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$$0.4 \, a + 150 \, b = 209 \tag{1}$$ $$0.1 \, a - 250 \, b = 35 \tag{2}$$

Attempt: the determinant of the co-efficient matrix

[ 0.4 150] [0.1 -250]

= (0.4)(-250) - (150)(0.1) = -115 $$D = -115$$

I don't know what to do from here. How do I solve for a and b?

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Assuming you have a non-singular matrix $A$, Cramer's rule says that the system $Ax = b$ can be solved by the formula $x_j = \frac{\det A_j}{\det A}$, where $A_j$ represents the matrix obtained by replacing the $j$-th column of $A$ by the column matrix $b$. In this way you obtain the solution $x = (x_1, x_2, \ldots , x_n)$ component-wise. Since you have a $2 \times 2$ system, this method should be computationally simple. As an example, I provide $x_1$ below: \begin{equation} x_1 = \frac{ \det A_1}{ \det A} = \frac{-57500}{-115} = 500, \quad A_1 = \left( \begin{array}[cc] 2209 & 150\\ 35 & -250 \end{array}\right) \end{equation}

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  • $\begingroup$ Alright, thanks. I tried solving for b following your steps and I got 0.06. Is that right? $\endgroup$ – john scott Sep 24 '15 at 18:33
  • $\begingroup$ Yes that is correct. In the future please add what you have tried or thought about in your post. You can also check your answer by plugging in your $ x $ into your system and see if you get $ b $. $\endgroup$ – Kevin Sheng Sep 24 '15 at 18:35
  • $\begingroup$ Okay I will. I'm new here. Just getting used to the system. Thanks Mr. Kevin. $\endgroup$ – john scott Sep 24 '15 at 18:38

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