"Hannah and Tim both think of a number.
Hannah's number is negative. Tim's number is one more than Hannah's. They each take the reciprocal of their numbers. The sum of the reciprocals is $\frac{5}{6}$.
Use algebra to work out Hannah's original number."
I'm struggling with this question. It clearly makes a simultaneous equation but after that I'm unsure of how to solve it.
So far: Let $x$ be Hannah's number and let $y$ be Tim's number. From the question I get the two equations: $$x+1=y$$ $$\frac{1}{x} + \frac{1}{y} = \frac{5}{6}$$ I tried taking the reciprocal of the last equation but I found out that $(x+y)^{-1}$ is not the same as $x^{-1} + y^{-1}$ and as such I can't figure out how to take the reciprocal.
Any help would be greatly appreciated.
We introduce unknowns $h$ and $t$.
$$ h < 0 $$
$$ t = h + 1 $$
$$ \frac{1}{h} + \frac{1}{t} = \frac{5}{6} $$
And implicitly it gives $$ h \ne 0 \\ t \ne 0 $$
$$ \frac{5}{6} = \frac{t + h}{h t} \Rightarrow \\ \frac{5}{6} h t = h + t $$ The equations are not equivalent, because the second equation would also allow for $h = t = 0$.
If we now insert the equation of $t$ in terms of $h$ we get: $$ \frac{5}{6} h (h+1) = 2 h + 1 \iff \\ h^2 + h = \frac{12}{5} h + \frac{6}{5} \iff \\ h^2 - \frac{7}{5} h - \frac{6}{5} = 0 \iff \\ \left( h - \frac{7}{10} \right)^2 - \frac{7^2}{10^2} - \frac{12}{10} = 0 \iff \\ \left( h - \frac{7}{10} \right)^2 = \frac{169}{100} \iff \\ h = \frac{7}{10} \pm \sqrt{\frac{169}{100}} = \frac{7 \pm 13}{10} \in \{ -0.6, 2 \} $$ This means $h = -0.6 \in \mathbb{Q}$.