Chapter 7.2, Problem 80E

(a)

To determine

To write: a system of linear equations in which one equation represents the amount of final mixture required and the other represents the percent of acid in the final mixture.

Answer to Problem 80E

The system of linear equations in which one equation represents the amount of final mixture required and the other represents the percent of acid in the final mixture are:

  $\begin{array}{l}x+y=30\\ x+2y=48\end{array}$

Explanation of Solution

Given information:

Thirty liters of an $40\%$ acid solution is obtained by mixing a $25\%$ solution with a $50\%$ solution.

Let $x$ and $y$ represent the amounts of the $25\%$ and $50\%$ solutions, respectively.

Calculation:

The equation represents the amount of final mixture required is:

  $x+y=30$

And

The equation represents the percent of acid in the final mixture is:

  $\begin{array}{l}\left(25\%\text{ of }x+50\%\text{ of }y\right)=40\%\text{ of }30\\ \left(\frac{25}{100}\times x+\frac{50}{100}\times y\right)=\frac{40}{100}\times 30\\ 25x+50y=1200\\ x+2y=48\end{array}$

Therefore, the system of linear equations can be written as:

  $\begin{array}{l}x+y=30\\ x+2y=48\end{array}$

(b)

To determine

To plot: a system of linear equations on graph using graphing utility.

Answer to Problem 80E

The solution of system of linear equations using graphing utility is:

  $\left(x,y\right)=\left(12,18\right)$

Explanation of Solution

Given information:

Thirty liters of an $40\%$ acid solution is obtained by mixing a $25\%$ solution with a $50\%$ solution.

Let $x$ and $y$ represent the amounts of the $25\%$ and $50\%$ solutions, respectively.

Calculation:

The system of linear equations in which one equation represents the amount of final mixture required and the other represents the percent of acid in the final mixture are:

  $\begin{array}{l}x+y=30\\ x+2y=48\end{array}$

Plot the given system of linear equations using graphing utility:

  

  $P(12,18)$ is the graphical solution of the given system of linear equations.

Therefore,the solution of system of linear equations using graphing utility is:

  $\left(x,y\right)=\left(12,18\right)$

(c)

To determine

To state:what will happen to $50\%$ solution when the amount of $25\%$ solution increases.

Answer to Problem 80E

When the amount of $25\%$ solution increases, the amount of $50\%$ solution decreases.

Explanation of Solution

Given information:

Thirty liters of an $40\%$ acid solution is obtained by mixing a $25\%$ solution with a $50\%$ solution.

Let $x$ and $y$ represent the amounts of the $25\%$ and $50\%$ solutions, respectively.

Calculation:

The equation represents the percent of acid in the final mixture is:

  $\begin{array}{l}x+2y=48\\ 2y=48-x\\ y=\frac{48-x}{2}\end{array}$

Analyze the above equation by table method, put different values of $x$ to get $y$ :

$x$	$y=\frac{48-x}{2}$
$0$	24
$2$	23
$4$	$22$
8	$20$

From the table, it is clear that when the amount of $25\%$ solution increases, the amount of $50\%$ solution decreases.

(d)

To determine

To solve: a system of linear equations and determine the amount of each solution to obtained the specified concentration of the final mixture.

Answer to Problem 80E

  $12\text{ litres}$ of $25\%$ solution is required and $18\text{ litres}$ of $50\%$ solution is required.

Explanation of Solution

Given information:

Thirty liters of an $40\%$ acid solution is obtained by mixing a $25\%$ solution with a $50\%$ solution.

Let $x$ and $y$ represent the amounts of the $25\%$ and $50\%$ solutions, respectively.

Calculation:

Since, the system of linear equations can be written as:

  $x+y=30$  ...(1)

  $x+2y=48$  ...(2)

Solve equation (1) and (2) by elimination method:

  $\underset{\_}{\begin{array}{l}\cancel{x}+y=30\\ \text{- }\cancel{x}+2y=48\end{array}}$

  $\begin{array}{l}\text{ -y}=-18\\ \text{ y}=18\end{array}$

Calculate the value of $x$ by putting value of $y$ in equation (1):

  $\begin{array}{l}x+18=30\\ x=30-18\\ x=12\end{array}$

Therefore, $12\text{ litres}$ of $25\%$ solution is required and $18\text{ litres}$ of $50\%$ solution is required.