Experimental

Algebra Level 4

a a and b b are two real numbers that are not equal to 6 6 such that a b − 6 ( a + b ) + n = 0 ab - 6(a + b) + n = 0 and

a ( a − 6 ) + n b ( b − 6 ) + n = a − 6 b − 6 \large \dfrac{a(a - 6) + n}{b(b - 6) + n} = \dfrac{a - 6}{b - 6}

Calculate the value of n n .

This is part of the series: " It's easy, believe me! "


The answer is 18.

Thành Đạt Lê by Thành Đạt Lê , 17, Singapore

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1 solution

Sudeep Salgia
Jun 1, 2018

We have been given the equation, a ( a − 6 ) + n b ( b − 6 ) + n = a − 6 b − 6 \displaystyle \frac{a(a-6) + n}{b(b-6) + n} = \frac{a-6}{b-6} . Cross multiplying, we get ( a ( a − 6 ) + n ) ( b − 6 ) = ( b ( b − 6 ) + n ) ( a − 6 ) ⟹ a ( a − 6 ) ( b − 6 ) + n ( b − 6 ) = b ( a − 6 ) ( b − 6 ) + n ( a − 6 ) \left(a(a-6) + n \right) (b-6) = \left(b(b-6) + n \right) (a-6) \implies a(a-6)(b-6) + n(b-6) = b(a-6)(b-6) + n(a-6) On rearranging, we have ( a − b ) ( a − 6 ) ( b − 6 ) = n ( a − b ) \displaystyle (a-b)(a-6)(b-6) = n(a-b) . Since a ≠ b a \neq b , we have, ( a − 6 ) ( b − 6 ) = n ⟹ a b − 6 ( a + b ) + 36 − n = 0 (a-6)(b-6) = n \implies ab - 6(a+b) + 36 -n = 0 . But, we know that a b − 6 ( a + b ) = − n ab -6(a+b) = -n . Therefore, we can write, − n + 36 − n = 0 ⟹ n = 18 - n + 36 -n = 0 \implies \boxed{n = 18} .

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In this question. If we set a=b=12 and n=0. How about it? It is true. Because in question ,nothing explanation about (a equal to b) or ( a not equal to b). We just know that a and b not equal to 6

Rofiud Darojad - 2 years ago

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