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a chemist has a container of 12% peroxide solution and a container of 2…

Question

a chemist has a container of 12% peroxide solution and a container of 20% peroxide solution. she needs to mix the two solutions to create 100 ml of a 14% solution. if x represents the amount of 12% peroxide solution and y represents the amount of 20% peroxide solution she needs, which matrix equation can determine the amount of each solution she needs to make the 14% solution?
\begin{bmatrix}0.12&0.2\\1&1end{bmatrix}\begin{bmatrix}x\yend{bmatrix}=\begin{bmatrix}0.14\\100end{bmatrix}
\begin{bmatrix}0.12&0.2\\1&1end{bmatrix}\begin{bmatrix}x\yend{bmatrix}=\begin{bmatrix}14\\100end{bmatrix}
\begin{bmatrix}1&1\\0.12&0.2end{bmatrix}\begin{bmatrix}x\yend{bmatrix}=\begin{bmatrix}14\\100end{bmatrix}
\begin{bmatrix}1&1\\0.12&0.2end{bmatrix}\begin{bmatrix}x\yend{bmatrix}=\begin{bmatrix}0.14\\100end{bmatrix}

Explanation:

Step1: Set up amount - of - solution equation

The total volume of the final solution is 100 mL. So, $x + y=100$, which in matrix - form gives the first row of the coefficient matrix

$$\begin{bmatrix}1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$$

and the first element of the right - hand side vector

$$\begin{bmatrix}100\end{bmatrix}$$

.

Step2: Set up amount - of - peroxide equation

The amount of peroxide in the 12% solution is $0.12x$, the amount of peroxide in the 20% solution is $0.2y$, and the amount of peroxide in the 14% solution of 100 mL is $0.14\times100 = 14$. So, $0.12x+0.2y = 14$, which gives the second row of the coefficient matrix

$$\begin{bmatrix}0.12&0.2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$$

and the second element of the right - hand side vector

$$\begin{bmatrix}14\end{bmatrix}$$

.
Combining these two equations into a matrix equation, we get

$$\begin{bmatrix}0.12&0.2\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}14\\100\end{bmatrix}$$

.

Answer:

$$\begin{bmatrix}0.12&0.2\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}14\\100\end{bmatrix}$$

(corresponding to the second option in the multiple - choice list)