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Question Number 59201 by otchereabdullai@gmail.com last updated on 05/May/19

make x the subject of  x=m+x

$$\mathrm{make}\:\mathrm{x}\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}\:\:\mathrm{x}=\mathrm{m}+\mathrm{x} \\ $$

Commented by Forkum Michael Choungong last updated on 05/May/19

x−x=m  x(1−1)=m  ⇒ x= (m/((1−1)))    generally thats not true since  1−1=0   x= ∞

$${x}−{x}={m} \\ $$$${x}\left(\mathrm{1}−\mathrm{1}\right)={m} \\ $$$$\Rightarrow\:{x}=\:\frac{{m}}{\left(\mathrm{1}−\mathrm{1}\right)} \\ $$$$ \\ $$$${generally}\:{thats}\:{not}\:{true}\:{since} \\ $$$$\mathrm{1}−\mathrm{1}=\mathrm{0}\: \\ $$$${x}=\:\infty \\ $$

Commented by otchereabdullai@gmail.com last updated on 05/May/19

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

Commented by Senior Sun last updated on 06/May/19

what do you mean in your solution?

$${what}\:{do}\:{you}\:{mean}\:{in}\:{your}\:{solution}? \\ $$

Commented by Forkum Michael Choungong last updated on 06/May/19

i mean (m/0) is undefined

$${i}\:{mean}\:\frac{{m}}{\mathrm{0}}\:{is}\:{undefined} \\ $$

Answered by MJS last updated on 05/May/19

it′s impossible  x=m+x ⇒ ∀x: m=0

$$\mathrm{it}'\mathrm{s}\:\mathrm{impossible} \\ $$$${x}={m}+{x}\:\Rightarrow\:\forall{x}:\:{m}=\mathrm{0} \\ $$

Commented by otchereabdullai@gmail.com last updated on 05/May/19

thanks prof

$$\mathrm{thanks}\:\mathrm{prof}\: \\ $$

Commented by otchereabdullai@gmail.com last updated on 05/May/19

I thought there is a different approach  to such qustion

$$\mathrm{I}\:\mathrm{thought}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{different}\:\mathrm{approach} \\ $$$$\mathrm{to}\:\mathrm{such}\:\mathrm{qustion} \\ $$

Answered by malwaan last updated on 06/May/19

x−x=m  x(1−1)=m  x×0=m  there is no solution  solution set = ∅

$${x}−{x}={m} \\ $$$${x}\left(\mathrm{1}−\mathrm{1}\right)={m} \\ $$$${x}×\mathrm{0}={m} \\ $$$${there}\:{is}\:{no}\:{solution} \\ $$$${solution}\:{set}\:=\:\emptyset \\ $$

Commented by otchereabdullai@gmail.com last updated on 06/May/19

thanks

$$\mathrm{thanks} \\ $$

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