Q. 28 - Precalculus Enhanced with Graphing Utilities

Question

Megan went clothes shopping and bought $2$ pairs of flare jeans, $2$ camisoles, and $4$ T-shirts for $\backslash (90.00$. At the same store, Paige bought one pair of flare jeans and $3$ T-shirts for $\backslash )42.50$, while Kara bought $1$ pair of flare jeans, $3$ camisoles, and $2$ T-shirts for $\$62.00$. Determine the price of each clothing item. 

Step-by-Step Solution

Answer

Therefore the price of a pair of jeans, a camisole and a T-shirt is $x=\$29,y=\$8 \& \$4.5$ respectively.

Step 1. Given information

Different cost prices are given for different goods are given.

Step 2. Forming of equations

Let the cost price of jeans, camlsole and T-shirts are $x,y \& z$ respectively.

Therefore the system of equations is,

$\left\{\begin{array}{r}2x+2y+4z=92.00\\ x+3z=42.50\\ x+3y+2z=62.00\end{array}\right.$

Now take determinant of coefficients of the equations,

$$\begin{gathered} D=\left|\begin{array}{lll}2& 2& 4\\ 1& 0& 3\\ 1& 3& 2\end{array}\right| \\ =2(0\times 2-3\times 3)-2(1\times 2-3\times 1)+4(1\times 3-0\times 1) \\ =2(-9)-2(-1)+4\left(3\right) \\ =-4 \end{gathered}$$

Step 3. Finding determinants

Now for D_x replace the first column with constants and for D_y replace the second column and for D_z replace third column.

$$\begin{gathered} D_x=\left|\begin{array}{ccc}92& 2& 4\\ 42.5& 0& 3\\ 62& 3& 2\end{array}\right| \\ =92(0\times 2-3\times 3)-2(42.5\times 2-3\times 62)+4(42.5\times 3-0\times 62) \\ =92(-9)-2(-101)+4(127.5) \\ =-116 \\ {D}_y=\left|\begin{array}{ccc}2& 92& 4\\ 1& 42.5& 3\\ 1& 62& 2\end{array}\right| \\ =2(42.5\times 2-3\times 62)-92(1\times 2-3\times 1)+4(1\times 62-42.5\times 1) \\ =2(-101)-92(-1)+4(19.5) \\ =-32 \\ {D}_z=\left|\begin{array}{ccc}2& 2& 92\\ 1& 0& 42.5\\ 1& 3& 62\end{array}\right| \\ =2(0\times 62-42.5\times 3)-2(1\times 62-42.5\times 1)+92(1\times 3-0\times 1) \\ =2(-127.5)-2(19.5)+92\left(3\right) \\ =-18 \end{gathered}$$

Step 4. Finding values

Substitute the values of the determinants as,

$$\begin{gathered} x=\frac{D_x}{D} \\ =\frac{-116}{-4} \\ =29 \\ y=\frac{D_y}{D} \\ =\frac{-32}{-4} \\ =8 \\ z=\frac{D_z}{D} \\ =\frac{-18}{-4} \\ =4.5 \end{gathered}$$