Solutions to Algebra Questions and Problems for Grade 6

Detailed solutions to Algebra Questions and Problems for Grade 6 are presented along with explanations.

Solutions


  1. Like terms have the same variable with equal exponents. Also all numbers are like terms.
    A)     The terms 6 x and 12 x have same variable x with exponents equal to 1 and are therefore like terms.
    The terms 5 and -6 are numbers and therefore like terms
    B)   The terms 2 x 2 and 9 x 2 have same variable x with exponents equal to 2 , they are like terms.
    The terms - 4 and + 9 are numbers and therefore like terms.
    C)   The terms x / 5 and x / 7 have the same variable with exponents equal to 1, they are like terms
    D)   The terms 0.2 x , 1.2 x and x / 2 have the same variable x with exponents equal to 1; they are like terms
    E)   The terms 5 x and 7 x are like terms
    The terms - 8 and - 4 are like terms
    The terms - 2 x 2 and + 9 x 2 are like terms
    F)   There are no like terms in this expression.
    G)   The terms 5 a b and 6 b a are like terms

  2. Substitute x by the given numerical value and simplify.
    A)   6 (2) + 5 = 12 + 5 = 17
    B)   12 (1) 2 + 5 (1) - 2 = 12(1) + 5 - 2 = 12 + 5 - 2 = 15
    C)   2( 0 + 7) + 0 = 2 (7) = 14
    D)   2 (2) + 3 (4) - 7 = 4 + 12 - 7 = 9

  3. Notes
    1) We may add or subtract like terms as follows:
    2 x + 6 x = (2 + 6) x by factoring first
    = 8 x simplifying the terms in brackets.
    2) We use distributive property to expand an expression as follows: a(b + c) = a b + a c

    A)   3 x + 5 x
    = (3 + 5) x , factor x out
    = 8 x , simplify

    B)   2( x + 7) + x
    = 2(x) + 2(7) + x = 2x + 14 + x , expand and simplify
    = (2x + x) + 14 , group like terms
    = (2 + 1) x + 14 = 3 x + 14 , simplify

    C)   2(x + 3) + 3(x + 5) + 3
    = 2(x) + 2(3) + 3(x) + 3(5) + 3 = 2x + 6 + 3x + 15 + 3 , expand and simplify
    = (2x + 3x) + (6 + 15 + 3) , group like terms
    = (2 + 3) x + 24 = 5 x + 24 , simplify

    D)   2 (a + 1) + 5 b + 3(a + b) + 3
    = 2(a) + 2(1) + 5 b + 3(a) + 3(b) + 3 = 2 a + 2 + 5 b + 3 a + 3 b + 3 , expand and simplify
    = (2 a + 3 a) + (5 b + 3 b) + (2 + 3) , group like terms
    = (2 + 3) a + (5 + 3) b + 5 = 5 a + 8 b + 5 , simplify

  4. Notes
    1) To factor an expression is to write it as a product.
    2) The distributive property may be used to expand an algebraic expression : a ( x + y) = a x + a y
    3) The same distributive property may be used as : a x + a y = a ( x + y) to factor an expression where a is called the common factor

    A)   3 x + 3
    = 3 (x) + 3 (1) , 3 is a common factor
    = 3(x + 1) , factored form

    B)   8 x + 4
    = 4 (2) (x) + 4 , write 8 as 4 (2)
    = 4 (2 x ) + 4 (1) , 4 is a common factor
    = 4 (2x + 1) , factored form

    C)   a x + 3 a , a is a common factor
    = a(x + 3) , factored form

    D)   (x +1) y + 4 (x + 1)
    = (x + 1) (y) + (x + 1) (4) , x + 1 is a common factor
    = (x + 1)(y + 4) , factored form

    E)   x + 2 + b x + 2 b
    = (x + 2) + b(x + 2) , factor b out in b x + 2 b
    = (x + 2)(1) + (x + 2) b , x + 2 is now a common factor
    = (x + 2)(1 + b) , factored form

    Notes
    In order to check factoring, expand the factored form, simplify and make sure it is equivalent to the given expression.

