Exercise 30 Page 208

Introduce variables, then set up and solve a system of equations.

Hot dog: $3.25
Popcorn: $2.25
Soda: $2.50

Practice makes perfect

Let's introduce variables for the price of each snack.

Description	Variable
Price of a hot dog	h
Price of a popcorn	p
Price of a soda	s

Let's use these variables and the information in the given table to write an equation system.

Name	Hot Dogs	Popcorn	Soda	Price
Dustin	1	2	3	$15.25
Luis	2	0	3	$14.00
Marci	1	2	1	$10.25

1h+ 2p+ 3s=15.25 & (I) 2h+ 0p+ 3s=14.00 & (II) 1h+ 2p+ 1s=10.25 & (III) To find the price of each snack we need to solve this system of equations. Notice that in Equations (I) and (III) the coefficients of h and p are the same. This means that if we take the difference of these two equations, variables h and p will be eliminated. This allows us to find the value of s.

h+2p+3s=15.25 & (I) 2h+3s=14.00 & (II) h+2p+s=10.25 & (III)

SysEqnSub

(I): Subtract (III)

3s- s=15.25- 10.25 2h+3s=14.00 h+2p+s=10.25

▼

(I): Solve for s

SubTerms

(I): Subtract terms

2s=5.00 2h+3s=14.00 h+2p+s=10.25

DivEqn

(I): .LHS /2.=.RHS /2.

s=2.50 2h+3s=14.00 h+2p+s=10.25

Once we have the value of s, we can substitute it into Equations (II) and (III). Notice that Equation (II) only has variables h and s, so substituting s=2.50 into this equation allows us to find h.

s=2.50 & (I) 2h+3s=14.00 & (II) h+2p+s=10.25 & (III)

(II), (III): s= 2.50

s=2.50 2h+3( 2.50)=14.00 h+2p+ 2.50=10.25

▼

(II): Solve for h

Multiply

(II): Multiply

s=2.50 2h+7.50=14.00 h+2p+2.50=10.25

SubEqn

(II): LHS-7.50=RHS-7.50

s=2.50 2h=6.50 h+2p+2.50=10.25

DivEqn