How do you solve the system of equations #x+4y = 14# and #3x+7y=22#?

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Answer 1 · 2017-08-03T03:48:51

See a solution process below:

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

#x + 4y = 14#

#x + 4y - color(red)(4y) = 14 - color(red)(4y)#

#x + 0 = 14 - 4y#

#x = 14 - 4y#

Step 2) Substitute #(14 - 4y)# for #x# in the second equation and solve for #y#:

#3x + 7y = 22# becomes:

#3(14 - 4y) + 7y = 22#

#(3 * 14) - (3 * 4y) + 7y = 22#

#42 - 12y + 7y = 22#

#42 + (-12 + 7)y = 22#

#42 + (-5)y = 22#

#42 - 5y = 22#

#-color(red)(42) + 42 - 5y = -color(red)(42) + 22#

#0 - 5y = -20#

#-5y = -20#

#(-5y)/color(red)(-5) = (-20)/color(red)(-5)#

#(color(red)(cancel(color(black)(-5)))y)/cancel(color(red)(-5)) = 4#

#y = 4#

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

#x = 14 - 4y# becomes:

#x = 14 - (4 * 4)#

#x = 14 - 16#

#x = -2#

The solution is: #x = -2# and #y = 4# or #(-2, 4)#