  5. A)   x + 5 = 8
    x + 5 - 5 = 8 - 5 , subtract 5 from both sides of the equation
    x = 3 , simplify and solve for x
    Substitute x by 3 (solution found above) in both sides of the given equation to check the answer
    Right side: (3) + 5 = 8
    Left side = 8
    x = 3 is the solution to the given equation.

    B)   2 x = 4
    2 x / 2 = 4 / 2 , divide both sides by 2
    x = 2 , simplify and solve for x
    Substitute x by 2 (solution found above) in both sides of the given equation to check the answer
    Right side: 2(2) = 4
    Left side = 4
    x = 2 is the solution to the given equation.

    C)   x / 3 = 2
    3 (x / 3) = 3 (2) , multiply both sides by 3
    x = 6 , simplify and solve for x
    Substitute x by 6 (solution found above) in both sides of the given equation to check the answer
    Right side: 6 / 3 = 2
    Left side = 2
    x = 6 is the solution to the given equation.

    D)   0.2 x = 1
    0.2 x / 0.2 = 1 / 0.2 , divide both sides by 0.2
    x = 5 , simplify and solve for x
    Substitute x by 5 (solution found above) in both sides of the given equation to check the answer
    Right side: 0.2(5) = 1
    Left side = 1
    x = 5 is the solution to the given equation.

    E)   3 x + 6 = 12
    3 x + 6 - 6 = 12 - 6 , subtract 6 from both sides of the equation
    3 x = 6 , simplify
    3 x / 3 = 6 / 3 , divide both sides of the equation by 300
    x = 2 , simplify and solve for x
    Substitute x by 2 (solution found above) in both sides of the given equation to check the answer
    Right side: 3(2) + 6 = 6 + 6 = 12
    Left side = 12
    x = 2 is the solution to the given equation.

    F)   3 (x + 2) + 2 = 8
    3 (x ) + 3 (2) + 2 = 8 , expand 3 (x + 2)
    3 x + 6 + 2 = 8 , simplify
    3x + 8 = 8 , group like terms
    3 x + 8 - 8 = 8 - 8 , subtract 8 from both sides
    3x = 0 , simplify
    3x / 3 = 0 / 3 , divide both sides by 300
    x = 0 , simplify to solve
    Substitute x by 0 (solution found above) in both sides of the given equation to check the answer
    Right side: 3 (0 + 2) + 2 = 3(2) + 2 = 6 + 2 = 8
    Left side = 8
    x = 0 is the solution to the given equation.

  6. We use examples with 1, 2 , 3 ... boxes and then generalize for n boxes.
    1 box contains     m = 1 × m = m toys
    2 boxes contain     m + m = 2 × m = 2 m toys
    3 boxes contain     m + m + m = 3 × m = 3 m toys
    .
    .
    n boxes contain     m + m + m + .... + m = n × m = n m toys

  7. In exponential form, the expressions a × a × a and b × b are written as
    a × a × a = a 3 and b × b = b 2
    and the given expression a × a × - b × b is written as
    a × a × - b × b = a 3 - b 2

  8. The A area of a rectangle is given by the product width × length.
    A = width × length = 3 × (x + 2) = 3 (x) + 3(2) , use distributive property
    = 3 x + 6 , simplify

  9. Let x be the total number of students in the class. If 2 / 3 study math, then
    3/3 - 2 / 3 = 1 / 3 of the total number of students do not study math
    1 / 3 of the total number of students = (1 / 3) of x = x (1 / 3) = x / 3
    The number of students who do not study math is known and equal to 8. Hence
    x / 3 = 8
    Multiply both sides of the above equation by 3
    3 (x / 3) = 3 (8) Simplify and solve for x
    x = 24 students are in this class

