Math, asked by mela4babu, 1 month ago

please solved this question​

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Answered by RvChaudharY50
11

Solution :-

Let ,

→ 9 + a + b = 15 => a + b = 6 ------- Eqn.(1)

→ c + 6 + d = 9 => c + d = 3 -------- Eqn.(2)

→ e + f + 3 = 29 => e + f = 26 -------- Eqn.(3)

and,

→ 9 + c + e = 19 => c + e = 10 -------- Eqn.(4)

→ a + 6 + f = 11 => a + f = 5 -------- Eqn.(5)

→ b + d + 3 = 15 => b + d = 12 -------- Eqn.(6)

adding all six equations we get,

→ (a + b) + (c + d) + (e + f) + (c + e) + (a + f) + (b + d) = 6 + 3 + 26 + 10 + 5 + 12

→ a + a + b + b + c + c + d + d + e + e + f + f = 62

→ 2a + 2b + 2c + 2d + 2e + 2f = 62

→ 2(a + b + c + d + e + f) = 62

→ (a + b + c + d + e + f) = 31 -------- Eqn.(7)

but, adding first three equations,

→ (a + b + c + d + e + f) = 6 + 3 + 26 = 35 ------- Eqn.(8)

and, adding Eqn.(4) , Eqn.(5) and Eqn.(6),

→ (c + e) + (a + f) + (b + d) = 10 + 5 + 12

→ (a + b + c + d + e + f) = 27 ------- Eqn.(9)

as we can see that,

→ 31 ≠ 35 ≠ 27

→ Eqn.(7) ≠ Eqn.(8) ≠ Eqn.(9)

therefore, given data is incorrect or inconsistent .

Hence, we can conclude that, no solution is possible .

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Answered by vinayraj989542
5

please solved this questions find the answers

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