Answer
(a) $\begin{cases} x+y+z=18 \\ 75x +90y + 60z=1380 \\ -75x +90y + 60z=180 \end{cases}$
(b) $8$ pounds of apples, $6$ pounds of peaches, and $4$ pounds of pears.
Work Step by Step
(a) Assume she bought $x$ pounds of apples, $y$ pounds of peaches, and $z$ pounds of pears, we can set up the following equations:
$\begin{cases} x+y+z=18 \\ 75x +90y + 60z=1380 \\ -75x +90y + 60z=180 \end{cases}$
where the first equation is the total weight, 2nd is the total cost, and 3rd reflect the price difference given in the conditions.
(b) Set up the following determinants and use the Cramer's Rule:
$\begin{array}( \\|D|= \\ \\ \end{array}\begin{vmatrix} 1&1&1\\ 75&90&60\\ -75&90&60 \end{vmatrix},
\begin{array}( \\|D_x|= \\ \\ \end{array}\begin{vmatrix} 18&1&1\\ 1380&90&60\\ 180&90&60 \end{vmatrix}, \begin{array}( \\|D_y|= \\ \\ \end{array}\begin{vmatrix} 1&18&1\\ 75&1380&60\\ -75&180&60 \end{vmatrix}, \begin{array}( \\|D_z|= \\ \\ \end{array}\begin{vmatrix} 1&1&18\\ 75&90&1380\\ -75&90&180 \end{vmatrix}$
Do the operation $R_2-R_3\to R_2$ to get
$\begin{array}( \\|D|= \\ \\ \end{array}\begin{vmatrix} 1&1&1\\150&0&0\\ -75&90&60 \end{vmatrix}$
Expand with row2 to get: $|D|=-150(60-90)=4500$
Do similar operations for other determinants to create some zeros and calculate the determinant values, we can get:
$\begin{array}( \\|D_x|= \\ \\ \end{array}\begin{vmatrix} 18&1&1\\ 1200&0&0\\ 180&90&60 \end{vmatrix}$
Expand with row2 to get: $|D_x|=-1200(60-90)=36000$
Similarly, do $R_3+R_2\to R_3, R_2-75R_1\to R_2$, we can get:
$\begin{array}( \\|D_y|= \\ \\ \end{array}\begin{vmatrix} 1&18&1\\ 0&30&-15\\ 0&1560&120 \end{vmatrix}, \begin{array}( \\|D_z|= \\ \\ \end{array}\begin{vmatrix} 1&1&18\\ 0&15&30\\ 0&180&1560 \end{vmatrix}$
Expand using column1 to get:
$|D_y|=30(120)+15(1560)=27000$ and $|D_z|=15(1560)-30(180)=18000$
We can finally get the solutions:
$x=\frac{|D_x|}{|D|}=\frac{36000}{4500}=8$,
$y=\frac{|D_y|}{|D|}=\frac{27000}{4500}=6$,
$z=\frac{|D_z|}{|D|}=\frac{18000}{4500}=4$
