A Long Equation I

$\begin{aligned} X & = \left(\frac 12 + {\color{#3D99F6}\frac 13 + \frac 14 + \frac 15} \right) \left({\color{#3D99F6}\frac 13 + \frac 14 + \frac 15} + \frac 16 \right) - \left(\frac 12 + {\color{#3D99F6}\frac 13 + \frac 14 + \frac 15} + \frac 16 \right) \left({\color{#3D99F6}\frac 13 + \frac 14 + \frac 15} \right) & \small \color{#3D99F6} \text{Let }a = \frac 13 + \frac 14 + \frac 15 \\ & = \left(\frac 12 + {\color{#3D99F6}a} \right) \left({\color{#3D99F6}a} + \frac 16 \right) - \left(\frac 12 + {\color{#3D99F6}a} + \frac 16 \right) {\color{#3D99F6}a} \\ & = \frac 12 a + \frac 1{12} + a^2 + \frac 16 a - \frac 12 a - a^2 - \frac 16a \\ & = \boxed{\frac 1{12}} \end{aligned}$

$\begin{aligned} S &=\left(\blue{\frac 12 + \frac 13 + \frac 14 + \frac 15 }\right) \left(\red{\frac 13 + \frac 14 + \frac 15 }+ \frac 16 \right) - \left(\blue{\frac 12 + \frac 13 + \frac 14 + \frac 15 }+ \frac 16 \right) \left(\red{\frac 13 + \frac 14 + \frac 15 }\right) \\ &=a\left(b+\dfrac{1}{6}\right)-b\left(a+\dfrac{1}{6}\right) \small\text{ Let } a=\blue {\frac12+\frac13+\frac14+\frac15}\text{, } b= \red{\frac13+\frac14+\frac15} \\ &=ab+\dfrac{a}{6}-ab-\dfrac{b}{6} \\ &=\dfrac{a-b}{6} \\ &=\dfrac{1}{6}\times \dfrac{1}{2} \\ &=\dfrac{1}{12} \end{aligned}$

$\begin{aligned} &~ \left(\frac12+\frac13+\frac14+\frac15\right)\left(\color{#D61F06}{\frac13+\frac14+\frac15}\color{#3D99F6}{+\frac16}\right)-\left(\color{magenta}{\frac12+\frac13+\frac14+\frac15}\color{#20A900}{+\frac16}\right)\left(\frac13+\frac14+\frac15\right) \\ &=\left[\cancel{\left(\frac12+\frac13+\frac14+\frac15\right)\left(\color{#D61F06}{\frac13+\frac14+\frac15}\right)}+\left(\frac12+\frac13+\frac14+\frac15\right)\left(\color{#3D99F6}{\frac16}\right)\right]-\left[\cancel{\left(\color{magenta}{\frac12+\frac13+\frac14+\frac15}\right)\left(\frac13+\frac14+\frac15\right)}+\left(\color{#20A900}{\frac16}\right)\left(\frac13+\frac14+\frac15\right)\right] \\ &=\left(\color{#3D99F6}{\frac16}\right)\left(\color{cyan}{\frac12}\color{#EC7300}{+\frac13+\frac14+\frac15}\right)-\left(\color{#20A900}{\frac16}\right)\left(\frac13+\frac14+\frac15\right) \\ &=\left[\left(\color{#3D99F6}{\frac16}\right)\left(\color{cyan}{\frac12}\right)+\cancel{\left(\color{#3D99F6}{\frac16}\right)\left(\color{#EC7300}{\frac13+\frac14+\frac15}\right)}\right]-\cancel{\left(\color{#20A900}{\frac16}\right)\left(\frac13+\frac14+\frac15\right)} \\ &=\left(\color{#3D99F6}{\frac16}\right)\left(\color{cyan}{\frac12}\right) \\ &=\frac1{12} \end{aligned}$

Multiply the individual fractions each by 60 (least common multiplier) and than divide by 60^2 as the multiplier is present twice in each product of that result.

((30+20+15+12)(20+15+12+10)-(30+20+15+12+10)(20+15+12))/(60*60)

$\frac{77\times57-87\times47}{60^2}$

300/3600

1/12

This is not the way I actually solved the problem. This is the way I would have done it in second grade (about age eight years).

I started by saying $x= \left( \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \right)$ then our sum becomes $\left( x + \frac{1}{2} \right) \left( x + \frac{1}{6} \right) -\left( x + \frac{4}{6} \right) x$ We expand to get $x^2 + \frac{4}{6}x + \frac{1}{12}-x^2 -\frac{4}{6}x= \frac{1}{12}$ so our answer is $\frac{1}{12}$

Let $S$ be a sum of the numbers. Say $a = \frac {1}{3} + \frac {1}{4} + \frac {1}{5}, b = \frac {1}{2}, c = \frac {1}{6}$ . Thus, $S = (a + b)(a + c) - (a + b + c)(a)$ . By distribution law, we have $S = a^2 + ab + ac + bc - a^2 - ab - ac = bc = (\frac {1}{2})(\frac {1}{6}) = \frac {1}{12}$ .

2x6 =12 3x4=12 4x3=12 6x2=12 no idea about 5

If 1/3 + 1/4 + 1/5 = a, 1/2 = x and 1/6 = y then the equation becomes
(a+x) (b+x) - (a+b+x) (x)
By solving it we get
ax + bx + ab + x^2 - (ax + bx + x^2) = ab which is 1/2 x 1/6 = 1/12

Another way to solve this is by assigning a variable to each fraction. For example: $x={\frac{1}{2}}$ , $y={\frac{1}{3}}$ , $z={\frac{1}{4}}$ , $u={\frac{1}{5}}$ , $v={\frac{1}{6}}$ .

So then, after expanding each of the multiplied sets, you just cancel out the common terms. Doing so leaves you with xv = ${\frac{1}{12}}$ .