# Thread: solving simultaneous linear eqns

1. ## solving simultaneous linear eqns

Solve, using matrices, the pair of simultaneous linear eqns,

3x= 2y + 8
14= 5x -4y

2. Originally Posted by maybeline9216
Solve, using matrices, the pair of simultaneous linear eqns,

3x= 2y + 8
14= 5x -4y
Re-write:

3x - 2y = 8

5x - 4y = 14

Can you construct the matrix equation from this?

3. yea but my answer is so big x=30,y=41 and my solution just works for the first eqn.

the paper's answer is x=2 y=-1

i dunno where went wrong but from these 2 simultaneous qns :

3x-2y=8 -(1)
5x-4y=14

im wondering do u nid to do something to them[like (1)x2?] or u can just straight-away write out the matrix from there??

4. Originally Posted by maybeline9216
yea but my answer is so big x=30,y=41 and my solution just works for the first eqn.

the paper's answer is x=2 y=-1

i dunno where went wrong but from these 2 simultaneous qns :

3x-2y=8 -(1)
5x-4y=14

im wondering do u nid to do something to them[like (1)x2?] or u can just straight-away write out the matrix from there??
$\begin{bmatrix}3 & -2\\5& -4\end{bmatrix} \, \begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}8 \\ 14 \end{bmatrix}$

which has the form AX = B. Left multiply both sides by the inverse of matrix A to solve for X. The solution to the equations is the matrix elements of X.

5. I did the same mtd as you.... my matrix is same as yours too....but my answer is different

6. Originally Posted by maybeline9216
I did the same mtd as you.... my matrix is same as yours too....but my answer is different
What do you get for the inverse matrix $A^{-1}$? What do you get for the product $A^{-1} B$?

7. Originally Posted by mr fantastic
What do you get for the inverse matrix $A^{-1}$? What do you get for the product $A^{-1} B$?
i got [ 2 -1 ] as my inverse matrix
[ 2.5 -1.5 ]

P.S i dunno hw to use latex, pls bear with me...

and i got [30] as my( A^-1)(B)
[41]

8. Originally Posted by maybeline9216
i got [ 2 -1 ] as my inverse matrix
[ 2.5 -1.5 ]

P.S i dunno hw to use latex, pls bear with me...

and i got [30] as my( A^-1)(B)
[41]
Your inverse $A^{-1}$ is correct, your product $A^{-1} B$ is wrong.

The first entry in the product is 16 - 14 = 2, NOT 16 + 14 = 30 ....

9. thanks lots=) i found out where im careless already!!!