we know that
In a Rhombus all sides are congruent
so
\(MJ=JK=KL=ML\)
we have
\(MJ=3x\ units\\JK=(2x+4)\ units\)
equate MJ and JK
\(3x=2x+4\)
Solve for x
\(3x-2x=4\)
\(x=4\ units\)
Find the length side of the rhombus
\(MJ=3x\)
\(MJ=3*4=12\ units\)
therefore
the answer is
The length side of the rhombus is \(12\ units\)
