Chapter 7.2, Problem 50E

a.

To determine

The system of equations of the fuel mixture.

Answer to Problem 50E

  $\{\begin{array}{l}x+y=500\\ .87x+.92y=445\end{array}$

Explanation of Solution

Given information:

Five hundred gallons of $89$ -octane gasoline is obtained by mixing $87$ -octane gasoline with $92$ -octane gasoline.

  $(a)$ Write a system of equations in which one equation represents the amount of final mixture required and the other represents the amount of $87$ -and $92$ -octane gasolines in the final mixture.

Calculation:

Let $x$ represent the number of gallons of $87$ -octane

Let $y$ represent the number of gallons of $92$ -octane

Sum of two types of gasoline equalling $500$ gallons.

According to the given information, the equations are as shown below:

  $\begin{array}{l}x+y=500\\ .87x+.92y=(.89)(500)\end{array}$

Simplify by the following the order of operations:

  $\begin{array}{l}x+y=500\\ .87x+.92y=445\end{array}$

Therefore, the system of equations are $\{\begin{array}{l}x+y=500\\ .87x+.92y=445\end{array}$ .

b.

To determine

The solution of the system of equations by graphically.

Answer to Problem 50E

  $(300,200)$ .

Explanation of Solution

Given information:

Five hundred gallons of $89$ -octane gasoline is obtained by mixing $87$ -octane gasoline with $92$ -octane gasoline.

  $(b)$ Use a graphing utility to graph the two equations in part(a) in the same viewing window. As the amount of $87$ -octane gasoline increases, how does the amount of $92$ -octane gasoline change?

Calculation:

The two equations in part (a) are:

  $\{\begin{array}{l}x+y=500\\ .87x+.92y=445\end{array}$

Byusing a graphing utility the two equations are as shown below:

  

The amount of $92$ -octane gasoline change, because the $87$ -octane gasoline increases than $y$ decreases.

Thus, the solution is $(300,200)$ .

c.

To determine

To explain:The type of gasoline required to obtain the gallons.

Answer to Problem 50E

  $300$ Gallons of $87$ -octane gasoline and $200$ gallons of $92$ -octane gasoline

Explanation of Solution

Given information:

Five hundred gallons of $89$ -octane gasoline is obtained by mixing $87$ -octane gasoline with $92$ -octane gasoline.

  $(c)$ How much of each type of gasoline is required to obtain the $500$ gallons of $89-$ octane gasoline?

Calculation:

Let $x$ represent the number of gallons of $87$ -octane

Let $y$ represent the number of gallons of $92$ -octane

Sum of two types of gasoline equalling $500$ gallons.

According to the given information, the equations are as shown below:

  $\begin{array}{l}x+y=500\\ .87x+.92y=(.89)(500)\end{array}$

Simplify by the following the order of operations:

  $\begin{array}{l}x+y=500\\ .87x+.92y=445\end{array}$

Multiply the first equation by $-9.2$ so that the $y$ variables in the system of equations are opposites.

  $\begin{array}{l}-.92\left(x+y=500\right)\\ -.92x-.92y=-460\end{array}$

Add the equations together to eliminate the $y$ variables:

  $\begin{array}{l}-.92x-.92y=-460\text{ (1)}\\ \text{.87}x\text{+}\text{.92}y\text{=445 }\text{}\text{(2)}\end{array}$

Divide both sides by $-0.5$ to solve for $x$ :

  $\begin{array}{l}-0.5x=-15\\ x=300\end{array}$

Plug $x=300$ into the first equation of the system to get the value of $y$ :

  $\begin{array}{l}300+y=500\\ y=200\end{array}$

Thus, $300$ gallons of $87$ -octane gasoline and $200$ gallons of $92$ -octane gasoline are required to obtain $500$ gallons of $89$ -octane gasoline.