Question 1174799
given: 

{{{PT=y}}}
{{{TR=2x+1}}}
{{{QT=4​y}}}
{{{TS=5x+13}}}

The diagonals of a parallelogram bisect each other 

{{{PT = TR}}}    and  {{{QT = TS}}}

{{{y = 2x+1 }}}    and {{{4y = 5x+13}}}

We have a system of equations
{{{y = 2x+1}}} .....eq.1
{{{4y = 5x+13}}}.....eq.2
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substitute {{{y}}} from eq.1 in eq.2

{{{4(2x+1 ) = 5x+13}}}.....eq.2,...solve for {{{x}}}

{{{8x+4 = 5x+13}}}

{{{8x-5x = 13-4}}}

{{{3x = 9}}}

{{{x=3}}}

Now solve for {{{y}}}

{{{y = 2x+1}}} .....eq.1

{{{y = 2*3+1}}} 

  {{{y = 7}}}