Let A = {1, 2, 3, 4} and R be a relation on t | vedclass

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(A) Given $A = \{1, 2, 3, 4\}$. The relation $R$ is defined on $A 	imes A$ such that $2a + 3b = 4c + 5d$,where $a, b, c, d \in A$.Let $S_1 = 2a + 3b$

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$6$

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(A) Given $A = \{1, 2, 3, 4\}$. The relation $R$ is defined on $A 	imes A$ such that $2a + 3b = 4c + 5d$,where $a, b, c, d \in A$.Let $S_1 = 2a + 3b$ and $S_2 = 4c + 5d$.Possible values for $S_1$ for $a, b \in \{1, 2, 3, 4\}$:If $a=1: 2+3=5, 2+6=8, 2+9=11, 2+12=14$If $a=2: 4+3=7, 4+6=10, 4+9=13, 4+12=16$If $a=3: 6+3=9, 6+6=12, 6+9=15, 6+12=18$If $a=4: 8+3=11, 8+6=14, 8+9=17, 8+12=20$Possible values for $S_2$ for $c, d \in \{1, 2, 3, 4\}$:If $c=1: 4+5=9, 4+10=14, 4+15=19, 4+20=24$If $c=2: 8+5=13, 8+10=18, 8+15=23, 8+20=28$If $c=3: 12+5=17, 12+10=22, 12+15=27, 12+20=32$If $c=4: 16+5=21, 16+10=26, 16+15=31, 16+20=36$We look for common values $\alpha = S_1 = S_2$:For $\alpha = 9$: $(a,b)=(3,1)$ and $(c,d)=(1,1) \implies ((3,1),(1,1))$For $\alpha = 13$: $(a,b)=(2,3)$ and $(c,d)=(2,1) \implies ((2,3),(2,1))$For $\alpha = 14$: $(a,b)=(1,4)$ and $(c,d)=(1,2) \implies ((1,4),(1,2))$For $\alpha = 14$: $(a,b)=(4,2)$ and $(c,d)=(1,2) \implies ((4,2),(1,2))$For $\alpha = 17$: $(a,b)=(4,3)$ and $(c,d)=(3,1) \implies ((4,3),(3,1))$For $\alpha = 18$: $(a,b)=(3,4)$ and $(c,d)=(2,2) \implies ((3,4),(2,2))$Total number of elements is $6$.

Let $A = \{1, 2, 3, 4\}$ and $R$ be a relation on the set $A \times A$ defined by $R = \{((a, b), (c, d)) : 2a + 3b = 4c + 5d\}$. Then the number of elements in $R$ is:

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