Question 1042658
.
<pre>
How to Solve 2^x . 3^y=5^y-x,10^x+y=2??
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
{{{2^x * 3^y}}} = {{{5^(y-x)}}},    (1)
{{{10^(x+y)}}} = {{{2}}}        (2)

is equivalent to

x*ln(2) + y*ln(3) = (y-x)*ln(5),    (1')
(x+y)*ln(10) = ln(2).               (2')

It is a system of two linear equations in two unknowns x and y.

Solve it by any method (Substitution, Elimination, Determinant).
</pre>