  10. We find the distance covered in 1, 2, 3 ... hours and then generalize for x hours to find distance d.
    for 1 hour     d = 60 km = 1 × 60 km
    for 2 hours     d = 60 km + 60 km = 2 × 60 km
    for 3 hours     d = 60 km + 60 km + 60 km= 3 × 60 km
    .
    .
    for x hours     d = 60 km + 60 km +. . .+ 60 km = x × 60 km = 60 x km

  11. A)   2 3 + 3 2 = 2 × 2 × 2 + 3 × 3 = 8 + 9 = 17
    B)   0.1 3 = 0.1 × 0.1 × 0.1 = 0.001
    C)   6 × (2 / 3) = (6 / 1) × (2 / 3) = (6 × 2) / (1 × 3) = 12 / 3 = 4

  12. The total number of marbles is
    3 + 5 + 7 = 15
    Ratio of blue marbles to the total number of marbles is
    5 / 15
    Divide numerator and denominator by the common factor 5 to reduce the above fraction
    (5 ÷ 5) / (15 ÷ 5) = 1 / 3
    The ratio of blue marbles to the total number of marbles is
    1:3

  13. Use the cross product on the given equation with fractions
    a × 18 = 3 × 5
    Divide both sides by 18
    a × 18 / 18 = 3 × 5 / 18
    Simplify
    a = 15 / 18
    Divide numerator and denominator by 5 to reduce the fraction
    a = (15 3) / (18 3)
    Simplify
    a = 5 / 6

  14. A)   Substitute x by 0 and y by 0 in the given equation
    2 (0) + 3 (0) = 8
    Simplify the left side of the equation
    0 = 8 , false statement and therefore (0,0) is not a solution to the given equation

    B)   Substitute x by 4 and y by 0 in the given equation
    2 (4) + 3 (0) = 8
    Simplify the left side of the equation
    8 = 8 , true statement and therefore (4 , 0) is a solution to the given equation

    C)   Substitute x by 1 and y by 2 in the given equation
    2 (1) + 3 (2) = 8
    Simplify the left side of the equation
    8 = 8 , true statement and therefore (1 , 2) is a solution to the given equation

  15. A)   Factors of 4 are: 1 , 2 , 4
    B)   Factors of 12 are: 1 , 2 , 3 , 4 , 6 , 12
    C)   Factors of 50 are: 1 , 2 , 5 , 10 , 25 , 50

  16. Make a list of all factors for each number in the given pair and select the common factor to the two numbers that is the largest
    A)   Factors of 6 are: 1 , 2 , 3 , 6             Factors of 3 are: 1 , 3
    the greatest common factor of 6 and 3 is 3
    B)   Factors of 18 are: 1 , 2 , 3 , 6 , 9 , 18             Factors of 24 are: 1 , 2 , 3 , 4 , 6 , 8 , 12 , 24
    the greatest common factor of 18 and 24 is 6
    C)   Factors of 50 are: 1 , 2 , 5 , 10 , 25 , 50             Factors of 60 are: 1 , 2 , 3 , 4 , 5 , 6 , 10 , 12 , 15 , 20 , 30 , 60
    the greatest common factor of 50 and 60 is 10

  17. 672,000,259

  18. Expand and simplify the given expressions if necessary
    A)   2(x + 3) - 2 = 2(x) + 2(3) - 2 = 2 x + 6 - 2 = 2 x + 4
    B)   2x + 4
    C)   3(x + 3) - x - 5 = 3(x) + 3(3) - x - 5 = 3x + 9 - x - 5 = (3x - x) + (9 - 5) = 2 x + 4
    All three expressions are equivalent

