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Question Number 75606 by liki last updated on 13/Dec/19

Commented by mathmax by abdo last updated on 15/Dec/19

{ ((x−y=3)),((2x−2y =k we have Δ_s = determinant (((1 −1)),((2 −2))))) :}=−2+2=0 Δ_x = determinant (((3 −1)),((k −2)))=−6+k Δ_y = determinant (((1 3)),((2 k)))=k−6 ifk=6 infinity os solutions if k≠6 no solution. k=6 ⇒S ={(x,3+x)/x ∈R}

$$\begin{cases}{{x}−{y}=\mathrm{3}}\\{\mathrm{2}{x}−\mathrm{2}{y}\:={k}\:\:\:\:\:{we}\:{have}\:\Delta_{{s}} =\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:−\mathrm{2}}\end{vmatrix}}\end{cases}=−\mathrm{2}+\mathrm{2}=\mathrm{0} \\ $$$$\Delta_{{x}} =\begin{vmatrix}{\mathrm{3}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{{k}\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}}\end{vmatrix}=−\mathrm{6}+{k} \\ $$$$\Delta_{{y}} =\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:{k}}\end{vmatrix}={k}−\mathrm{6} \\ $$$${ifk}=\mathrm{6}\:\:\:\:{infinity}\:{os}\:{solutions} \\ $$$${if}\:{k}\neq\mathrm{6}\:\:\:\:{no}\:{solution}. \\ $$$${k}=\mathrm{6}\:\Rightarrow{S}\:=\left\{\left({x},\mathrm{3}+{x}\right)/{x}\:\in{R}\right\} \\ $$

Answered by vishalbhardwaj last updated on 13/Dec/19

(a) for no solution : (1/2) = ((−1)/(−2)) ≠ (3/k) ⇒ k≠6 ⇒ k ∈ R−{6} (b) unique solution does not exist (c) for infinitely many solution k = 6

$$\left(\mathrm{a}\right)\:\mathrm{for}\:\mathrm{no}\:\mathrm{solution}\:: \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{−\mathrm{1}}{−\mathrm{2}}\:\neq\:\frac{\mathrm{3}}{\mathrm{k}}\:\Rightarrow\:\mathrm{k}\neq\mathrm{6}\:\Rightarrow \\ $$$$\mathrm{k}\:\in\:\mathrm{R}−\left\{\mathrm{6}\right\} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{unique}\:\mathrm{solution}\:\mathrm{does}\:\mathrm{not} \\ $$$$\mathrm{exist} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{for}\:\mathrm{infinitely}\:\mathrm{many}\:\mathrm{solution} \\ $$$$\:\mathrm{k}\:=\:\mathrm{6} \\ $$

Commented by liki last updated on 13/Dec/19

... sory sir , can you explain how to get those answers because yet not get you. ...

$$...\:{sory}\:{sir}\:,\:\boldsymbol{{can}}\:\boldsymbol{{you}}\:\boldsymbol{{explain}}\:\boldsymbol{{how}} \\ $$$$\:\:\boldsymbol{{to}}\:\boldsymbol{{get}}\:\boldsymbol{{those}}\:\boldsymbol{{answers}}\:\:\boldsymbol{{because}}\:\boldsymbol{{yet}}\:\boldsymbol{{not}}\:\boldsymbol{{get}}\:\boldsymbol{{you}}. \\ $$$$\:\:\:...\: \\ $$

Commented by vishalbhardwaj last updated on 13/Dec/19

we have conditions regarding this like : Let we have lines a_1 x+b_1 y+c_1 =0 and a_2 x+b_2 y+c_2 =0 then (i) If (a_1 /a_2 )≠(b_1 /b_2 ) then lines will have unique solution (ii) If (a_1 /a_2 )=(b_1 /b_2 )≠(c_1 /c_2 ) then no solution (iii) If (a_1 /a_2 )=(b_1 /b_2 )=(c_1 /c_2 ) then infinitely many solution

$$\mathrm{we}\:\mathrm{have}\:\mathrm{conditions}\:\mathrm{regarding} \\ $$$$\mathrm{this}\:\mathrm{like}\::\: \\ $$$$\:\mathrm{Let}\:\mathrm{we}\:\mathrm{have}\:\mathrm{lines}\:{a}_{\mathrm{1}} {x}+{b}_{\mathrm{1}} {y}+{c}_{\mathrm{1}} =\mathrm{0} \\ $$$$\mathrm{and}\:{a}_{\mathrm{2}} {x}+{b}_{\mathrm{2}} {y}+{c}_{\mathrm{2}} =\mathrm{0}\:\mathrm{then} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{If}\:\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{2}} }\neq\frac{{b}_{\mathrm{1}} }{{b}_{\mathrm{2}} }\:{then}\:{lines}\:{will} \\ $$$${have}\:{unique}\:{solution} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{If}\:\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{2}} }=\frac{{b}_{\mathrm{1}} }{{b}_{\mathrm{2}} }\neq\frac{{c}_{\mathrm{1}} }{{c}_{\mathrm{2}} }\:{then}\: \\ $$$${no}\:{solution} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{If}\:\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{2}} }=\frac{{b}_{\mathrm{1}} }{{b}_{\mathrm{2}} }=\frac{{c}_{\mathrm{1}} }{{c}_{\mathrm{2}} }\:{then}\: \\ $$$$\mathrm{infinitely}\:\mathrm{many}\:\mathrm{solution} \\ $$$$ \\ $$

Commented by liki last updated on 13/Dec/19

...thank so much sir ,

$$...{thank}\:{so}\:{much}\:{sir}\:,\: \\ $$$$ \\ $$

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