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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

2x1x2=h6x1+3x2=3h(a)k3h(b)k=3h

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.

theeqn.systemis6x1+3x2=3h6x1+3x2=kifk3hthesystemwillhavenosolution.ifk=3hthesystemisinfactonlyoneeqn.6x1+3x2=kwhichinfinitesolutions.forbothcasethereareinfinitevaluesforhandk.

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

more explaination please

moreexplainationplease

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

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

wereaskedtofindthevalueofhandk

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

thanx i gat you

thanxigatyou

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