  19. Make a list of the first few multiples for each number in the given pair and select the common multiple to the two numbers that is the lowest. Multiples of a number are obtained by multiplying the number by 1 ,2 , 3 ...
    A)   First few multiples of 2 are: 2 , 4 , 6 , 8 ...            First few multiples of 3 are: 3 , 6 , 9 , ...
    lowest common multiple of 2 and 3 is 6
    B)   First few multiples of 7 are: 7 , 14 , 21 ...            First few multiples of 14 are: 14 , 28 , 42 ...
    lowest common multiple of 7 and 14 is 14
    C)   First few multiples of 25 are: 25 , 50 , 75 ...             First few multiples of 15 are: 15 , 30 , 45 , 60 , 75 ...
    lowest common multiple of 25 and 15 is 75

  20. A)   The two fractions 1 / 3 and 2 / 3 have same (common) denominators 3 and can therefore be added by adding the numerators and keep the common denominator.
    1 / 3 + 2 / 3 = (1 + 2) / 3 = 3 / 3 = 1
    B)   The two fractions 2 / 5 and 1 / 7 do not have same (common) denominators and we therefore need to first find a common denominator by finding the lowest common multiple of the denominators 5 and 7.
    First few multiples of 5 are: 5 , 10 , 15 , 20 , 25 , 30 , 35 ...            First few multiples of 7 are: 7 , 14 , 21, 18 , 35 , ...
    lowest common multiple of 5 and 7 is 35
    We now convert the fractions 2 / 5 and 1 / 7 to equivalent fractions with common denominator equal to 35.
    To covert 2 / 5 to an equivalent fraction with a denominator equal to 35, we need to multiply its numerator and denominator by 7. Hence
    2 / 5 = (7 × 2) / (7 × 5) = 14 / 35
    To covert 1 / 7 to an equivalent fraction with a denominator equal to 35, we need to multiply numerator and denominator by 5. Hence
    1 / 7= (5 × 1) / (5 × 7) = 5 / 35
    We now substitute the fractions in the expressions to simplify by their equivalent fractions with common denominator.
    2/5 - 1 / 7 = 14 / 35 - 5 / 35 = (14 - 5) / 35 = 9 / 35

  21. 2 / 3 of 21 is mathematically written as
    (2 / 3) × 21
    Simplify the above
    = (2 / 3) × (21 / 1) = (2 × 21) / (3 × 1) = 42 / 3 = 14

  22. 40% of 1 / 4 is mathematically written as
    40% × (1 / 4)
    Write percentage as a fraction and simplify
    = (40 / 100) × 1 / 4 = (40 × 1)/ (100 × 4) = 40 / 400 = 1 / 10 = 0.1

  23. 20% of 50% is mathematically written as
    20% × 50%
    Write percentages as fractions and simplify
    = (20 / 100) × (50 / 100) = (20 × 50) / (100 × 100) = 1000 / 10000 = 1 / 10 = 0.1

  24. There are 60 minutes in one hour
    1 hour = 60 minutes
    There 60 seconds in 1 minute; hence
    1 hour = 60 minutes = 60 × 60 seconds = 3600 seconds

  25. There are 31 days in the month of January
    January = 31 days
    There 24 hours in one day; hence
    January = 31 days = 31 × 24 hours
    In one hour there are 60 minutes; hence
    January = 31 × 24 × 60 minutes = 44640 minutes

  26. (-2 , 0) is on the x axis 2 units to the left of the origin (0 , 0)
    (0 , 3) is on the y axis 3 units above the origin (0 , 0)
    (-2 , - 3) is in quadrant 3

  27. We need to convert all the given fractions to equivalent fractions the same denominator 100
    7 / 5 = (20 × 7) / (20 × 5) = 140 / 100
    12 / 10 = (10 × 12) / (10 × 10) = 120 / 100
    21 / 20 = (5 × 21) / (5 × 20) = 110 / 100
    111% = 111 / 100
    We now have all fractions with common denominator and we can compare them by comparing the numerators.
    smallest to largest: 110 / 100 ; 111 / 100 ; 120 / 100 , 140 / 100
    which are the equivalent of the original fractions
    21 / 20 ; 111% ; 12 / 10 ; 7 / 5     order from smallest to largest.

More References and links

Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers
Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers

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