Menu Close

Question-170568




Question Number 170568 by mathlove last updated on 27/May/22
Answered by Rasheed.Sindhi last updated on 27/May/22
 determinant (((f(2x+2)+2g(4x+7)=x−1_( f(x−1)+g(2x+1)=2x          ) )))  (1):    2x+2=y⇒x=((y−2)/2) :      f(y)+2g(4∙((y−2)/2)+7)=((y−2)/2)−1          ⇒f(y)+2g(2y+3)=((y−4)/2)...(3)  (2):    x−1=y⇒x=y+1      f(y)+g(2y+3)=2(y+1)...(4)  (3)−(4): g(2y+3)=((y−4)/2)−2y−2=((y−4−4y−4)/2)  g(2y+3)=((−3y−8)/2)  2y+3=x⇒y=((x−3)/2)  g(x)=((−3(((x−3)/2))−8)/2)=((−3x+9−16)/4)  g(x)=((−3x−7)/4)  (3): f(y)+2g(2y+3)=((y−4)/2)          f(x)+2g(2x+3)=((x−4)/2)         f(x)=((x−4)/2)−2g(2x+3)         f(x)=((x−4)/2)−2(((−3(2x+3)−7)/4))                =((x−4)/2)−((−6x−9−7)/2)                =((x−4)/2)+((6x+16)/2)               =((7x+12)/2)                       f(x)=((7x+12)/2) , g(x)=((−3x−7)/4)  Pl confirm the answers
$$\begin{array}{|c|}{\underset{\:\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{g}\left(\mathrm{2x}+\mathrm{1}\right)=\mathrm{2x}\:\:\:\:\:\:\:\:\:\:} {\mathrm{f}\left(\mathrm{2x}+\mathrm{2}\right)+\mathrm{2g}\left(\mathrm{4x}+\mathrm{7}\right)=\mathrm{x}−\mathrm{1}}}\\\hline\end{array} \\ $$$$\left(\mathrm{1}\right):\:\:\:\:\mathrm{2x}+\mathrm{2}=\mathrm{y}\Rightarrow\mathrm{x}=\frac{\mathrm{y}−\mathrm{2}}{\mathrm{2}}\:: \\ $$$$\:\:\:\:\mathrm{f}\left(\mathrm{y}\right)+\mathrm{2g}\left(\mathrm{4}\centerdot\frac{\mathrm{y}−\mathrm{2}}{\mathrm{2}}+\mathrm{7}\right)=\frac{\mathrm{y}−\mathrm{2}}{\mathrm{2}}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\mathrm{f}\left(\mathrm{y}\right)+\mathrm{2g}\left(\mathrm{2y}+\mathrm{3}\right)=\frac{\mathrm{y}−\mathrm{4}}{\mathrm{2}}…\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right):\:\:\:\:\mathrm{x}−\mathrm{1}=\mathrm{y}\Rightarrow\mathrm{x}=\mathrm{y}+\mathrm{1} \\ $$$$\:\:\:\:\mathrm{f}\left(\mathrm{y}\right)+\mathrm{g}\left(\mathrm{2y}+\mathrm{3}\right)=\mathrm{2}\left(\mathrm{y}+\mathrm{1}\right)…\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{3}\right)−\left(\mathrm{4}\right):\:\mathrm{g}\left(\mathrm{2y}+\mathrm{3}\right)=\frac{\mathrm{y}−\mathrm{4}}{\mathrm{2}}−\mathrm{2y}−\mathrm{2}=\frac{\mathrm{y}−\mathrm{4}−\mathrm{4y}−\mathrm{4}}{\mathrm{2}} \\ $$$$\mathrm{g}\left(\mathrm{2y}+\mathrm{3}\right)=\frac{−\mathrm{3y}−\mathrm{8}}{\mathrm{2}} \\ $$$$\mathrm{2y}+\mathrm{3}=\mathrm{x}\Rightarrow\mathrm{y}=\frac{\mathrm{x}−\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{−\mathrm{3}\left(\frac{\mathrm{x}−\mathrm{3}}{\mathrm{2}}\right)−\mathrm{8}}{\mathrm{2}}=\frac{−\mathrm{3x}+\mathrm{9}−\mathrm{16}}{\mathrm{4}} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{−\mathrm{3x}−\mathrm{7}}{\mathrm{4}} \\ $$$$\left(\mathrm{3}\right):\:\mathrm{f}\left(\mathrm{y}\right)+\mathrm{2g}\left(\mathrm{2y}+\mathrm{3}\right)=\frac{\mathrm{y}−\mathrm{4}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)+\mathrm{2g}\left(\mathrm{2x}+\mathrm{3}\right)=\frac{\mathrm{x}−\mathrm{4}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}−\mathrm{4}}{\mathrm{2}}−\mathrm{2g}\left(\mathrm{2x}+\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}−\mathrm{4}}{\mathrm{2}}−\mathrm{2}\left(\frac{−\mathrm{3}\left(\mathrm{2x}+\mathrm{3}\right)−\mathrm{7}}{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{x}−\mathrm{4}}{\mathrm{2}}−\frac{−\mathrm{6x}−\mathrm{9}−\mathrm{7}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{x}−\mathrm{4}}{\mathrm{2}}+\frac{\mathrm{6x}+\mathrm{16}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{7x}+\mathrm{12}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{7x}+\mathrm{12}}{\mathrm{2}}\:,\:\mathrm{g}\left(\mathrm{x}\right)=\frac{−\mathrm{3x}−\mathrm{7}}{\mathrm{4}} \\ $$$$\mathrm{Pl}\:\mathrm{confirm}\:\mathrm{the}\:\mathrm{answers} \\ $$
Commented by Rasheed.Sindhi last updated on 27/May/22
VERIFICATION:   determinant (((f(2x+2)+2g(4x+7)=x−1_( f(x−1)+g(2x+1)=2x          ) )))   •f(2x+2)+2g(4x+7)=x−1  ((7(2x+2)+12)/2)+2(((−3(4x+7)−7)/4))=x−1  ((14x+14+12)/2)+2(((−12x−21−7)/4))=x−1  ((14x+26)/2)+((−12x−28)/2)=x−1  7x+13−6x−14=x−1  x−1=x−1✓   •f(x−1)+g(2x+1)=2x  ((7(x−1)+12)/2)+((−3(2x+1)−7)/4)=2x  ((7x−7+12)/2)+((−6x−3−7)/4)=2x  ((7x+5)/2)+((−6x−10)/4)=2x  ((7x+5)/2)+((−3x−5)/2)=2x  ((4x)/2)=2x  2x=2x✓
$$\mathcal{VERIFICATION}: \\ $$$$\begin{array}{|c|}{\underset{\:\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{g}\left(\mathrm{2x}+\mathrm{1}\right)=\mathrm{2x}\:\:\:\:\:\:\:\:\:\:} {\mathrm{f}\left(\mathrm{2x}+\mathrm{2}\right)+\mathrm{2g}\left(\mathrm{4x}+\mathrm{7}\right)=\mathrm{x}−\mathrm{1}}}\\\hline\end{array}\: \\ $$$$\bullet\mathrm{f}\left(\mathrm{2x}+\mathrm{2}\right)+\mathrm{2g}\left(\mathrm{4x}+\mathrm{7}\right)=\mathrm{x}−\mathrm{1} \\ $$$$\frac{\mathrm{7}\left(\mathrm{2x}+\mathrm{2}\right)+\mathrm{12}}{\mathrm{2}}+\mathrm{2}\left(\frac{−\mathrm{3}\left(\mathrm{4x}+\mathrm{7}\right)−\mathrm{7}}{\mathrm{4}}\right)=\mathrm{x}−\mathrm{1} \\ $$$$\frac{\mathrm{14x}+\mathrm{14}+\mathrm{12}}{\mathrm{2}}+\mathrm{2}\left(\frac{−\mathrm{12x}−\mathrm{21}−\mathrm{7}}{\mathrm{4}}\right)=\mathrm{x}−\mathrm{1} \\ $$$$\frac{\mathrm{14x}+\mathrm{26}}{\mathrm{2}}+\frac{−\mathrm{12x}−\mathrm{28}}{\mathrm{2}}=\mathrm{x}−\mathrm{1} \\ $$$$\mathrm{7x}+\mathrm{13}−\mathrm{6x}−\mathrm{14}=\mathrm{x}−\mathrm{1} \\ $$$$\mathrm{x}−\mathrm{1}=\mathrm{x}−\mathrm{1}\checkmark \\ $$$$\:\bullet\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{g}\left(\mathrm{2x}+\mathrm{1}\right)=\mathrm{2x} \\ $$$$\frac{\mathrm{7}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{12}}{\mathrm{2}}+\frac{−\mathrm{3}\left(\mathrm{2x}+\mathrm{1}\right)−\mathrm{7}}{\mathrm{4}}=\mathrm{2x} \\ $$$$\frac{\mathrm{7x}−\mathrm{7}+\mathrm{12}}{\mathrm{2}}+\frac{−\mathrm{6x}−\mathrm{3}−\mathrm{7}}{\mathrm{4}}=\mathrm{2x} \\ $$$$\frac{\mathrm{7x}+\mathrm{5}}{\mathrm{2}}+\frac{−\mathrm{6x}−\mathrm{10}}{\mathrm{4}}=\mathrm{2x} \\ $$$$\frac{\mathrm{7x}+\mathrm{5}}{\mathrm{2}}+\frac{−\mathrm{3x}−\mathrm{5}}{\mathrm{2}}=\mathrm{2x} \\ $$$$\frac{\mathrm{4x}}{\mathrm{2}}=\mathrm{2x} \\ $$$$\mathrm{2x}=\mathrm{2x}\checkmark \\ $$
