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Systems of Linear Equations Worksheet

Systems of Linear Equations Worksheet
  • Page 1
 1.  
A two-digit number is 9 less than the number obtained by reversing its digits. Find the original number if it is 5 times the sum of its digits
a.
46
b.
45
c.
54
d.
64


Solution:

Let t represent the tens digit.

Let u represent the units number.

Then, 10t + u represents the original number and 10u + t represents the number obtained by reversing the digits.
[Number = 10 × number in tens place + number in units place.]

5(tens digit + units digit) = original number

5(t + u) = 10t + u

- 5t + 4u = 0 - - - (1)
[Simplify.]

Reversed number - 9 = original number

10u + t - 9 = 10t + u

9u - 9t = 9

u - t = 1 - - - (2)
[Divide throughout by 9.]

Solve the system of linear equations by using the addition method:

- 5t + 4u = 0; - t + u = 1;

- 5t + 4u = 0
 5t - 5 u = - 5       [Multiply second equation by - 5.]
-------------------
        - u = - 5            [Add.]

u = 5
[Solve for u.]

- 5t + 4(5) = 0
[Substitute the value of u in the equation (1).]

t = 4

So, the original number is: 10t + u = 10(4) + 5 = 45.


Correct answer : (2)
 2.  
The weight of 4 brown eggs and 2 white eggs together is 16 ounces, 2 brown eggs and 4 white eggs together weigh 23 ounces. Find the weight of 1 brown egg and 1 white egg together.
a.
15 oz
b.
6.50 oz
c.
5 oz
d.
1.50 oz


Solution:

Let b represent the weight of brown egg in ounces.

Let w represent the weight of white egg in ounces.

4b + 2w = 16 - - - (1);
2b + 4w = 23 - - - (2)
[As per the data.]

Use the addition method to solve the system of linear equations:
- 2b - w = - 8      [Divide equation (1) by - 2.]
 2b + 4w = 23
------------------
        3w = 15      [Add.]

w = 5
[Solve for w.]

4b + 2(5) = 16
[Substitute 5 for w in equation (1).]

4b + 10 = 16
[Substitute the values.]

b = 1.50

So, the weight of a brown egg is 11 / 2 ounces and the weight of a white egg is 5 ounces.

The weight of 1 brown egg and 1 white egg together = 5 + 1.50 = 6.50 ounces.


Correct answer : (2)
 3.  
5 years ago, Ursula was thrice as old as Alice. After 8 years Ursula will be twice as old as Alice. What are their present ages?
a.
Ursula: 44 years, Alice: 18 years
b.
Ursula: 8 years, Alice: 5 years
c.
Ursula: 49 years, Alice: 26 years
d.
Ursula: 39 years, Alice: 23 years


Solution:

Let b represent Ursula’s present age and c represent Alice’s present age.

5 years ago: b - 5 = 3 (c - 5)
8 years later: b + 8 = 2 (c + 8)

Simplify and write as simultaneous linear equations. Use the addition method to solve the system.

b - 3c = - 10 - - - (1)
b - 2c = 8 - - - (2)

- b + 3c = 10     [Multiply equation (1) by - 1.]
  b - 2c = 8
-----------------
         c = 18          [Add.]

b - 3(18) = - 10
[Substitute the value of c in equation (1).]

b - 54 = - 10

b = 44

Therefore, Ursula’s present age is 44 years and Alice’s present age is 18 years.


Correct answer : (1)
 4.  
When the digits of a two-digit number are reversed, the new number is 7 more than twice the original number. Find the original number if the sum of its digits is 11.
a.
30
b.
77
c.
83
d.
38


Solution:

Let t represent the tens digit and u represent the units digit.

Then, the original number is 10t + u and the new number after reversing the digits is 10u + t.
[Two digit number = 10 × number in tens place + number in unit place.]

Tens digit + units digit = 11

t + u = 11 - - - (1)

New number = 7 + 2(original number).

10u + t = 7 + 2(10t + u)

10u + t = 7 + 20t + 2u

8u - 19t = 7 - - - (2)
[Simplify.]

Write as a system of linear equations and use the addition method to solve them:

 8t + 8u =  88           [Multiply equation (1) by 8.]
19t - 8u = - 7           [Multiply equation (2) by - 1.]
--------------------
 27t      =   81         [Add.]

t = 3

3 + u = 11
[Substitute the values.]

u = 8

Original number = 10t + u = 10(3) + 8 = 38

So, the original number is 38.


Correct answer : (4)
 5.  
A steamer travels the distance between the two ports in 9 hours, when it goes downstream. When it goes upstream, it covers the same distance, 396 km in 11 hours. Find the speed of the steamer and the current. Assume that both are constant.
a.
36, 44
b.
40, 4
c.
40, 80
d.
36, 40


Solution:

Let b represent the speed of the boat in km/h.

Let c represent the speed of the current in km/h.

(b - c) × 11 = 396 - - - (1);
(b + c) × 9 = 396 - - - (2);
[Use the formula rate × time = distance.]

