US Common Core K-12 Curriculum
Algebra : Word Problems on Systems of Linear Equations

Problem 1 :

Amy and Jose each bought plants from the same store. Amy spent $188 on 7 cherry trees and 11 rose bushes. Jose spent $236 on 13 cherry trees and 11 rose bushes. Find the coset of each cherry tree and each rose bush.

Solution :

Let x be the cost of each cherry tree and y be the cost of each rose bush.

Then, we can have the following equations for Amy and Jose from the given information.

Amy :

7x + 11y = 188 ----(1)

Jose :

13x + 11y = 236 ----(2)

Let's subtract (1) from (2) to eliminate y.

6x = 48

Divide both sides by 6.

x = 8

Plug in x = 8 in (1)

7(8) + 11y = 188

56 + 11y = 188

Subtract 56 from both sides.

11y = 132

Divide both sides by 11.

y = 12

(x, y) = (8, 12)

Therefore, the cost of each cherry tree is $8 and that of of each rose bush is $12.

Problem 2 :

A woman has 21 coins in her pocket, all of which are dimes and quarters. If the total value of change is $3.90, how many dimes and how many quarters does she have?

Solution :

Let x be the number of dimes and y be the quarters.

Then, we can have the following equations from the given information.

x + y = 21 ----(1)

10x + 25y = 390 ----(2)

(In (2), $3.90 is converted to 390 cents, because 1 dime = 10 cents and 1 quarter = 25 cents)

In (2), since each term is a multiple of 5, divide each term by 5.

2x + 5y = 78 ----(2)

Let's multiply the (1) by -2 and add it to (2) to eliminate x.

3y = 36

Divide both sides by 3.

y = 12

Plug in y = 12 in (1).

x + 12 = 21

Subtract 12 from both sides.

x = 9

(x, y) = (9, 12)

Therefore, the woman has 9 dimes and 12 quarters.

Problem 3 :

A professional organizer charges a $45 consultation fee, plus $20 per hour. Her competitor charges a $30 consultation fee, plus $26 per hour. Find the number of hours that the total cost will be the same.

Solution :

Let x be the number of hours and y be the total cost.

Then, we can have the following equations from the given information.

Professional organizer :

y = 45 + 20x ----(1)

Her competitor :

y = 30 + 26x ----(2)

When the total cost is same,

y = y

45 + 20x = 30 + 26x

Subtract 20x and 30 from both sides.

15 = 6x

Divide both sides by 6.

2.5 = x

When the total cost is same for both, the number of hours is 2.5.

Problem 4 :

The cost of 8 apples and 6 oranges is $33. If the cost of 4 apples is $4.50 more than 5 oranges, determine the cost of each apple and orange.

Solution :

Let x be the cost of each apple and y be the cost of each orange.

Then, we can have the following equations from the given information.

8x + 6y = 33 ----(1)

4x = 5y + 4.50 ----(2)

In (2), we have 4x = 5y + 4.50. So, plug in 5y + 4.50 for 4x in (1).

8x + 6y = 33

2(4x) + 6y = 33

2(5y + 4.50) + 6y = 33

10y + 9 + 6y = 33

16y + 9 = 33

Subtract 9 from both sides.

16y = 24

Divide both sides by 16.

y = 1.50

Plug in y = 1.5 in (2). 

4x = 5(1.50) + 4.50

4x = 7.50 + 4.50

4x = 12

Divide both sides by 4.

x = 3

(x, y) = (3, 1.5)

Therefore, the cost of each apple is $3 and that of orange is $1.50.

Problem 5 :

Peter is older than James by 7 years. 15 years back James' age was ¾ of Peter’s age. determine their present ages.

Solution :

Let p be the present age of Peter and j be the present age of James.

Then, we can have the following equations from the given information.

p = j + 7 ----(1)

j - 15 = ¾(p - 15) ----(2)

In (1), we have p = j + 7. So, plug in j + 7 for p in (2).

j - 15 = ¾(j + 7 - 15)

j - 15 = ¾(j - 8)

Multiply both sides by 4.

4(j - 15) = 3(j - 8)

4j - 60 = 3j - 24

Subtract 3j from both sides.

j - 60 = -24

Add 60 to both sides.

j = 36

Plug in j = 36 in (1).

p = 36 + 7

p = 43

(p, j) = (43, 36)

Therefore, the present age of Peter is 43 years and that of James is 36 years.

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