McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
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Exercise 48 Page 210

Introduce variables for the price of a pound of nuts and a pound of raisin.

Nuts: $5.25
Raisins: $6.50

Practice makes perfect

There are several ways of solving this question. Let's use inverse matrices to find the solution. To do this, we need to introduce variables and we set up an equation system.

Description Variable
Price of a pound of nuts n
Price of a pound of raisins r

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

Description Equation
For 2 pounds of nuts and 2 pounds of raisins Sonia paid $23.50. 2n + 2r=23.50
For 3 pounds of nuts and 1 pound of raisins Drew paid $22.25. 3n + 1r=22.25
We found a system of equations in the two unknowns n and r. We can write this system as a matrix equation. 2n+ 2r=23.50 3n+1r=22.25 ⇕ 2& 2 3& 1 n r = 23.50 22.25 A matrix equation like this can be solved using the inverse matrix. MX=P ⟹ X=M^(-1)P We can use the determinant to find the inverse of a 2* 2 matrix. M= 2& 2 3& 1 ⇓ |M|= 2(1)- 2( 3)=-4 ⇓ M^(-1)=1/-4 1& - 2 - 3& 2 Let's use this inverse to find the answer to the question.

n r =M^(-1)P
n r =1/-4 1& - 2 - 3& 2 23.50 22.25
â–¼
Evaluate right-hand side
n r =1/-4 1(23.50)+(- 2)(22.25) - 3(23.50)+ 2(22.25)
n r =1/-4 23.50+(-44.50) -70.50+44.50
n r =1/-4 -21 -26

Multiply matrix by 1/-4

n r = 5.25 6.5

This result means that 1 pound of nuts costs $5.25, and 1 pound of raisins costs $6.50.