Solve the system of equations using addition method.
 b - c = 36          [Multiply equation (1) by 1 / 11.]
b + c = 44            [Multiply equation (2) by 1 / 9.]
-----------
 2b    =  80              [Add.]

b = 40
[Solve for b.]

(40 - c) × 11 = 396
[Substitute 40 for b in equation (1).]

40 - c = 36
[Substitute the values.]

c = 4

Therefore the speed of the steamer is 40 km/h and the speed of the current is 4 km/h.


Correct answer : (2)
 6.  
Joe invested $7,400, part at an annual interest rate of 4% and the rest at 5%. He received a total annual interest of $350. How much did he invest at each rate?
a.
$3,000, $7,435
b.
$4,050, $7,420
c.
$2,000, $5,400
d.
$3,025, $7,460


Solution:

Let x represent the amount invested at 4 %, in dollars.

Let y represent the amount invested at 5 %, in dollars.

x + y = 7400 - - - (1)

Total annual interest = (4 / 100)x + (5 / 100)y = 350 or 4x + 5y = 35000 - - - (2)
[Use the formula: rate × amount = annual interest.]

Solve the system of linear equations using the addition method:

 - 5x - 5y = - 37000        [Multiply equation (1) by - 5 .]
   4x + 5y =  35000
-----------------------
 - x    =        -2000              [Add.]

x = 2000
[Solve for x.]

2000 + y = 7400
[Substitute the value of x in equation (1).]

y = 5400

So, Joe invested $2,000 at 4% and $5,400 at 5%.


Correct answer : (3)
 7.  
Tommy has some quarters and some cents in his piggy bank. The total number of coins in the bank is 142, and their value is $24.46. How many quarters and cents does he have?
a.
96, 46
b.
101, 41
c.
104, 38
d.
91, 51


Solution:

Let q represent the number of quarters.

Let c represent the number of cents.

q + c = 142 - - - (1)

Value of quarters + Value of cents = $24.46

(0.25)q + (0.01)c = 24.46

25q + c = 2446 - - - (2)
[Multiply from each of the side by 100.]

Solve the system of linear equations by using the addition method:

- q    - c = - 142          [Multiply equation (1) by -1.]
 25q + c =  2446            
-----------------
 24q        =   2304              [Add.]

q = 96
[Solve for q.]

96 + c = 142
[Substitute the value of q in equation (1).]

c = 46

So, Tommy has 96 quarters and 46 cents in his piggy bank.


Correct answer : (1)
 8.  
The length of a rectangle is 7 in more than its width. The perimeter of the rectangle is 110 in. Find its length and width.
a.
62, 55
b.
26, 115
c.
29, 26
d.
31, 24


Solution:

Let l represent the length of the rectangle.

Let w represent the width of the rectangle.

l - w = 7 ---------- (1)

The perimeter of a rectangle = 2 (length + width).

2(l + w) = 110

l + w = 55 --------------- (2)
[Divide throughout by 2.]

Solve the system of linear equations by using the addition method:

l - w = 7
l + w = 55
-----------
2l      = 62    [Add.]

l = 31
[Solve for l.]

31 - w = 7
[Substitute the value of l in equation (1).]

w = 24

So, the length of the rectangle is 31 in and its width is 24 in


Correct answer : (4)
 9.  
The cost of 8 erasers and 10 pens is $24. 2 erasers and 12 pens cost $25. Find the cost of an eraser and the cost of a pen.
a.
$2, $0.5
b.
$2.5, $0.5
c.
$12, $14
d.
$8, $10


Solution:

Let p represent the cost of a pen, in dollars.

Let e represent the cost of an eraser, in dollars.

10p + 8e = 24 - - - (1); 12p + 2e = 25 - - - (2)
[As per the data.]

Use the addition method to solve the system.

    10p + 8e = 24
 - 48p - 8e = - 100            [Multiply equation (2) by - 4.]
--------------------
 - 38p        = - 76              [Add.]

p = 2
[Solve for p.]

10(2) + 8e = 24
[Substitute 2 for p in equation (1).]

20 + 8e = 24

e = 0.5

So, a pen costs $2 and an eraser costs $0.5.


Correct answer : (1)
 10.  
2 years hence, the age of a father will be 5 times the age of his son. The sum of their present ages is 38. Find their present ages.
a.
31, 7
b.
33, 5
c.
28, 10
d.
34, 4


Solution:

Let f represent the father’s present age.

Let s represent the son’s present age.

f + s = 38 - - - (1)

2 years hence, f + 2 = 5 (s + 2)

That is: f - 5s = 8 - - - (2)
[Simplify.]

Solve the system of linear equations by using the addition method:

   f +  s  =    38
- f + 5s = - 8         [Multiply equation (2) by - 1.]
--------------------
         6s  = 30              [Add.]

s = 5
[Solve for s.]

f + 5 = 38
[Substitute the value of s in equation (1).]

f = 33

Therefore, the father’s present age is 33 years and the son’s present age is 5 years.


Correct answer : (2)

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