Question

Question #060f9

Answer 1

`x=3`

Explanation:

Remember that:

`(""_b^a) = (a!)/(b!(a-b)!)`

Therefore, we can change our equation to say:

`(x!)/(2!(x-2)!) + ((x+1)!)/(2!(x-1)!) = x+6`

Simplifying both fractions, we get:

`((x)(x-1))/2 + ((x+1)(x))/2 = x+6`

`(x^2-x)/2 + (x^2 + x)/2 = x+6`

`(x^2-x+x^2+x)/2 = x+6`

`(2x^2)/2 = x+6`

`x^2 = x+6`

`x^2 - x - 6 = 0`

`(x-3)(x+2) = 0`

Therefore, `x = -2 " " or " " x=3`

Since `(-2)!` is undefined, we can eliminate that solution.

The only answer left is `x=3`

Final Answer

Answer 2

The answer is `x=3`

Explanation:

The notation used is that of a combination of 2 objects chosen from a set of `x` objects where order does not matter.

I will first simplify the left side of the equation, then will combine this with the right side in order to solve.

The first combination is

`(x!)/((2!)(x-2!))` which simplifies to `((x)(x-1)(x-2)!)/(2!(x-2)!)`

or simply `(x(x-1))/2`

The second combination is

`((x+1)!)/((2!)(x-1)!)` which simplifies to `((x+1)(x)(x-1)!)/(2!(x-1)!)`

or simply `((x+1)(x))/2`

Now, add the two expressions

`(x(x-1))/2+((x+1)(x))/2`

`((x^2-x)+(x^2+x))/2`

So, the left side of the equation simplifies to `x^2`

Now, the equation becomes

`x^2=x+6`

`x^2-x-6=0`

This factors as `(x-3)(x+2)=0`

The roots of this equation are `x=3` and `x=-2`

Since a combination having `n` = -2 makes no sense, we reject this result.

The answer is `x=3`