# Question 060f9

Oct 30, 2017

$x = 3$

#### 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:

$\frac{\left(x\right) \left(x - 1\right)}{2} + \frac{\left(x + 1\right) \left(x\right)}{2} = x + 6$

$\frac{{x}^{2} - x}{2} + \frac{{x}^{2} + x}{2} = x + 6$

$\frac{{x}^{2} - x + {x}^{2} + x}{2} = x + 6$

$\frac{2 {x}^{2}}{2} = x + 6$

${x}^{2} = x + 6$

${x}^{2} - x - 6 = 0$

$\left(x - 3\right) \left(x + 2\right) = 0$

Therefore, $x = - 2 \text{ " or " } x = 3$

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

The only answer left is $x = 3$

Oct 30, 2017

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 $\frac{x \left(x - 1\right)}{2}$

The second combination is

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

or simply $\frac{\left(x + 1\right) \left(x\right)}{2}$

$\frac{x \left(x - 1\right)}{2} + \frac{\left(x + 1\right) \left(x\right)}{2}$

$\frac{\left({x}^{2} - x\right) + \left({x}^{2} + x\right)}{2}$

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 $\left(x - 3\right) \left(x + 2\right) = 0$

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 $x = 3$