How do you solve the system of equations #5x - 3y = 0# and #- 5x + 12y = 0#?

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Answer 1 · 2018-03-29T10:33:40

x=0
y=0

Just add the two linear equations together

#5x-3y=0#
#-5x+12y=0#

#0+9y=0#
#y=0#

Put the y value into the first equation to figure out x
#5x-3(0)=0#
#5x=0#
#x=0#

Answer 2 · 2018-03-29T10:40:18

#color(blue)(x=0)#

#color(blue)(y=0)#

#5x-3y=0 \ \ \ \ \ \ \ \ \ \[1]#

#-5x+12y=0 \ \ \ \ [2]#

Add #[1]# and #[2]#

# \ \ \ \ 5x-3y=0#
#\-5x+12y=0#
# \ \ \ \ \ \ \ \0+9y=0#
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y=0#

Substituting this value of y in #[1]#

#5x-3(0)=0#

#5x=0#

#x=0#

So solutions are:

#color(blue)(x=0)#

#color(blue)(y=0)#

This is an example of a homogeneous system. #(0,0)# is always a solution to these systems and is known as the trivial solution.