How do you solve the system of equations #2x - 2y = 0# and #4x + 9y = 0#?

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Answer 1 · 2017-07-23T22:07:44

See a solution process below:

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

#2x - 2y = 0#

#2x - 2y + color(red)(2y) = 0 + color(red)(2y)#

#2x - 0 = 2y#

#2x = 2y#

#(2x)/color(red)(2) = (2y)/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = (color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2))#

#x = y#

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

#4x + 9y = 0# becomes:

#4y + 9y = 0#

#(4 + 9)y = 0#

#13y = 0#

#(13y)/color(red)(13) = 0/color(red)(13)#

#(color(red)(cancel(color(black)(13)))y)/cancel(color(red)(13)) = 0#

#y = 0#

Step 3) Substitute #0# for #y# in the solution to the first equation at the end Step 1 to determine #x#:

#x = y# becomes:

#x = 0#

The Solution Is: #x = 0# and #y = 0# or #(0, 0)#

Answer 2 · 2017-07-23T22:18:43

See below

#2x - 2y = 0#
When #x = 0, -2y = 0#
#-2y/-2y# = #0/-2y#
#y=0#

When #y = 0, 2x = 0#
#2x/2x# = #0/2x#
#x = 0#
To check if it is right, you substitute these values
(#2 × 0) - (2 × 0#) = #0#

You do the same for #4x + 9y = 0#