Question #a5043

1 Answer
Mar 9, 2016

#a=9,b=4#

Explanation:

Let me make sure I've understood this question correctly:

By placing the comma within #a,a# like so, are you trying to denote a two-digit decimal number like #5,5#, or #5 1/2#, where #a=5#?

If so, I've answered that question:

#(a,a+b,b)b=128,7#

First, we must find a way to write the decimals algebraically.

#(a+1/10a+b+1/10b)b=128.7#

We can do this since whatever follows the comma must be in the tenths place of the decimal. The first variable is in the ones place, so we can leave it as is.

Distribute the #b# and simplify the fractions.

#11/10ab+11/10b^2=128,7#

Multiply both sides by #10/11#.

#ab+b^2=117#

#b(a+b)=117#

We must now find the factors of #117#:

#{:(1,,,117),(3,,,39),(9,,,13):}#

The only pair that will work here is #9,13# because they yield single digit #a# and #b# values.

We must set #b=4# so that #a+b=13#, which means that #a=9#.

Thus,

#(9,9+4,4)9=128,7#