Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 212598 by MATHEMATICSAM last updated on 18/Oct/24

$$\mathrm{Help}\mathrm{me}\mathrm{to}\mathrm{solve}\mathrm{pls}$$$$\mathrm{Q}212576$$

Answered by A5T last updated on 18/Oct/24

$$Lemma:\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a}{b}=\frac{c}{d}=\frac{a\underset{-}{+}c}{b\underset{-}{+}d}$$$$Proof:Let\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk\wedge c=dk$$$$\Rightarrow a\underset{-}{+}c=\left(b\underset{-}{+}d\right)k\Rightarrow \frac{a\underset{-}{+}c}{b\underset{-}{+}d}=k=\frac{a}{c}=\frac{b}{d}◻$$$$\Rightarrow \frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}=$$$$\frac{(x-y)(x+y+z)}{(a-b)(a+b+c)}=\frac{(y-z)(x+y+z)}{(b-c)(a+b+c)}=\frac{(z-x)(x+y+z)}{(c-a)(a+b+c)}$$$$\Rightarrow \frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}\Rightarrow ...$$

Commented by MATHEMATICSAM last updated on 18/Oct/24

$$\mathrm{I}\mathrm{want}\mathrm{to}\mathrm{know}\mathrm{that}\mathrm{is}\mathrm{there}\mathrm{any}\mathrm{process}$$$$\mathrm{to}\mathrm{get}\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\mathrm{from}$$$$\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}$$$$\mathrm{I}\mathrm{want}\mathrm{to}\mathrm{reversely}\mathrm{prove}\mathrm{from}$$$$\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}\mathrm{to}\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$$$$\mathrm{By}\mathrm{holding}$$$$\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}=k\mathrm{or}\mathrm{something}$$

Commented by A5T last updated on 18/Oct/24

$$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\Rightarrow \frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}butthe$$$$converseisnotnecessarilytrue.$$$$Forexample:(x,y,z)=(5,3,1);(a,b,c)=(3,2,1)$$$$\frac{5-3}{3-2}=\frac{3-1}{2-1}=\frac{1-5}{1-3}=2but\frac{5}{3}\ne \frac{3}{2}\ne \frac{1}{1}.$$$$So,we{can}^{\prime }t‘‘{reverse}^{″}it.$$$$\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}\nRightarrow \frac{x}{a}=\frac{y}{b}=\frac{z}{c}$$

Commented by Rasheed.Sindhi last updated on 19/Oct/24

$$\frac{(x-y)(x+y+z)}{(a-b)(a+b+c)}=\frac{(y-z)(x+y+z)}{(b-c)(a+b+c)}=\frac{(z-x)(x+y+z)}{(c-a)(a+b+c)}$$$$\stackrel{why}{\Rightarrow }\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}$$

Commented by MATHEMATICSAM last updated on 19/Oct/24

$$\mathrm{He}\mathrm{cancelled}\frac{x+y+z}{a+b+c}\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{equal}$$$$\mathrm{to}0.$$

Commented by Rasheed.Sindhi last updated on 19/Oct/24

$$Ok,thanks!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com