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(B) Given the system of equations:$ax - by = 2a - b$  $(1)$$(c + 1)x + cy = 10 - a + 3b$  $(2)$Since $(x = 1, y = 3)$ is a solution,substitute these v

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exactly two

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(B) Given the system of equations:$ax - by = 2a - b$  $(1)$$(c + 1)x + cy = 10 - a + 3b$  $(2)$Since $(x = 1, y = 3)$ is a solution,substitute these values into the equations:From $(1)$: $a(1) - b(3) = 2a - b \implies a - 3b = 2a - b \implies a = -2b$.From $(2)$: $(c + 1)(1) + c(3) = 10 - a + 3b \implies c + 1 + 3c = 10 - a + 3b \implies 4c + 1 = 10 - a + 3b$.Substitute $a = -2b$ into the second result: $4c + 1 = 10 - (-2b) + 3b \implies 4c + 1 = 10 + 5b \implies 4c = 9 + 5b \implies c = \frac{9 + 5b}{4}$.For infinitely many solutions,the ratio of coefficients must be equal:$\frac{a}{c + 1} = \frac{-b}{c} = \frac{2a - b}{10 - a + 3b}$.Using $\frac{a}{c + 1} = \frac{-b}{c}$ and substituting $a = -2b$: $\frac{-2b}{c + 1} = \frac{-b}{c}$.This gives $b = 0$ or $c = 1$.Case $1$: If $b = 0$,then $a = -2(0) = 0$. Substituting into $c = \frac{9 + 5(0)}{4}$,we get $c = \frac{9}{4}$. This gives the triplet $(0, 0, 9/4)$.Case $2$: If $c = 1$,then $1 = \frac{9 + 5b}{4} \implies 4 = 9 + 5b \implies 5b = -5 \implies b = -1$. Then $a = -2(-1) = 2$. This gives the triplet $(2, -1, 1)$.Thus,there are exactly two such triplets.

Number of triplets of $(a, b, c)$ for which the system of equations $ax - by = 2a - b$ and $(c + 1)x + cy = 10 - a + 3b$ has infinitely many solutions and $(x = 1, y = 3)$ is one of the solutions,is:

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