Miss Croft made snack bags for the picnic that contain three types of snacks: packages of crackers, packages of cookies, and candy bars. A snack bag containing 6 of each type of snack costs $21.00. A snack bag containing 8 packages of crackers, 5 packages of cookies, and 10 candy bars costs $26.00. A snack bag containing 5 packages of crackers, 4 packages of cookies, and 7 candy bars costs $18.50. Solve a system of three linear equations in three variables to determine the cost of one candy bar.

The cost of one candy bar is $1.5 and this can be determined by forming the linear equation with the help of given data.

Given :

Form the linear equation in order to determine the cost of one candy bar. Let 'x' be the cost of crackers, 'y' be the cost of cookies, and 'z' be the cost of candy bars.

So, the linear equation that represents the situation "A snack bag containing 6 of each type of snack costs $21.00" is:

6x + 6y + 6z = 21  --- (1)

The linear equation that represents the situation "A snack bag containing 8 packages of crackers, 5 packages of cookies, and 10 candy bars costs $26.00" is:

8x + 5y + 10z = 26  --- (2)

The linear equation that represents the situation "A snack bag containing 5 packages of crackers, 4 packages of cookies, and 7 candy bars costs $18.50" is:

5x + 4y + 7z = 18.50  --- (3)

Simplify all three equations in order to determine the value of 'x', 'y', and 'z'.

z = $1.25

y = $1.5

x = $0.75

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$1.25

Step-by-step explanation:

Let the cost of cracker be a.

Let the cost of cookies be b.

Let the cost of candy bars be c.

From the question,

6a + 6b + 6c = $21

8a + 5b + 10c = $26

5a + 4b + 7c = $18.50

To solve this, we first pick any two pairs of equation. This will be:

6a + 6b + 6c = $21 i

8a + 5b + 10c = $26 ii

Multiply equation i by 8

Multiply equation ii by 6

48a + 48b + 48c = $168 iii

48a + 30b + 60c = $156 iv

Subtract equation iv from iii

18b - 12c = 12

We then pick another two pairs

8a + 5b + 10c = $26

5a + 4b + 7c = $18.50

Multiply equation i by 5

Multiply equation ii by 8

40a + 25b + 50c = $130

40a + 32b + 56c = $148

Subtract the equations

-7b - 6c = -18

Then, solve the new equations formed

18b - 12c = 12 v

-7b - 6c = -18 vi

Multiply equation i by -7

Multiply equation ii by 18

-126b + 84c = -84

-126b - 108c = -324

Subtract the equations

192c = 240

c = 240/129

c = $1.25

From equation v, put the value of c

18b - 12c = 12

18b - 12($1.25) = 12

18b - $15 = $12

18b = $27

b = $27/18

b = $1.5

Since,

6a + 6b + 6c = $21

6a + 6($1.5) + 6($1.25) = $21

6a + $9 + $7.5 = $21

6a + $16.5 = $21

6a = $21 - $16.5

6a = $4.5

a = $4.5/6

a = $0.75

One candy bar cost $1.25

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