Exercise 12 Page 285

Practice makes perfect

a To understand whether the given information is enough to determine the cost of 1 gallon of gasoline and 1 quart of oil, we will write an equation for each case. Let x be the cost of 1 gallon of gasoline and y be the cost of 1 quart of oil

Verbal Expression	My Receipt	My Friend's Receipt
Cost of gasoline ($)	10 x	5 x
Cost of oil ($)	2 y	y
Total cost ($)	10 x+2 y=45.50	5 x+ y=22.75

With this information we can write a system of two equations.

10x+2y=45.50 & (I) 5x+y=22.75 & (II)

Notice that multiplying Equation (II) by 2 results in Equation (I). This means that the equations are the same, so the given information is not enough to determine the costs of gasoline and the oil.

b On the receipt we can see that 8 gallons of gasoline and 2 quarts of oil cost $38.40 in total. We can write an equation for the receipt by proceeding in the same way as we did in Part A.

8x+2y=38.40 If we combine it with one of the previous equations into a system, we can see that there is no proportion between coefficients of equations.

10x+2y=45.50 & (I) 8x+2y=38.40 & (II)

Therefore, we can conclude that these equations are different and we can determine the cost of 1 gallon of gasoline and 1 quart of oil.

c In order to determine the cost of 1 gallon of gasoline and 1 quart of oil, we will use the system in Part B.

10x+2y=45.50 & (I) 8x+2y=38.40 & (II)
Notice that the y-variable has the same coefficient. Therefore, we can use Elimination Method to solve the system. Let's start!

10x+2y=45.50 & (I) 8x+2y=38.40 & (II)

SysEqnSub

(I): Subtract (II)

10x+2y-( 8x+2y)=45.50- 38.40 8x+2y=38.40

Distr

(I): Distribute -1

10x+2y-8x-2y=45.50-38.40 8x+2y=38.40

SubTerms

(I): Subtract terms

2x=7.10 8x+2y=38.40

DivEqn

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