McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
Study Guide and Review
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Exercise 42 Page 210

Introduce variables for the price of a necklace and a pair of earrings.

Necklace: $15.75
Pair of earrings: $25.50

Practice makes perfect

We are asked to use Cramer's Rule to find the price of a necklace and a pair of earrings. To do this, we need to introduce variables and set up an equation system.

Description Variable
Price of a necklace n
Price of a pair of earrings e

Let's use these variables to write equations expressing the information given in the question.

Description Equation
For 3 necklaces and 2 pairs of earrings Alana paid $98.25. 3n + 2e=98.25
For 2 necklaces and 4 pairs of earrings Petra paid $133.50. 2n + 4e=133.50
We found a system of equations in the two unknowns n and e. 3n+ 2e= 98.25 2n+ 4e= 133.50 To use Cramer's Rule, we need the coefficient matrix of the system. C= 3 & 2 2 & 4 If the determinant of the coefficient matrix |C| is different from zero, Cramer's Rule tells us how to find the solution to our system by using determinants. n=98.25 & 2 133.50 & 4/|C| and e= 3 & 98.25 2 & 133.50/|C| Let's start by calculating |C|.

|C|= 3 & 2 2 & 4
â–¼
Calculate determinant

a & b c & d =ad-bc

|C|=3(4)-2(2)
|C|=12-4
|C|=8

This is not 0, so we can now substitute 8 for |C| and calculate the value of n.

n=98.25 & 2 133.50 & 4/|C|
n=98.25 & 2 133.50 & 4/8
â–¼
Simplify right-hand side

a & b c & d =ad-bc

n=98.25(4)-2(133.50)/8
n=393-267/8
n=126/8
n=15.75

This result means that the price of a necklace is $15.75. We can use similar calculations to find the price of a pair of earrings.

e=3 & 98.25 2 & 133.50/|C|
e=3& 98.25 & 2& 133.50/8
â–¼
Simplify right-hand side

a & b c & d =ad-bc

e=3(133.50)-98.25(2)/8
e=400.50-196.50/8
e=204/8
e=25.50

This means that the price of a pair of earrings is $25.50. Let's summarize the results in a table.

Item Price
Necklace $15.75
Pair of earrings $25.50