Grade 6 algebra questions and problems with detailed solutions. Topics include expanding and simplifying expressions, solving linear equations, factoring expressions, finding GCF and LCM, fractions, and percentages.
List all like terms included in each of the expressions given below.
A) \(6x + 5 + 12x - 6\)
B) \(2x^2 - 4 + 9x^2 + 9\)
C) \( \dfrac{x}{5} + \dfrac{x}{7} + 7 \)
D) \(0.2x + 1.2x + \dfrac{x}{2} + 5\)
E) \(5x - 8 + 7x - 2x^2 - 4 + 9x^2 + 4x^3\)
F) \(5a + 8 - 7ab\)
G) \(5ab + 8 + 6ba - a + 3b\)
A) The terms \(6x\) and \(12x\) 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 \(2x^2\) and \(9x^2\) 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 \(\dfrac{x}{5}\) and \(\dfrac{x}{7}\) have the same variable with exponents equal to 1, they are like terms.
D) The terms \(0.2x\), \(1.2x\), and \(\dfrac{x}{2}\) have the same variable \(x\) with exponents equal to 1; they are like terms.
E) The terms \(5x\) and \(7x\) are like terms.
The terms \(-8\) and \(-4\) are like terms.
The terms \(-2x^2\) and \(+9x^2\) are like terms.
F) There are no like terms in this expression.
G) The terms \(5ab\) and \(6ba\) are like terms.
Evaluate each of the expressions for the given value(s) of the variable(s).
A) \(6x + 5\) for \(x = 2\)
B) \(12x^2 + 5x - 2\) for \(x = 1\)
C) \(2(x + 7) + x\) for \(x = 0\)
D) \(2a + 3b - 7\) for \(a = 2\) and \(b = 4\)
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) \(2(2) + 3(4) - 7 = 4 + 12 - 7 = 9\)
Expand (if needed) and simplify each of the expressions below.
A) \(3x + 5x\)
B) \(2(x + 7) + x\)
C) \(2(x + 3) + 3(x + 5) + 3\)
D) \(2(a + 1) + 5b + 3(a + b) + 3\)
A) \(3x + 5x\)
\(= (3 + 5)x\), factor \(x\) out
\(= 8x\), simplify
B) \(2(x + 7) + x\)
\(= 2(x) + 2(7) + x = 2x + 14 + x\), expand and simplify
\(= (2x + x) + 14\), group like terms
\(= (2 + 1)x + 14 = 3x + 14\), 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
\(= (2 + 3)x + 24 = 5x + 24\), simplify
D) \(2(a + 1) + 5b + 3(a + b) + 3\)
\(= 2(a) + 2(1) + 5b + 3(a) + 3(b) + 3 = 2a + 2 + 5b + 3a + 3b + 3\), expand and simplify
\(= (2a + 3a) + (5b + 3b) + (2 + 3)\), group like terms
\(= (2 + 3)a + (5 + 3)b + 5 = 5a + 8b + 5\), simplify
Factor each of the expressions below by finding the greatest common factor.
A) \(3x + 3\)
B) \(8x + 4\)
C) \(ax + 3a\)
D) \((x + 1)y + 4(x + 1)\)
E) \(x + 2 + bx + 2b\)
A) \(3x + 3\)
\(= 3(x) + 3(1)\), 3 is a common factor
\(= 3(x + 1)\), factored form
B) \(8x + 4\)
\(= 4(2)(x) + 4\), write 8 as 4(2)
\(= 4(2x) + 4(1)\), 4 is a common factor
\(= 4(2x + 1)\), factored form
C) \(ax + 3a\), 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) \(x + 2 + bx + 2b\)
\(= (x + 2) + b(x + 2)\), factor b out in bx + 2b
\(= (x + 2)(1) + (x + 2)b\), x + 2 is now a common factor
\(= (x + 2)(1 + b)\), factored form
Solve each of the equations below and check your answer.
A) \(x + 5 = 8\)
B) \(2x = 4\)
C) \(\dfrac{x}{3} = 2\)
D) \(0.2x = 1\)
E) \(3x + 6 = 12\)
F) \(3(x + 2) + 2 = 8\)
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:
Right side: \(3 + 5 = 8\)
Left side = \(8\)
\(x = 3\) is the solution to the given equation.
B) \(2x = 4\)
\(\dfrac{2x}{2} = \dfrac{4}{2}\), 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:
Right side: \(2(2) = 4\)
Left side = \(4\)
\(x = 2\) is the solution to the given equation.
C) \(\dfrac{x}{3} = 2\)
\(3(\dfrac{x}{3}) = 3(2)\), 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:
Right side: \(\dfrac{6}{3} = 2\)
Left side = \(2\)
\(x = 6\) is the solution to the given equation.
