How do you solve the following system: #2x+y=5, -29=5y-3x #?

1 Answer
Jul 27, 2017

Answer:

See a solution process below:

Explanation:

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

#2x + y = 5#

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

#0 + y = -2x + 5#

#y = -2x + 5#

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

#-29 = 5y - 3x# becomes:

#-29 = 5(-2x + 5) - 3x#

#-29 = (5 xx -2x) + (5 xx 5) - 3x#

#-29 = -10x + 25 - 3x#

#-29 = -10x - 3x + 25#

#-29 = (-10 - 3)x + 25#

#-29 = -13x + 25#

#-29 - color(red)(25) = -13x + 25 - color(red)(25)#

#-54 = -13x + 0#

#-54 = -13x#

#(-54)/color(red)(-13) = (-13x)/color(red)(-13)#

#54/13 = (color(red)(cancel(color(black)(-13)))x)/cancel(color(red)(-13))#

#54/13 = x#

#x = 54/13#

Step 3) Substitute #54/13# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = -2x + 5# becomes:

#y = (-2 xx 54/13) + 5#

#y = -108/13 + 5#

#y = -108/13 + (13/13 xx 5)#

#y = -108/13 + 65/13#

#y = -43/13#

The Solution Is: #x = 54/13# and #y = -43/13# or #(54/13, -43/13)#