Question Number 68591 by TawaTawa last updated on 14/Sep/19
![](https://www.tinkutara.com/question/9274.png)
Commented by kaivan.ahmadi last updated on 14/Sep/19
![102+2p+3q=0 17+3p−4q=0 ⇒ { ((2p+3q=−102)),((3p−4q=−17)) :}⇒ { ((−6p−9q=306)),((6p−8q=−34)) :}⇒ −17q=272⇒q=−16 and 2p−48=−102⇒2p=−54⇒p=−27 ⇒(a) is answer](https://www.tinkutara.com/question/Q68599.png)
$$\mathrm{102}+\mathrm{2}{p}+\mathrm{3}{q}=\mathrm{0} \\ $$$$\mathrm{17}+\mathrm{3}{p}−\mathrm{4}{q}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\begin{cases}{\mathrm{2}{p}+\mathrm{3}{q}=−\mathrm{102}}\\{\mathrm{3}{p}−\mathrm{4}{q}=−\mathrm{17}}\end{cases}\Rightarrow\begin{cases}{−\mathrm{6}{p}−\mathrm{9}{q}=\mathrm{306}}\\{\mathrm{6}{p}−\mathrm{8}{q}=−\mathrm{34}}\end{cases}\Rightarrow \\ $$$$−\mathrm{17}{q}=\mathrm{272}\Rightarrow{q}=−\mathrm{16} \\ $$$${and} \\ $$$$\mathrm{2}{p}−\mathrm{48}=−\mathrm{102}\Rightarrow\mathrm{2}{p}=−\mathrm{54}\Rightarrow{p}=−\mathrm{27} \\ $$$$\Rightarrow\left({a}\right)\:{is}\:{answer} \\ $$
Commented by TawaTawa last updated on 14/Sep/19
![God bless you sir](https://www.tinkutara.com/question/Q68604.png)
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by TawaTawa last updated on 14/Sep/19
![How did you get 102 and 17 sir](https://www.tinkutara.com/question/Q68605.png)
$$\mathrm{How}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{102}\:\mathrm{and}\:\mathrm{17}\:\mathrm{sir} \\ $$
Commented by TawaTawa last updated on 14/Sep/19
![I mean how did he get equation (i) and (ii) from the question](https://www.tinkutara.com/question/Q68607.png)
$$\mathrm{I}\:\mathrm{mean}\:\mathrm{how}\:\mathrm{did}\:\mathrm{he}\:\mathrm{get}\:\mathrm{equation}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right)\:\mathrm{from}\:\mathrm{the}\:\mathrm{question} \\ $$
Commented by kaivan.ahmadi last updated on 14/Sep/19
![read inner product of vectors.](https://www.tinkutara.com/question/Q68638.png)
$${read}\:{inner}\:{product}\:{of}\:{vectors}. \\ $$
Commented by kaivan.ahmadi last updated on 14/Sep/19
![6×17+2×p+3×q=0 1×17+3×p+(−4)×q=0](https://www.tinkutara.com/question/Q68639.png)
$$\mathrm{6}×\mathrm{17}+\mathrm{2}×{p}+\mathrm{3}×{q}=\mathrm{0} \\ $$$$\mathrm{1}×\mathrm{17}+\mathrm{3}×{p}+\left(−\mathrm{4}\right)×{q}=\mathrm{0} \\ $$