College Algebra, Chapter 7, 7.4, Section 7.4, Problem 60

A roadside fruit stand sells apples at $75$ cents a pound, peaches at $90$ cents a pound, and pears at $60$ cents a pound. Muriel buys $18$ pounds of fruit at a total cost of $\$ 13.80$. Her peaches and pears together cost $\$ 1.80$ more than her apples.

a.) Set up a linear system for the number of pounds of apples, peaches and pears that she bought.

b.) Solve the system using Cramer's Rule.

a.) Make a linear system for the number of pounds of apples, peaches and pears that she brought.

If we let $x, y$ and $z$ be the number of pounds of apples, peaches and pears, respectively. Then we have

$x + y + z = 18$

In terms of the cost,

$0.75 x + 0.90 y + 0.60z = 13.80$, remember that $100$ cents = $\$ 1$

If the peaches and pears together, cost $\$ 1.80$ more than the apples, then we have

$0.90y + 0.60z = 1.80 + 0.75x$

Thus, the system is


$
\left\{
\begin{equation}
\begin{aligned}

x + y + z =& 18
\\
0.75x + 0.90y + 0.60z =& 13.80
\\
-0.75x + 0.90y + 0.60z =& 1.80

\end{aligned}
\end{equation}
\right.
$


b.) Solve for the system using Cramer's Rule

We can write the system as


$
\left\{
\begin{equation}
\begin{aligned}

x + y + z=&18
&&
\\
75x+90y+60z =& 1380 && 100 \times \text{ Equation 2}
\\
-75x + 90y+60z =& 180
&& 100 \times \text{ Equation 2}

\end{aligned}
\end{equation}
\right.
$


For this system we have


$
\begin{equation}
\begin{aligned}

|D| =& \left| \begin{array}{ccc}
1 & 1 & 1 \\
75 & 90 & 60 \\
-75 & 90 & 60
\end{array} \right|
\\
\\
|D| =& 1 \left| \begin{array}{cc}
90 & 60 \\
90 & 60
\end{array} \right| -1 \left| \begin{array}{cc}
75 & 60 \\
-75 & 60
\end{array} \right| + \left| \begin{array}{cc}
75 & 90 \\
-75 & 90
\end{array} \right|
\\
\\
|D| =& (90 \cdot 60 - 60 \cdot 90) - (75 \cdot 60 - 60 \cdot (-75)) + (75 \cdot 90 - 90 \cdot (-75))
\\
\\
|D| =& 4500
\\
\\
|D_x| =& \left| \begin{array}{ccc}
18 & 1 & 1 \\
1380 & 90 & 60 \\
180 & 90 & 60
\end{array} \right|
\\
\\
|D_x| =& 18 \left| \begin{array}{cc}
90 & 60 \\
90 & 60
\end{array} \right| - 1 \left| \begin{array}{cc}
1380 & 60 \\
180 & 60
\end{array} \right| + 1 \left| \begin{array}{cc}
1380 & 90 \\
180 & 90
\end{array} \right|
\\
\\
|D_x| =& 18 (90 \cdot 60 - 60 \cdot 90) - (1380 \cdot 60 - 60 \cdot 180) + (1380 \cdot 90 - 90 \cdot 180)
\\
\\
|D_x| =& 36,000
\\
\\
|D_y| =& \left| \begin{array}{ccc}
1 & 18 & 1 \\
75 & 1380 & 60 \\
-75 & 180 & 60
\end{array} \right|
\\
\\
|D_y| =& 1 \left| \begin{array}{cc}
1380 & 60 \\
180 & 60
\end{array} \right| - 18 \left| \begin{array}{cc}
75 & 60 \\
-75 & 60
\end{array} \right| + 1 \left| \begin{array}{cc}
75 & 60 \\
-75 & 60
\end{array} \right|
\\
\\
|D_y| =& (1380 \cdot 60 - 60 \cdot 180) - 18 (75 \cdot 60 - 60 \cdot (-75)) + (75 \cdot 60 - 60 \cdot (-75))
\\
\\
|D_y| =& 27,000
\\
\\
|D_z| =& \left| \begin{array}{ccc}
1 & 1 & 18 \\
75 & 90 & 1380 \\
-75 & 90 & 180
\end{array} \right|
\\
\\
|D_z| =& 1 \left| \begin{array}{cc}
90 & 1380 \\
90 & 180
\end{array} \right| - 1 \left| \begin{array}{cc}
75 & 1380 \\
-75 & 180
\end{array} \right| + 18 \left| \begin{array}{cc}
75 & 90 \\
-75 & 90
\end{array} \right|
\\
\\
|D_z| =& 18,000

\end{aligned}
\end{equation}
$


The solution is


$
\begin{equation}
\begin{aligned}

x =& \frac{|D_x|}{|D|} = \frac{36000}{4500} = 8
\\
\\
y =& \frac{|D_y|}{|D|} = \frac{27000}{4500} = 6
\\
\\
z =& \frac{|D_z|}{|D|} = \frac{18000}{4500} = 4

\end{aligned}
\end{equation}
$


This means that Muriel buys 8 pounds of apples, 6 pounds of peaches and 4 pounds of pears.