Question Number 179284 by Agnibhoo98 last updated on 27/Oct/22
![Solve for x (a/(ax − 1)) + (b/(bx − 1)) = a + b [x ≠ (1/a), (1/b)]](https://www.tinkutara.com/question/Q179284.png)
$$\mathrm{Solve}\:\mathrm{for}\:\boldsymbol{{x}} \\ $$$$\frac{{a}}{{ax}\:−\:\mathrm{1}}\:+\:\frac{{b}}{{bx}\:−\:\mathrm{1}}\:=\:{a}\:+\:{b}\:\left[{x}\:\neq\:\frac{\mathrm{1}}{{a}},\:\frac{\mathrm{1}}{{b}}\right] \\ $$
Answered by FelipeLz last updated on 27/Oct/22
![((a(bx−1)+b(ax−1))/((ax−1)(bx−1))) = a+b ((2abx−(a+b))/(abx^2 −(a+b)x+1)) = a+b { ((a+b = s)),((ab = p)) :} ((2px−s)/(px^2 −sx+1)) = s psx^2 −(2p+s^2 )x+2s = 0 ⇒ x = ((2p+s^2 ±(√([−(2p+s^2 )]^2 −8ps^2 )))/(2ps)) x = ((2p+s^2 ±(√(4p^2 −4ps^2 +s^4 )))/(2ps)) x = ((2p+s^2 ±(2p−s^2 ))/(2ps)) { ((x_1 = ((2p+s^2 +2p−s^2 )/(2ps)) = ((4p)/(2ps)) = (2/s) = (2/(a+b)))),((x_2 = ((2p+s^2 −2p+s^2 )/(2ps)) = ((2s^2 )/(2ps)) = (s/p) = ((a+b)/(ab)))) :}](https://www.tinkutara.com/question/Q179293.png)
$$\frac{{a}\left({bx}−\mathrm{1}\right)+{b}\left({ax}−\mathrm{1}\right)}{\left({ax}−\mathrm{1}\right)\left({bx}−\mathrm{1}\right)}\:=\:{a}+{b} \\ $$$$\frac{\mathrm{2}{abx}−\left({a}+{b}\right)}{{abx}^{\mathrm{2}} −\left({a}+{b}\right){x}+\mathrm{1}}\:=\:{a}+{b} \\ $$$$\begin{cases}{{a}+{b}\:=\:{s}}\\{{ab}\:=\:{p}}\end{cases} \\ $$$$\frac{\mathrm{2}{px}−{s}}{{px}^{\mathrm{2}} −{sx}+\mathrm{1}}\:=\:{s} \\ $$$${psx}^{\mathrm{2}} −\left(\mathrm{2}{p}+{s}^{\mathrm{2}} \right){x}+\mathrm{2}{s}\:=\:\mathrm{0}\:\Rightarrow\:{x}\:=\:\frac{\mathrm{2}{p}+{s}^{\mathrm{2}} \pm\sqrt{\left[−\left(\mathrm{2}{p}+{s}^{\mathrm{2}} \right)\right]^{\mathrm{2}} −\mathrm{8}{ps}^{\mathrm{2}} }}{\mathrm{2}{ps}} \\ $$$${x}\:=\:\frac{\mathrm{2}{p}+{s}^{\mathrm{2}} \pm\sqrt{\mathrm{4}{p}^{\mathrm{2}} −\mathrm{4}{ps}^{\mathrm{2}} +{s}^{\mathrm{4}} }}{\mathrm{2}{ps}} \\ $$$${x}\:=\:\frac{\mathrm{2}{p}+{s}^{\mathrm{2}} \pm\left(\mathrm{2}{p}−{s}^{\mathrm{2}} \right)}{\mathrm{2}{ps}} \\ $$$$\begin{cases}{{x}_{\mathrm{1}} \:=\:\frac{\mathrm{2}{p}+{s}^{\mathrm{2}} +\mathrm{2}{p}−{s}^{\mathrm{2}} }{\mathrm{2}{ps}}\:=\:\frac{\mathrm{4}{p}}{\mathrm{2}{ps}}\:=\:\frac{\mathrm{2}}{{s}}\:=\:\frac{\mathrm{2}}{{a}+{b}}}\\{{x}_{\mathrm{2}} \:=\:\frac{\mathrm{2}{p}+{s}^{\mathrm{2}} −\mathrm{2}{p}+{s}^{\mathrm{2}} }{\mathrm{2}{ps}}\:=\:\frac{\mathrm{2}{s}^{\mathrm{2}} }{\mathrm{2}{ps}}\:=\:\frac{{s}}{{p}}\:=\:\frac{{a}+{b}}{{ab}}}\end{cases} \\ $$
Answered by som(math1967) last updated on 28/Oct/22
$$\:\frac{{a}}{{ax}−\mathrm{1}}\:−{b}={a}−\frac{{b}}{{bx}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{a}+{b}−{abx}}{{ax}−\mathrm{1}}=\frac{{abx}−{a}−{b}}{{bx}−\mathrm{1}} \\ $$$$\Rightarrow\frac{\left({a}+{b}−{abx}\right)}{{ax}−\mathrm{1}}\:+\frac{{a}+{b}−{abx}}{{bx}−\mathrm{1}}=\mathrm{0} \\ $$$$\Rightarrow\left({a}+{b}−{abx}\right)\left(\frac{\mathrm{1}}{{ax}−\mathrm{1}}+\frac{\mathrm{1}}{{bx}−\mathrm{1}}\right)=\mathrm{0} \\ $$$${either}\:\left({a}+{b}−{abx}\right)=\mathrm{0} \\ $$$$\therefore\:{x}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}} \\ $$$${or}\:\frac{\mathrm{1}}{{ax}−\mathrm{1}}+\frac{\mathrm{1}}{{bx}−\mathrm{1}}=\mathrm{0} \\ $$$$\Rightarrow\:{bx}−\mathrm{1}+{ax}−\mathrm{1}=\mathrm{0} \\ $$$$\:\therefore{x}=\frac{\mathrm{2}}{{a}+{b}} \\ $$