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Question Number 186627 by Mingma last updated on 07/Feb/23

Commented by Frix last updated on 07/Feb/23

$$\mathrm{As}\mathrm{always}:\mathrm{The}\mathrm{answer}\mathrm{is}42$$$$\bullet \mathrm{We}\mathrm{have}2\mathrm{equations}\mathrm{for}6\mathrm{unknowns}\Rightarrow$$$$\mathrm{We}\mathrm{can}\mathrm{choose}4\mathrm{of}\mathrm{them}\mathrm{and}\mathrm{solve}\mathrm{for}\mathrm{the}$$$$\mathrm{remaining}2.$$$$\bullet \mathrm{We}\mathrm{can}\mathrm{set}\frac{a+b+c}{x+y+z}=42\Rightarrow \mathrm{We}\mathrm{now}\mathrm{have}$$$$3\mathrm{equations}\mathrm{for}6\mathrm{unknowns}\Rightarrow \mathrm{We}\mathrm{can}$$$$\mathrm{choose}3\mathrm{of}\mathrm{them}\mathrm{and}\mathrm{solve}\mathrm{for}\mathrm{the}\mathrm{remaining}3$$

Answered by ajfour last updated on 07/Feb/23

$$z=qx,y=px$$$$x(a+bp+cq)=20$$$$y(\frac{a}{p}+b+\frac{cq}{p})=20$$$$z(\frac{a}{q}+\frac{bp}{q}+c)=20$$$$ax+by+cz=\mathrm{\Sigma }\frac{20a}{a+bp+cq}=20$$$$\Rightarrow a(\frac{a}{p}+b+\frac{cq}{p})(\frac{a}{q}+\frac{bp}{q}+c)$$$$+b(\frac{a}{q}+\frac{bp}{q}+c)(a+bp+cq)$$$$+c(a+bp+cq)(\frac{a}{p}+b+\frac{cq}{p})=1$$$$let?=\frac{1}{Q}$$$$\Rightarrow x+y+z=Q(a+b+c)$$$$\Rightarrow \frac{1}{(a+bp+cq)}+\frac{1}{(\frac{a}{p}+b+\frac{cq}{p})}$$$$+\frac{1}{(\frac{a}{q}+\frac{bp}{q}+c)}=\frac{Q(a+b+c)}{20}$$$$\Rightarrow \frac{1+p+q}{s}=\frac{Q(a+b+c)}{20}$$$$\mathrm{\&}\frac{{as}^2}{pq}+\frac{{bs}^2}{q}+\frac{{cs}^2}{p}=1$$$$\Rightarrow s^3=pq$$$$\Rightarrow 20(1+p+q)=Q(a+b+c){\left(pq\right)}^{1/3}$$$${(a+b+c)}^2=16-2(ab+bc+ca)$$$$Q=\frac{20(1+p+q)}{{\left(pq\right)}^{1/3}(a+b+c)}$$$$Q=\frac{20(1+\frac{y}{x}+\frac{z}{x})}{{\left(\frac{yz}{x^2}\right)}^{1/3}(a+b+c)}=\frac{20Q}{{\left(xyz\right)}^{1/3}}$$$$\Rightarrow {\left(xyz\right)}^{1/3}=20$$$$bysymmetry$$$$a^2=b^2=c^2=\frac{16}{3}$$$$x+y+z=20+20+20$$$$\frac{1}{?}=Q=\frac{x+y+z}{a+b+c}$$$$Q=\pm \frac{60}{3\left(\frac{4}{\sqrt{3}}\right)}=\pm 5\sqrt{3}$$$$?=\frac{1}{Q}=\pm \frac{1}{5\sqrt{3}}$$$$$$$$$$

Answered by ajfour last updated on 07/Feb/23

$$orsimplyifatallanswerisunique$$$$thenwecangetthesameasaspecial$$$$casetooa=b=c$$$$a(x+y+z)=20$$$$a^2=\frac{16}{3}$$$$a=\pm \frac{4}{\sqrt{3}}$$$$\frac{a+b+c}{x+y+z}=\frac{3a}{20/a}=\frac{3a^2}{20}=\frac{16}{20}=\frac{4}{5}$$

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