Commented by SLVR last updated on 31/May/22
dear sir   how y variable has  2 notation y=2x+2and  y=x−1 at one go...kindly  explain
$${dear}\:{sir}\:\:\:{how}\:{y}\:{variable}\:{has} \\ $$$$\mathrm{2}\:{notation}\:{y}=\mathrm{2}{x}+\mathrm{2}{and} \\ $$$${y}={x}−\mathrm{1}\:{at}\:{one}\:{go}…{kindly} \\ $$$${explain} \\ $$
Commented by Rasheed.Sindhi last updated on 31/May/22
sir mr W can better explain.  •f(2x+2)+2g(4x+7)=x−1  • f(x−1)+g(2x+1)=2x  The two equations can be considerd  separately with respect to x. Actually  x in one equation is nothing to do  with x in other equation.(They′re  simultaneous only with respect  to f and g only).Albeit all x in one equation  are related  to each other  and all x in other equation are related to   to each other.
$$\boldsymbol{{sir}}\:\boldsymbol{{mr}}\:\boldsymbol{{W}}\:{can}\:{better}\:{explain}. \\ $$$$\bullet\mathrm{f}\left(\mathrm{2x}+\mathrm{2}\right)+\mathrm{2g}\left(\mathrm{4x}+\mathrm{7}\right)=\mathrm{x}−\mathrm{1} \\ $$$$\bullet\:\mathrm{f}\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{g}\left(\mathrm{2x}+\mathrm{1}\right)=\mathrm{2x} \\ $$$$\mathcal{T}{he}\:{two}\:{equations}\:{can}\:{be}\:{considerd} \\ $$$${separately}\:{with}\:{respect}\:{to}\:\mathrm{x}.\:{Actually} \\ $$$$\mathrm{x}\:{in}\:{one}\:{equation}\:{is}\:{nothing}\:{to}\:{do} \\ $$$${with}\:\mathrm{x}\:{in}\:{other}\:{equation}.\left(\mathcal{T}{hey}'{re}\right. \\ $$$${simultaneous}\:{only}\:{with}\:{respect} \\ $$$$\left.{to}\:\mathrm{f}\:{and}\:\mathrm{g}\:{only}\right).{Albeit}\:{all}\:{x}\:{in}\:{one}\:{equation} \\ $$$${are}\:{related}\:\:{to}\:{each}\:{other} \\ $$$${and}\:{all}\:{x}\:{in}\:{other}\:{equation}\:{are}\:{related}\:{to}\: \\ $$$${to}\:{each}\:{other}. \\ $$
Commented by Rasheed.Sindhi last updated on 31/May/22
These equatios are separate &   simultaneous simultaneously! :)
$$\mathrm{These}\:\mathrm{equatios}\:\mathrm{are}\:\boldsymbol{\mathrm{separate}}\:\&\: \\ $$$$\left.\boldsymbol{\mathrm{simultaneous}}\:\mathrm{simultaneously}!\::\right) \\ $$
Commented by SLVR last updated on 31/May/22
understood well  sir..So kind
$${understood}\:{well}\:\:{sir}..{So}\:{kind} \\ $$
Commented by Rasheed.Sindhi last updated on 31/May/22
You′re welcome sir!
$${You}'{re}\:{welcome}\:{sir}! \\ $$
Commented by mathlove last updated on 02/Jun/22
thanks sir
$${thanks}\:{sir} \\ $$
Commented by mathlove last updated on 02/Jun/22
thank sir
$${thank}\:{sir} \\ $$
Commented by Tawa11 last updated on 08/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *