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# System of equations watch

1. If a simultaneous system of equations can be solved using the elimination method, should you also be able to use substitution?

I have a question which involves finding solutions for the following:

A: 2y(y+3)=9x
B: y(y-2) =3x

Y could be found by dividing a/b, eliminating x.

Y = 12

Substituting this back into A, x is found to be 40.

I can't find any solutions when using the substitution method. Is it just an error on my part?
2. Doy ou mean solving 1 of the equations for x or y then subbing that into the 2nd? If so it should work fine. Post what you did so we can see what went wrong.
3. (Original post by supreme)
If a simultaneous system of equations can be solved using the elimination method, should you also be able to use substitution?

I have a question which involves finding solutions for the following:

A: 2y(y+3)=9x
B: y(y-2) =3x

Y could be found by dividing a/b, eliminating x.

Y = 12

Substituting this back into A, x is found to be 40.

I can't find any solutions when using the substitution method. Is it just an error on my part?
You have lost solutions through the process of division. What is to say x cannot be 0? This should be checked before dividing.

x=y=0 is also a solution.

From scratch using substitution:

Note that the 2nd equation can be multiplied through by 3 so that we have 9x expressed in 2 different ways. So you can equate 3y(y-2) with 2y(y+3). So we get (y^2)-12y=y(y-12)=0. We derive two possible solutions. Both of these should be checked to ensure they do satisfy the original equations, as the argument up until this point has been in one direction only.

Always be very careful when dividing, otherwise crazy things can occur!
4. (Original post by Magu1re)
You have lost solutions through the process of division. What is to say x cannot be 0? This should be checked before dividing.

x=y=0 is also a solution.

From scratch using substitution:

Note that the 2nd equation can be multiplied through by 3 so that we have 9x expressed in 2 different ways. So you can equate 3y(y-2) with 2y(y+3). So we get (y^2)-12y=y(y-12)=0. So we derive to possible solutions. Both of these should be checked to ensure they do satisfy the original equations, as the argument up until this point has been in one direction only.

Always be very careful when dividing, otherwise crazy things can occur!
Thanks

The website I got the question from and the solution was misleading. I wanted to explore and further consider how it could be answered.

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Updated: April 7, 2013
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