JEE Main 2026 — Mathematics Algebra

We need $2a+3b = 3c+4d$2a+3b=3c+4d where $a,b,c,d \in \{1,2,3,4\}$a,b,c,d∈{1,2,3,4}.

Possible values of $2a+3b$2a+3b: range from $5$5 to $20$20.

Possible values of $3c+4d$3c+4d: range from $7$7 to $28$28.

For each common value $k$k, the number of elements in $R$R equals (number of $(a,b)$(a,b) with $2a+3b=k$2a+3b=k) $\times$× (number of $(c,d)$(c,d) with $3c+4d=k$3c+4d=k).

$k=7$k=7: $(a,b)=(2,1)$(a,b)=(2,1), $(c,d)=(1,1)$(c,d)=(1,1) $\Rightarrow 1 \times 1=1$⇒1×1=1

$k=10$k=10: $(2,2)$(2,2), $(2,1)$(2,1) $\Rightarrow 1$⇒1

$k=11$k=11: $(1,3),(4,1)$(1,3),(4,1), $(1,2)$(1,2) $\Rightarrow 2$⇒2

$k=13$k=13: $(2,3)$(2,3), $(3,1)$(3,1) $\Rightarrow 1$⇒1

$k=14$k=14: $(1,4),(4,2)$(1,4),(4,2), $(2,2)$(2,2) $\Rightarrow 2$⇒2

$k=15$k=15: $(3,3)$(3,3), $(1,3)$(1,3) $\Rightarrow 1$⇒1

$k=16$k=16: $(2,4)$(2,4), $(4,1)$(4,1) $\Rightarrow 1$⇒1

$k=17$k=17: $(4,3)$(4,3), $(3,2)$(3,2) $\Rightarrow 1$⇒1

$k=18$k=18: $(3,4)$(3,4), $(2,3)$(2,3) $\Rightarrow 1$⇒1

$k=20$k=20: $(4,4)$(4,4), $(4,2)$(4,2) $\Rightarrow 1$⇒1

Total $= 1+1+2+1+2+1+1+1+1+1 = 12$=1+1+2+1+2+1+1+1+1+1=12