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Question-54659




Question Number 54659 by Gulay last updated on 08/Feb/19
Commented by Gulay last updated on 08/Feb/19
sir could you help me?
$$\mathrm{sir}\:\mathrm{could}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}? \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 08/Feb/19
(s) ⇔   { ((−2x+y =7)),((−kx +y =9)) :}  Δ_s = determinant (((−2         1)),((−k          1)))=k−2   so if k≠2   the sy stem have one solution  (x,y) with x =(Δ_x /Δ) =( determinant (((7      1)),((9       1)))/(k−2)) =((−2)/(k−2))  and y =( determinant (((−2      7)),((−k       9)))/(k−2)) =((7k−18)/(k−2))  if  k =2    we get   (s)    { ((−2x+y =7)),((−2x +y =9        and  the system haven t any solution .)) :}
$$\left({s}\right)\:\Leftrightarrow\:\:\begin{cases}{−\mathrm{2}{x}+{y}\:=\mathrm{7}}\\{−{kx}\:+{y}\:=\mathrm{9}}\end{cases} \\ $$$$\Delta_{{s}} =\begin{vmatrix}{−\mathrm{2}\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−{k}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{vmatrix}={k}−\mathrm{2}\:\:\:{so}\:{if}\:{k}\neq\mathrm{2}\:\:\:{the}\:{sy}\:{stem}\:{have}\:{one}\:{solution} \\ $$$$\left({x},{y}\right)\:{with}\:{x}\:=\frac{\Delta_{{x}} }{\Delta}\:=\frac{\begin{vmatrix}{\mathrm{7}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{9}\:\:\:\:\:\:\:\mathrm{1}}\end{vmatrix}}{{k}−\mathrm{2}}\:=\frac{−\mathrm{2}}{{k}−\mathrm{2}}\:\:{and}\:{y}\:=\frac{\begin{vmatrix}{−\mathrm{2}\:\:\:\:\:\:\mathrm{7}}\\{−{k}\:\:\:\:\:\:\:\mathrm{9}}\end{vmatrix}}{{k}−\mathrm{2}}\:=\frac{\mathrm{7}{k}−\mathrm{18}}{{k}−\mathrm{2}} \\ $$$${if}\:\:{k}\:=\mathrm{2}\:\:\:\:{we}\:{get}\:\:\:\left({s}\right)\:\:\:\begin{cases}{−\mathrm{2}{x}+{y}\:=\mathrm{7}}\\{−\mathrm{2}{x}\:+{y}\:=\mathrm{9}\:\:\:\:\:\:\:\:{and}\:\:{the}\:{system}\:{haven}\:{t}\:{any}\:{solution}\:.}\end{cases} \\ $$
Answered by peter frank last updated on 08/Feb/19
y−y−2x−(−kx)=7−9=−2  x(k−2)=−2  x=((−2)/(k−2))  y=((6k−14)/(k−2))
$${y}−{y}−\mathrm{2}{x}−\left(−{kx}\right)=\mathrm{7}−\mathrm{9}=−\mathrm{2} \\ $$$${x}\left({k}−\mathrm{2}\right)=−\mathrm{2} \\ $$$${x}=\frac{−\mathrm{2}}{{k}−\mathrm{2}} \\ $$$${y}=\frac{\mathrm{6}{k}−\mathrm{14}}{{k}−\mathrm{2}} \\ $$

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