How do you solve the following system: #2x+7y=1, 5x - 7y = 12 #?

1 Answer
Aug 13, 2017

Answer:

See a solution process below:

Explanation:

Step 1) Solve each equation for #7y#:

Equation 1:

#2x + 7y = 1#

#-color(red)(2x) + 2x + 7y = -color(red)(2x) + 1#

#0 + 7y = -2x + 1#

#7y = -2x + 1#

Equation 2:

#5x - 7y = 12#

#-color(red)(5x) + 5x - 7y = -color(red)(5x) + 12#

#0 - 7y = -5x + 12#

#-7y = -5x + 12#

#color(red)(-1) xx -7y = color(red)(-1)(-5x + 12)#

#7y = 5x - 12#

Step 2) Because the left side of each equation is #7y# we can equate the right side of each equation and solve for #x#:

#-2x + 1 = 5x - 12#

#color(red)(2x) - 2x + 1 + color(blue)(12) = color(red)(2x) + 5x - 12 + color(blue)(12)#

#0 + 13 = (color(red)(2) + 5)x - 0#

#13 = 7x#

#13/color(red)(7) = (7x)/color(red)(7)#

#13/7 = (color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7))#

#13/7 = x#

#x = 13/7#

*Step 3) Substitute #13/7# for #x# in the solution to either equation in Step 1 and calculate #y#:

#7y = -2x + 1# becomes:

#7y = (-2 xx 13/7) + 1#

#7y = -26/7 + 1#

#7y = -26/7 + (7/7 xx 1)#

#7y = -26/7 + 7/7#

#7y = -19/7#

#(7y)/color(red)(7) = (-19/7)/color(red)(7)#

#(color(red)(cancel(color(black)(7)))y)/cancel(color(red)(7)) = -19/49#

#y = -19/49#

The Solution Is: #x = 13/7# and #y = -19/49# or #(13/7, -19/49)#