How do you solve the following linear system: #x/a+y/b=2#, #bx-ay=0#?

Question

832 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-28T23:18:35

#{(x = a), (y = b):}#

#{(1/ax + 1/by = 2), (bx - ay = 0):}#

Multiply the second equation by #1/(ab)#

#{(1/ax + 1/by = 2), (1/ax - 1/by = 0):}#

Add the second equation to the first

#1/ax + 1/by + (1/ax-1/by) = 2+0#

#=> 2/ax = 2#

#=> x = a#

Substitute this back into #1/ax +1/by = 2#

#=> 1/a a + 1/by = 2#

#=> 1 + 1/by = 2#

#=> 1/by = 1#

#=> y = b#

Thus we have the solution

#{(x = a), (y = b):}#