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# Linear Systems Worksheet

Linear Systems Worksheet
• Page 1
1.
The cost of an apple is $1 and that of an orange is$0.4. Gary purchased 18 fruits for $15. If A represents the number of apples and O the number of oranges, which system of linear equations represents the situation?  a. A + O = 15 and A + 0.4O = 18 b. A + O = 15 and 0.4A + O = 18 c. A + O = 18 and 0.4A + O = 15 d. A + O = 18 and A + 0.4O = 15 #### Solution: A + O = 18 [Total number of fruits = 18.] Cost of A apples is$A.
[Cost of an apple = $1.] Cost of O oranges is$0.4O.
[Cost of an orange = $0.4.] Cost of 18 fruits is$15, so A + 0.4O = 15
[Total cost = $15.] So, the system of equations are A + O = 18 and A + 0.4O = 15 Correct answer : (4) 2. Choose the easiest method to solve the linear system. 5$x$ + $y$ = 6 - 6$x$ - 4$y$ = 4  a. linear combination method b. substitution method #### Solution: Substitution method, because the coefficient of y is 1 in the Equation 1. You can solve Equation 1 for y and substitute the result into Equation 2. Correct answer : (2) 3. Choose the easiest method to solve the linear system. #### Solution: Linear combination method, as no variable can be easily isolated because none of the variables has a coefficient of 1 or Ã¢â‚¬â€œ 1. Correct answer : (0) 4. Choose the easiest method to solve the linear system. 5$x$ + 2$y$ = 16 $x$ - 4$y$ = - 10  a. substitution method b. linear combination method #### Solution: Substitution method, because the coefficient of x is 1 in the Equation 2. You can solve Equation 2 for x and substitute the result into Equation 1. Correct answer : (1) 5. Choose the best method to solve the linear system. - 8$x$ + 3$y$ = 5 11$x$ - 5$y$ = - 16  a. substitution method b. linear combination method #### Solution: Linear combination method, as no variable can be easily isolated because none of the variables has a coefficient of 1 or - 1. Correct answer : (2) 6. Jim bought two types of movie tickets at$4 and $5 for 8 of his friends. Total cost of the tickets is$37. Let $x$ be the number of $4 tickets and $y$ be the number of$5 tickets. Which system of linear equations represents the situation?
 a. $x$ + $y$ = 37 and 4$x$ + 5$y$ = 8 b. $x$ + $y$ = 8 and 5$x$ + 4$y$ = 37 c. 4$x$ + 5$y$ = 37 d. $x$ + $y$ = 8 and 4$x$ + 5$y$ = 37

#### Solution:

x + y = 8
[Total number of tickets = 8.]

Cost of x tickets = 4x

Cost of y tickets = 5y

4x + 5y = 37
[total ticket cost is $37.] So, the system of equations are x + y = 8 and 4x + 5y = 37. Correct answer : (4) 7. A shopkeeper sold 36 softballs and basketballs for a sum of$396. The price of a softball is $9 and that of a basketball is$27. How many softballs and basketballs did the shopkeeper sell?

#### Solution:

Let x be the number of softballs and y be the number of basketballs.

x + y = 36
[Equation 1.]

9x + 27y = 396
[Equation 2.]

x = - y + 36
[Revise Equation 1.]

9(- y + 36) + 27y = 396
[Substitute x = - y + 36 in Equation 2.]

18y + 324 = 396
[Combine like terms.]

18y = 72
[Subtract 324 from each side.]

y = 4
[Divide each side by 18.]

x = - (4) + 36 = 32
[Substitute y = 4 in revised equation 1.]

The shopkeeper sold 32 softballs and 4 basketballs.

8.
Ed purchased 28 pens and markers for $88. The price of a marker is$5 and the price of a pen is $3. How many pens and markers did Ed buy?  a. 25 markers, 3 pens b. 26 markers, 2 pens c. 2 markers, 26 pens d. 6 markers, 27 pens #### Solution: Let x be the number of markers and y be the number of pens. x + y = 28 [Equation 1.] 5x + 3y = 88 [Equation 2.] y = 28 - x [Revise Equation 1.] 5x + 3(28 - x) = 88 [Substitute y = 28 - x in Equation 2.] 2x + 84 = 88 [Combine like terms.] 2x = 4 [Subtract 84 from each side.] x = 2 [Divide each side by 2.] y = 28 - 2 = 26 [Substitute x = 2 in the revised equation 1.] Ed bought 2 markers and 26 pens. Correct answer : (3) 9. The price of each marble is$7 and the price of each dice is $5. A total of 34 marbles and dice were purchased for$202. Find the number of marbles and the number of dice purchased.
 a. 16 marbles, 18 dice b. 18 marbles, 16 dice c. 14 marbles, 20 dice d. 12 marbles, 22 dice

#### Solution:

Let x be the number of marbles and y be the number of dice purchased.

x + y = 34
[Equation 1.]

7x + 5y = 202
[Equation 2.]

x = 34 - y
[Rearrange Equation 1.]

7(34 - y) + 5y = 202
[Substitute x = 34 - y in Equation 2.]

- 2y + 238 = 202
[Combine like terms.]

- 2y = - 36
[Subtract 238 from each side.]

y = 18
[Divide each side by -2.]

x = 34 - 18 = 16
[Substitute y = 18 in the revised equation 1.]

16 marbles and 18 dice are purchased.

10.
Valerie purchased a total of 18 books and toys for the Dugway play school. Each book costs $23 and each toy costs$12. How many books and toys did she buy for \$337?
 a. 11 books and 7 toys b. 10 books and 8 toys c. 7 books and 11 toys d. 8 books and 10 toys

#### Solution:

Let x be the number of books and y be the number of toys Valerie purchased.

x + y = 18 --- (1)
[Linear equation for the total books and toys.]

23x + 12y = 337 --- (2)
[Equation for the total cost of the books and toys.]

23x + 12(- x + 18) = 337
[From equation 1, y = - x + 18. Substitute it in equation 2.]

11x + 216 = 337
[Combine like terms.]

11x = 121
[Subtract 216 from each side.]

x = 11
[Divide each side by 11.]

y = - (11) + 18 = 7
[Substitute x = 11 in equation 1.]

Valerie bought 11 books and 7 toys.