Thread: System of Equations

1. System of Equations

I cant finish this problem but have to solve it today!!!
Solve for all real and b equations

My progress:
to make this simple the top is (1) the midle is (2) and the bottom is (3)
i add (2) with (1) to get 3x+2az=2a
then i add (2) and (3) and get
6x + az + 3z = a + b
now im lost and cant solve it plz i need fast help :/

2. Re: Need fast help!

Originally Posted by Petrus
I cant finish this problem but have to solve it today!!!
Solve for all real and b equations

My progress:
to make this simple the top is (1) the midle is (2) and the bottom is (3)
i add (2) with (1) to get 3x+2az=2a
then i add (2) and (3) and get
6x + az + 3z = a + b
now im lost and cant solve it plz i need fast help :/
Calculate 2(1) (1) is the equation number.
Then do 2(1) - (2) = 0x + 3y + a = 0

Notice that the result does not have an x in it. Now do the same thing for the 3rd equation:
4(1) - (3)
Again the x term is missing. Can you finish this process?

And please don't beg. It's rude.

-Dan

3. Re: Need fast help!

Originally Posted by topsquark
Calculate 2(1) (1) is the equation number.
Then do 2(1) - (2) = 0x + 3y + a = 0

Notice that the result does not have an x in it. Now do the same thing for the 3rd equation:
4(1) - (3)
Again the x term is missing. Can you finish this process?

And please don't beg. It's rude.

-Dan
i got confused on this Calculate 2(1) (1) is the equation number. do u mean multiplicate 3x+2az=2a with x+y+az?
sorry but got confused on ur post, can u write the equation insted of 2(1)im so panic right now have to solve this fast :S dead line is close :/ insted of

4. Re: System of Equations

Originally Posted by Petrus
I cant finish this problem but have to solve it today!!!
Solve for all real and b equations
I would use a matrix inverse.
$\displaystyle \left[ {\begin{array}{rrr} 1 & 1 & a \\ 2 & { - 1} & a \\ 4 & 1 & 3 \\ \end{array} } \right]^{ - 1} = \frac{1}{{9\left( {a - 1} \right)}}\left[ {\begin{array}{rrr} { - (a + 3)} & {(a - 3)} & {2a} \\ {2(2a - 3)} & { - (4a - 3)} & a \\ 6 & 3 & { - 3} \\ \end{array} } \right]$

5. Re: System of Equations

Originally Posted by Plato
I would use a matrix inverse.
$\displaystyle \left[ {\begin{array}{rrr} 1 & 1 & a \\ 2 & { - 1} & a \\ 4 & 1 & 3 \\ \end{array} } \right]^{ - 1} = \frac{1}{{9\left( {a - 1} \right)}}\left[ {\begin{array}{rrr} { - (a + 3)} & {(a - 3)} & {2a} \\ {2(2a - 3)} & { - (4a - 3)} & a \\ 6 & 3 & { - 3} \\ \end{array} } \right]$
i wanna solve this without matrix

6. Re: System of Equations

Originally Posted by Petrus
i wanna solve this without matrix
That is fine. It harder without it.

7. Re: System of Equations

Originally Posted by Plato
That is fine. It harder without it.
yeah but im not "good" in matrix i need to sit and watch and read alot about matrix but i wanna solve this so i understand can u help me :/?

8. Re: System of Equations

The first equation is x+ y+ az= a and the second equation is 2x- y+ az= a. Subtracting the second equation from the first, -x+ 2y= 0 or x= 2y. Replacing x with 2y in the equations gives 2y+ y+ az= 3y+ az= a, 2x- y+ az= 4y- y+ az or 3y+ az=a again, and 4x+ 3y+ 3z= 8y+ 3y+ 3z= 11y+ 3z= b. If you multiply 3y+ az= a by 11, you get 33y+ 11az= 11a and if you multiply 11y+ 3z= b by 3, you get 33y+ 9z= 3b.

So now you have 33y+ 11az= 11a and 33y+ 9z= 3b. If you subtract the second of those from the first, what do you get?

9. Re: System of Equations

Originally Posted by HallsofIvy
The first equation is x+ y+ az= a and the second equation is 2x- y+ az= a. Subtracting the second equation from the first, -x+ 2y= 0 or x= 2y. Replacing x with 2y in the equations gives 2y+ y+ az= 3y+ az= a, 2x- y+ az= 4y- y+ az or 3y+ az=a again, and 4x+ 3y+ 3z= 8y+ 3y+ 3z= 11y+ 3z= b. If you multiply 3y+ az= a by 11, you get 33y+ 11az= 11a and if you multiply 11y+ 3z= b by 3, you get 33y+ 9z= 3b.

So now you have 33y+ 11az= 11a and 33y+ 9z= 3b. If you subtract the second of those from the first, what do you get?
9z-11az=3b-11a

10. Re: System of Equations

Originally Posted by Petrus
9z-11az=3b-11a
$\displaystyle z(9-11a) = 3b-11a$

continue ...

11. Re: System of Equations

z = (3b-11a)/(9-11a) what shall i do next =S ?

12. Re: System of Equations

Originally Posted by Petrus
z = (3b-11a)/(9-11a) what shall i do next =S ?
solve for x and y ...

13. Re: System of Equations

y = (a*b-3a)/(11a-9)
x= (2a(b-3))/(11a-9)

is this correct?

14. Re: System of Equations

Originally Posted by Petrus
y = (a*b-3a)/(11a-9)
x= (2a(b-3))/(11a-9)

is this correct?
it's your problem, you check it ...