How do you solve the following system: #4x-5y= -1, 4x+3y= -1 #?

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How do you solve the following system: #4x-5y= -1, 4x+3y= -1 #?

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Answer 1 · 2016-01-14T08:40:16

I found:
#x=-1/4#
#y=0#

Explanation:

I would try multiplying the first equation by #-1# and then add (in columns) to the second to get:
#{(color(red)(-4x-5y=1)),(4x+3x=-1):}# add them
#0-2y=0#
So:
#y=0#
Substitute this back into the first equation:
#4x-0=-1#
#x=-1/4#

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How do you solve the following system: #4x-5y= -1, 4x+3y= -1 #?

1 Answer

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