Question Number 197683 by sulaymonnorboyev140 last updated on 26/Sep/23
$$\frac{{a}+\mathrm{3}{b}}{{a}+{b}−\mathrm{1}}+\frac{{a}+\mathrm{3}{b}−\mathrm{1}}{{a}+{b}−\mathrm{3}}=\mathrm{4} \\ $$$${a}+{b}=? \\ $$
Commented by mr W last updated on 26/Sep/23
$${no}\:{unique}\:{solution}\:{for}\:{a}+{b}. \\ $$$${please}\:{check}\:{your}\:{question}! \\ $$
Answered by Frix last updated on 26/Sep/23
$$\mathrm{This}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\frac{\mathrm{2}\left({a}−\frac{\mathrm{11}}{\mathrm{4}}\right)^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{2}\left({b}+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} }{\mathrm{3}}=\mathrm{1}\wedge{b}\neq\mathrm{1}−{a}\wedge{b}\neq\mathrm{3}−{a} \\ $$$$\mathrm{Which}\:\mathrm{is}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{hyperbola}\:\mathrm{without} \\ $$$$\mathrm{the}\:\mathrm{points}\:\left(\frac{\mathrm{3}}{\mathrm{2}},\:−\frac{\mathrm{1}}{\mathrm{2}}\right)\:\mathrm{and}\:\left(\mathrm{4},−\mathrm{1}\right) \\ $$$${a}+{b}\in\mathbb{R}\backslash\left\{\mathrm{2}\right\} \\ $$
Answered by ajfour last updated on 26/Sep/23
$${a}+{b}={p}\:\:\:{a}−{b}={q} \\ $$$$\frac{\mathrm{2}{p}−{q}}{{p}−\mathrm{1}}+\frac{\mathrm{2}{p}−{q}−\mathrm{1}}{{p}−\mathrm{3}}=\mathrm{4} \\ $$$$\Rightarrow\:\:\left(\mathrm{2}{p}−{q}\right)\left({p}−\mathrm{3}\right)+\left(\mathrm{2}{p}−{q}−\mathrm{1}\right)\left({p}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{4}\left({p}−\mathrm{1}\right)\left({p}−\mathrm{3}\right) \\ $$$$\left(\mathrm{16}−\mathrm{6}−{q}−\mathrm{2}−{q}−\mathrm{1}\right){p} \\ $$$$\:\:\:\:\:\:=\mathrm{12}+\mathrm{3}{q}+{q}+\mathrm{1} \\ $$$$\Rightarrow\:\:\left(\mathrm{7}−\mathrm{2}{q}\right){p}=\mathrm{13}+\mathrm{4}{q} \\ $$$${p}={a}+{b}=\frac{\mathrm{13}+\mathrm{4}{q}}{\mathrm{7}−\mathrm{2}{q}}=\frac{\mathrm{27}−\left(\mathrm{14}−\mathrm{4}{q}\right)}{\mathrm{7}−\mathrm{2}{q}} \\ $$$${a}+{b}=\frac{\mathrm{27}}{\mathrm{7}−\mathrm{2}{q}}−\mathrm{2} \\ $$$$ \\ $$