Question 893859
Use a substitution,
{{{u=1/a}}}, {{{v=1/b}}}
.
.
.
{{{6u-4v=1}}}
{{{9u-8v=1}}}
Multiply eq. 1 by {{{-2}}} and add to eq. 2,
{{{-12u+8v+9u-8v=-2+1}}}
{{{-3u=-1}}}
{{{u=1/3}}}
Then,
{{{6(1/3)-4v=1}}}
{{{2-4v=1}}}
{{{-4v=-1}}}
{{{v=1/4}}}
So then,
{{{1/a=1/3}}}
{{{a=3}}}
and
{{{1/b=1/4}}}
{{{b=4}}}