Question 1074818
No, the function is not one-to-one.
{{{f(a,b)}}}=({{{3a+7b}}},{{{2a+5b}}})
{{{f(c,d)}}}=({{{3c+7d}}},{{{2c+5b}}})
So if the function is one-to-one,
{{{f(a,b)=f(c,d)}}} implies that {{{a=c}}} and {{{b=d}}}.
Just looking at the first component,
{{{3a+7b=3c+7d}}}
Let's assume {{{b=c=0}}}
Then,
{{{3a=7d}}}
{{{a=(7/3)d}}}
So, {{{a<>c}}} and f is not one-to-one.