Given #a+3/4b=2# and #b+3/4a=6#. How do you find #b/a#?

1 Answer
Dec 26, 2015

Answer:

Just solving the system would make possible to calculate #b/a#.

Explanation:

Let us solve the system by the row reduction method.

  1. Let us multiply both equations by 4, in order to quit every fraction:
    #4a+3b=8#
    #3a+4b=24#

  2. Now, let us multiply the first equation by 3 and the second one by 4, in order to get the same coefficient for #a#:
    #12 a + 9b = 24#
    #12a+16b=96#

  3. Now, let us substract both equations:
    #-7b=-72 rightarrow b = 72/7#

  4. We get #a# from the first equation:
    #a = 2 - 3/4 b = 2 - 3/4 cdot 72/7 = -40/7#

This way, the division #b/a# equals #{72/7}/{-40/7} = -72/40 = -9/5#