What is the solution for x in the equation 1.4x + 9 = -2?

[karush]

View attachment 8750

Ok I'm doing 11 and 12

I kept trying row reduction but no

W|A

https://www.physicsforums.com/attachments/8751

 

[HOI]

Row reduction:
$\begin{bmatrix}1 & 3 & -4 \\ 1 & 5 & 2 \\ -3 & -7 & 6 \end{bmatrix}\begin{bmatrix}-2 \\ 4 \\ 12 \end{bmatrix}$

Subtract the first row from the second row and add 3 times the first row to the third row:
$\begin{bmatrix}1 & 3 & -4 \\ 0 & 2 & 6 \\ 0 & 2 & -6 \end{bmatrix}\begin{bmatrix}-2 \\ 6 \\ 6 \end{bmatrix}$

Subtract the second row from the third row:
$\begin{bmatrix}1 & 3 & -4 \\ 0 & 2 & 6 \\ 0 & 0 & -12 \end{bmatrix}\begin{bmatrix}-2 \\ 6 \\ 0 \end{bmatrix}$

This is now an "upper triangular" matrix which can be solved by "back substitution". The third row is equivalent to the equation -12z= 0 so z= 0. The second row is equivalent to the equation 2y+ 6z= 6 and since z= 0, 2y= 6, y= 3. The first row is equivalent to x+ 3y- 4z= 2 and since y= 3 and z= 0, x+ 9= 2 so x= -7.

 

[karush]

$\textsf{I think $x=-11$,}$

 

[Greg]

x + 9 = -2 so x = -11.