Answer: Choice D
[tex]\begin{cases}5x+2y+z = 16\\ 7x-5y+2z = 3\\-5x+3y+z = 12\end{cases}[/tex]
==============================================
Explanation:
The numbers in the first matrix represent the coefficients of the x,y and z variable terms.
Along the first row we have 5, 2 and 1 in that order. This means we have 5x+2y+1z. This is the same as 5x+2y+z. This is set equal to 16 due to the 16 in the first row of the third matrix on the right hand side. The first equation is therefore 5x+2y+z = 16.
The second row has 7, -5, 2 in the first matrix. So we have 7x-5y+2z. It is set equal to 3 because it is in the second row of the third matrix. We have 7x-5y+2z = 3 as the second equation.
Finally the third row has -5, 3, 1 in the first matrix and 12 in the second matrix. Therefore we end up with -5x+3y+1z = 12 which is the same as -5x+3y+z = 12.
Side note: choice A is a trick answer. Note the second copy of x in the first equation. That should be a z.
