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Question Number 12265 by frank ntulah last updated on 17/Apr/17
Determinant method can be used to solve  the system below?, if yes solve by determinant method and   if no solve by another method             x+y−z=8  2x+y−2z=3  (give clear reason for your answer)
$${Determinant}\:{method}\:{can}\:{be}\:{used}\:{to}\:{solve} \\ $$$${the}\:{system}\:{below}?,\:\mathrm{if}\:\mathrm{yes}\:\mathrm{solve}\:\mathrm{by}\:\mathrm{determinant}\:\mathrm{method}\:\mathrm{and} \\ $$$$\:\mathrm{if}\:\mathrm{no}\:\mathrm{solve}\:\mathrm{by}\:\mathrm{another}\:\mathrm{method} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$${x}+{y}−{z}=\mathrm{8} \\ $$$$\mathrm{2}{x}+{y}−\mathrm{2}{z}=\mathrm{3} \\ $$$$\left({give}\:{clear}\:{reason}\:{for}\:{your}\:{answer}\right) \\ $$
Commented by mrW1 last updated on 17/Apr/17
let u=x−z, the equation system can be  transformed into  u+y=8  2u+y=3  which can be solved by determinant  method or any other method.  y=13  u=−5  ⇒x=any real number  ⇒y=13  ⇒z=x−u=x+5
$${let}\:{u}={x}−{z},\:{the}\:{equation}\:{system}\:{can}\:{be} \\ $$$${transformed}\:{into} \\ $$$${u}+{y}=\mathrm{8} \\ $$$$\mathrm{2}{u}+{y}=\mathrm{3} \\ $$$${which}\:{can}\:{be}\:{solved}\:{by}\:{determinant} \\ $$$${method}\:{or}\:{any}\:{other}\:{method}. \\ $$$${y}=\mathrm{13} \\ $$$${u}=−\mathrm{5} \\ $$$$\Rightarrow{x}={any}\:{real}\:{number} \\ $$$$\Rightarrow{y}=\mathrm{13} \\ $$$$\Rightarrow{z}={x}−{u}={x}+\mathrm{5} \\ $$
Commented by frank ntulah last updated on 17/Apr/17
thanks, God bless you sir
$$\mathrm{thanks},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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