancel(color(black)(-14)))y)/cancel(color(red)(-14)) = 2#

#y = 2#

Step 3) Substitute #2# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = 5 - 4y# becomes:

#x = 5 - (4 * 2)#

#x = 5 - 8#

#x = -3#

The Solution Is: #x = -3# and #y = 2# or #(-3, 2)#