The number of elements in the set { (a, b) : | vedclass

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(C) The given set is defined by the equation $2a^2 + 3b^2 = 35$,where $a, b \in \mathbb{Z}$.We test integer values for $a$ and $b$:If $a = \pm 2$,then

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$8$

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(C) The given set is defined by the equation $2a^2 + 3b^2 = 35$,where $a, b \in \mathbb{Z}$.We test integer values for $a$ and $b$:If $a = \pm 2$,then $2(4) + 3b^2 = 35 \implies 8 + 3b^2 = 35 \implies 3b^2 = 27 \implies b^2 = 9 \implies b = \pm 3$.This gives the pairs: $(2, 3), (2, -3), (-2, 3), (-2, -3)$.If $a = \pm 4$,then $2(16) + 3b^2 = 35 \implies 32 + 3b^2 = 35 \implies 3b^2 = 3 \implies b^2 = 1 \implies b = \pm 1$.This gives the pairs: $(4, 1), (4, -1), (-4, 1), (-4, -1)$.Checking other values of $a$: if $a = 0, 3b^2 = 35$ (no integer solution); if $a = \pm 1, 2 + 3b^2 = 35 \implies 3b^2 = 33 \implies b^2 = 11$ (no integer solution); if $a = \pm 3, 18 + 3b^2 = 35 \implies 3b^2 = 17$ (no integer solution).Thus,the total number of elements is $4 + 4 = 8$.

The number of elements in the set $\{ (a, b) : 2a^2 + 3b^2 = 35, a, b \in \mathbb{Z} \}$,where $\mathbb{Z}$ is the set of all integers,is

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