How would solve equation 2(xC2)+6(xC3)=9x where xC2 means combination of x taken 2 , xC3 means combination of x taken 3?

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The equation `2*C(x, 3) + 6*C(x, 2) = 9*x` has to be solved.

`2*C(x, 3) + 6*C(x, 2) = 9*x`

=> `2*((x!)/(3!*(x - 3)!)) + 6*((x!)/(2!*(x - 2)!)) = 9*x`

=> `2*((x*(x - 1)*(x - 2))/(3!)) + 6*((x*(x - 1))/(2!)) = 9*x`

=> `2*(((x - 1)*(x - 2))/(3!)) + 6*(((x - 1))/(2!)) = 9`

=> `2*(x - 1)*(x - 2) + 18*(x - 1) = 54`

=> `2(x^2 - 3x + 2) + 18x - 18 = 54`

=> `2x^2 - 6x + 4 + 18x - 18 = 54`

=> `2x^2 + 12x - 68 = 0`

=> `x^2 + 6x - 34 = 0`

The solution of this equation is `-3 +- sqrt 43`

But C(x, n) is defined only if x and n are positive whole numbers.

The equation 2*C(x, 3) + 6*C(x, 2) = 9*x does not have a solution.

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