Ask your own question, for FREE!
Mathematics 9 Online
OpenStudy (anonymous):

simultaneous method

OpenStudy (anonymous):

i have FOUR equation 1=C1+C2 1=C2+C4 3e=2C3+C4e e=C4e

OpenStudy (wolf1728):

Maybe we could manipulate those equations a bit. 1=C1+C2 1=C2+C4 Therefore, C1 + C2 = C2 + C4 and so C1 = C4

OpenStudy (wolf1728):

I guess you also have to solve for e?

OpenStudy (wolf1728):

That reduces the equations to 3 1=C1+C2 3e=2C3+C1e e=C1e

OpenStudy (anonymous):

no for constants

OpenStudy (wolf1728):

You mean you have to solve for C1 C2 and C3

OpenStudy (anonymous):

yes

OpenStudy (wolf1728):

according to the equation e=C1*e That means C1 must equal 1 right?

OpenStudy (anonymous):

c4 =1

OpenStudy (wolf1728):

and since C1 = C4 then?

OpenStudy (anonymous):

CI , C2, C3

OpenStudy (wolf1728):

If C1 = 1 then from this equation 1=C1+C2 C2 would equal zero

OpenStudy (anonymous):

C1=0

OpenStudy (anonymous):

C2=0

OpenStudy (wolf1728):

No C1 = 1

OpenStudy (anonymous):

C3=e

OpenStudy (wolf1728):

3e=2C3+C4e 2C3 = 4e C3 = 2*e

OpenStudy (wolf1728):

None of this used the traditional methods for solving simultaneous equations.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!