Exercise 10 Page 434

To model and solve real-world problems, you need to start by defining the variables.

Define the variables and translate verbal statements into linear equations. Then, solve the system formed by those equations.

Practice makes perfect

To model and solve real-world problems, we follow two steps.

Define the variables and translate verbal statements into linear equations.
Solve the system formed by those equations.

We will consider an example to fully understand the process. Suppose that the cost of one sandwich and two chocolates is $ 2.50, and the cost of one sandwich and three chocolates is $ 3.00. Let's find the price of a sandwich and the price of one chocolate.

Define Variables and Translate Verbal Statements

Let s be the price of a sandwich and c the price of one chocolate.

Verbal Statement	Linear Equation
The cost of one sandwich and two chocolates is$ 2.50.	1s + 2c =2.5
The cost of one sandwich and three chocolates is$ 3.00.	1s + 3c =3

Solve the System

Now, we have to solve the system formed by the above linear equations. 1s+2c=2.5 & (I) 1s+3c=3 & (II) Note that the s-variable has the same coefficient in both equations. Thus, we can use the Elimination Method. Let's subtract Equation (I) from Equation (II).

1s+2c=2.5 1s+3c=3

SysEqnSub

(II): Subtract I

1s+2c=2.5 1s+3c-( 1s+2c)=3- 2.5

Distr

(II): Distribute - 1

1s+2c=2.5 1s+3c-1s-2c=3-2.5

SubTerms

(II): Subtract terms

1s+2c=2.5 c=0.5

We found that the price of one chocolate is $ 0.50. To find the value of the s-variable, we will substitute 0.5 for c in Equation (I), and solve for s.

1s+2c=2.5 c=0.5

Substitute

(I): c= 0.5

1s+2( 0.5)=2.5 c=0.5

▼

(I): Solve for s

Multiply

(I): Multiply

1s+1=2.5 c=0.5

SubEqn

(I): LHS-1=RHS-1

1s=1.5 c=0.5