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
Explanation:
Let us solve the system by the row reduction method.

Let us multiply both equations by 4, in order to quit every fraction:
#4a+3b=8#
#3a+4b=24# 
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# 
Now, let us substract both equations:
#7b=72 rightarrow b = 72/7# 
We get
#a# from the first equation:
#a = 2  3/4 b = 2  3/4 cdot 72/7 = 40/7#
This way, the division