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Question Number 30711 by mondodotto@gmail.com last updated on 24/Feb/18

Answered by mrW2 last updated on 24/Feb/18

2x_1 −x_2 =h  ⇔−6x_1 +3x_2 =−3h  (a)  k≠−3h  (b)  k=−3h

$$\mathrm{2}{x}_{\mathrm{1}} −{x}_{\mathrm{2}} ={h} \\ $$$$\Leftrightarrow−\mathrm{6}{x}_{\mathrm{1}} +\mathrm{3}{x}_{\mathrm{2}} =−\mathrm{3}{h} \\ $$$$\left({a}\right) \\ $$$${k}\neq−\mathrm{3}{h} \\ $$$$\left({b}\right) \\ $$$${k}=−\mathrm{3}{h} \\ $$

Commented by mrW2 last updated on 25/Feb/18

the eqn. system is  −6x_1 +3x_2 =−3h  −6x_1 +3x_2 =k  if k≠−3h the system will have no  solution.  if k=−3h the system is in fact only  one eqn. −6x_1 +3x_2 =k which infinite  solutions.  for both case there are infinite values  for h and k.

$${the}\:{eqn}.\:{system}\:{is} \\ $$$$−\mathrm{6}{x}_{\mathrm{1}} +\mathrm{3}{x}_{\mathrm{2}} =−\mathrm{3}{h} \\ $$$$−\mathrm{6}{x}_{\mathrm{1}} +\mathrm{3}{x}_{\mathrm{2}} ={k} \\ $$$${if}\:{k}\neq−\mathrm{3}{h}\:{the}\:{system}\:{will}\:{have}\:{no} \\ $$$${solution}. \\ $$$${if}\:{k}=−\mathrm{3}{h}\:{the}\:{system}\:{is}\:{in}\:{fact}\:{only} \\ $$$${one}\:{eqn}.\:−\mathrm{6}{x}_{\mathrm{1}} +\mathrm{3}{x}_{\mathrm{2}} ={k}\:{which}\:{infinite} \\ $$$${solutions}. \\ $$$${for}\:{both}\:{case}\:{there}\:{are}\:{infinite}\:{values} \\ $$$${for}\:{h}\:{and}\:{k}. \\ $$

Commented by mondodotto@gmail.com last updated on 25/Feb/18

more explaination please

$$\mathrm{more}\:\mathrm{explaination}\:\mathrm{please} \\ $$

Commented by mondodotto@gmail.com last updated on 25/Feb/18

we′re asked to find  the value of h and k

$$\mathrm{we}'\mathrm{re}\:\mathrm{asked}\:\mathrm{to}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{h}\:\mathrm{and}\:\mathrm{k} \\ $$

Commented by mondodotto@gmail.com last updated on 25/Feb/18

thanx i gat you

$$\mathrm{thanx}\:\mathrm{i}\:\mathrm{gat}\:\mathrm{you} \\ $$

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