Chapter 4, Problem 28SGR

To determine

To find: The cost of 1 pair of necklace and one pair of earrings.

Answer to Problem 28SGR

The cost of the necklace is $\$15.75$ and cost of pair of earring is $\$25.50$ .

Explanation of Solution

Given:

The cost of 3 necklace and 4 pair of earrings is $\$98.25$ .

The cost of 2 necklace and 4 pairs of earrings is $\$133.50$ .

Calculation:

Consider the cost of the necklace is $x$ and the cost of the pair of earring is $y$ .

Then the equation for the cost of 3 necklace and 4 pair of earrings is,

  $3x+2y=98.25$

The equation for the cost of 2 necklace and 4 pairs of earrings is,

  $2x+4y=133.50$

Then, the determinant of the equation is,

  $\begin{array}{c}\Delta =\left|\begin{array}{cc}3& 2\\ 2& 4\end{array}\right|\\ =\left(3\right)\left(2\right)-\left(2\right)\left(2\right)\\ =12-4\\ =8\end{array}$

Solve for the value of $x$ as,

  $\begin{array}{c}x=\frac{\Delta x}{\Delta }\\ =\frac{\left|\begin{array}{cc}98.25& 2\\ 133.50& 4\end{array}\right|}{8}\\ =\frac{\left(98.25\right)4-2\left(133.50\right)}{8}\\ =\$15.75\end{array}$

Solve for the value of $y$ as,

  $\begin{array}{c}y=\frac{\Delta y}{\Delta }\\ =\frac{\left|\begin{array}{cc}3& 98.25\\ 2& 133.50\end{array}\right|}{8}\\ =\frac{3\left(133.50\right)-2\left(98.25\right)}{8}\\ =\$25.50\end{array}$