We need to count the number of ordered pairs $((a,b),(c,d))$ such that $a,b,c,d \in A=\{1,2,3,4\}$ and $2a+3b=4c+5d.$

Let us first list all possible values of $2a+3b$ for $a,b\in\{1,2,3,4\}$, and count how many times each value occurs.

For each fixed $b$:

• If $b=1$, then $2a+3=5,7,9,11$ for $a=1,2,3,4$.
• If $b=2$, then $2a+6=8,10,12,14.$
• If $b=3$, then $2a+9=11,13,15,17.$
• If $b=4$, then $2a+12=14,16,18,20.$

So the frequency table for $2a+3b$ is:

Now list all possible values of $4c+5d$ for $c,d\in\{1,2,3,4\}$.

For each fixed $d$:

• If $d=1$, then $4c+5=9,13,17,21.$
• If $d=2$, then $4c+10=14,18,22,26.$
• If $d=3$, then $4c+15=19,23,27,31.$
• If $d=4$, then $4c+20=24,28,32,36.$

So the frequency table for $4c+5d$ is:

For equality $2a+3b=4c+5d,$ only common values in the two tables matter.

The common values are: $9,13,14,17,18.$

For each common value, multiply the number of ways it occurs on the left and right.

• Value $9$: left frequency $1$, right frequency $1$ gives $1\cdot 1=1$
• Value $13$: left frequency $1$, right frequency $1$ gives $1\cdot 1=1$
• Value $14$: left frequency $2$, right frequency $1$ gives $2\cdot 1=2$
• Value $17$: left frequency $1$, right frequency $1$ gives $1\cdot 1=1$
• Value $18$: left frequency $1$, right frequency $1$ gives $1\cdot 1=1$

Therefore total number of elements in $R$ is $1+1+2+1+1=6.$

Hence, $|R|=6.$