we know that
If NO II KJ
then
triangles LKJ and LNO are similar and the ratio of the corresponding sides are equals
so
[tex]\frac{LO}{LN}= \frac{LJ}{LK}[/tex]
we have
[tex]LO=(x-4)\ in[/tex]
[tex]LN=(x-3)\ in[/tex]
[tex]LJ=LO+OJ=(x-4)+x=2x-4\ in[/tex]
[tex]LK=LN+NK=(x-3)+(x-2)=2x-1\ in[/tex]
substitute
[tex]\frac{x-4}{x-3}=\frac{2x-4}{2x-1} \\ \\(x-4)*(2x-1)=(2x-4)*(x-3)\\ \\2x^{2} -x-8x+4=2x^{2} -6x-4x+12\\ \\-9x+10x=12-4\\ \\ x=8\ units[/tex]
therefore
the answer is the option
[tex]x=8\ units[/tex]
