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

1 Answer
Dec 28, 2015

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

Explanation:

{(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):}