Question

Solve the following equations.

$\frac{{}^{x+1}{C}_2}{{}^x{C}_3}=\frac{4}{5},xϵN$

Answer 1

$\frac{{}^{x+1}{C}_2}{{}^x{C}_3}=\frac{4}{5},xϵN$

$\Rightarrow \frac{\frac{(x+1)!}{(x+1-2)!2!}}{\frac{x!}{(x-3)!3!}}=\frac{4}{5}$

$\Rightarrow \frac{\frac{(x+1)\left(x\right)(x-1)!}{(x-1)!2!}}{\frac{x(x-1)(x-2)(x-3)!}{(x-3)!3!}}=\frac{4}{5}$

$\Rightarrow \frac{x(x+1)}{2!}\times 5=\frac{4}{3\times 2}x(x-1)(x-2)$

or, $x\left[\frac{5}{2}\right(x+1)-\frac{2}{3}(x-1\left)\right(x-2\left)\right]=0$

$x=0$ as $x\in N$

So, $\frac{5}{2}(x+1)-\frac{2}{3}(x^2-3x+2)=0$

or, $15(x+1)-4(x^2-3x+2)=0$

$\Rightarrow 15x+15-4x^2+12x-8=0$

$\Rightarrow 27x+7-4x^2=0$

$\Rightarrow 4x^2-27x-7=0$

$\Rightarrow 4x^2-28x+x-7=0$

$\Rightarrow 4x(x-7)+1(x-2)=0$

$\Rightarrow x=7$ or $x\ne -\frac{1}{4}$

$\therefore x=7$