Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com