Thread: 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 :/

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

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

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.

Originally Posted by Plato

I would use a matrix inverse.

i wanna solve this without matrix

Originally Posted by Petrus

i wanna solve this without matrix

That is fine. It harder without it.

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 :/?

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?

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

Originally Posted by Petrus

9z-11az=3b-11a

continue ...

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

Originally Posted by Petrus

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

solve for x and y ...

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

is this correct?

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 ...