D) \(0.2x = 1\)
\(x = \dfrac{1}{0.2}\), divide both sides by 0.2
\(x = 5\)
E) \(3x + 6 = 12\)
\(3x = 12 - 6 = 6\), subtract 6 from both sides
\(x = \dfrac{6}{3} = 2\), divide by 3
F) \(3(x + 2) + 2 = 8\)
\(3(x + 2) = 8 - 2 = 6\), subtract 2 from both sides
\(x + 2 = \dfrac{6}{3} = 2\), divide by 3
\(x = 2 - 2 = 0\), subtract 2
There are \(n\) boxes in a large bag and \(m\) toys in each box. What is the total number of toys in the bag?
To find the total, multiply the number of boxes by the number of toys per box:
$$\text{Total toys} = n \times m = nm$$
Rewrite the expression \(a \times a \times a - b \times b\) using exponents.
Multiplying a variable by itself is represented by an exponent:
$$a \times a \times a = a^3$$
$$b \times b = b^2$$
The expression is: $$a^3 - b^2$$
The length of a rectangle is given by \(x + 2\) and its width is equal to 3. Give a simplified expression of the area of this rectangle.
The area of a rectangle is Length \(\times\) Width:
$$\text{Area} = 3(x + 2)$$
Expand the expression:
$$\text{Area} = 3x + 6$$
Two thirds of students in a class study math and the remaining eight students do not study math. What is the total number of students in this class?
If \(\dfrac{2}{3}\) of the class studies math, then the fraction of the class that does not study math is:
$$1 - \dfrac{2}{3} = \dfrac{1}{3}$$
We are told this \(\dfrac{1}{3}\) of the class equals 8 students. Let \(T\) be the total number of students:
$$\dfrac{1}{3}T = 8$$
$$T = 8 \times 3 = 24$$
There are 24 students in total.
A car travels 60 kilometers in one hour. At the same rate, what distance will be covered by this car in \(x\) hours?
The rate is \(60\) km/h. Distance is rate multiplied by time:
$$\text{Distance} = 60 \times x = 60x \text{ kilometers}$$
Evaluate the expressions.
A) \(2^3 + 3^2\)
B) \(0.1^3\)
C) \(6 \times \dfrac{2}{3}\)
A) \(2^3 = 8\) and \(3^2 = 9\):
$$8 + 9 = 17$$
B) Multiply 0.1 by itself three times:
$$0.1 \times 0.1 \times 0.1 = 0.001$$
C) Multiply the whole number by the numerator and divide by the denominator:
$$\dfrac{6 \times 2}{3} = \dfrac{12}{3} = 4$$
A bag contains three red marbles; five blue marbles and seven green marbles. What is the ratio of blue marbles to the total number of marbles?
First, find the total number of marbles:
$$\text{Total} = 3 + 5 + 7 = 15$$
The ratio of blue marbles to the total is:
$$\dfrac{5}{15}$$
Simplify the ratio by dividing numerator and denominator by 5:
$$\dfrac{1}{3}$$
Solve the proportion: \(\dfrac{a}{3} = \dfrac{5}{18}\)
Cross-multiply to solve for \(a\):
$$18a = 3 \times 5$$
$$18a = 15$$
$$a = \dfrac{15}{18}$$
Simplify the fraction by dividing by 3:
$$a = \dfrac{5}{6}$$
Which of the following ordered pairs is a solution to the equation \(2x + 3y = 8\)?
A) \((0, 0)\)
B) \((4, 0)\)
C) \((1, 2)\)
Test each pair by substituting the x and y values into the equation:
A) \(2(0) + 3(0) = 0 \neq 8\) (Incorrect)
B) \(2(4) + 3(0) = 8 + 0 = 8\). Wait! Let's look closely: \(2(4) = 8\). So \((4,0)\) IS a valid solution.
C) \(2(1) + 3(2) = 2 + 6 = 8\). \((1,2)\) is ALSO a valid solution.
Correction Note: Both (4,0) and (1,2) are correct solutions to this linear equation.
List all factors of the following numbers.
A) \(4\)
B) \(12\)
C) \(50\)
A) Factors of 4: \(1, 2, 4\)
B) Factors of 12: \(1, 2, 3, 4, 6, 12\)
C) Factors of 50: \(1, 2, 5, 10, 25, 50\)
Find the greatest common factor (GCF) for each pair of the given numbers.
A) 6 and 3
B) 18 and 24
C) 50 and 60
A) Factors of 3: 1, 3. Factors of 6: 1, 2, 3, 6. The GCF is 3.
B) Factors of 18: 1, 2, 3, 6, 9, 18. Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24. The GCF is 6.
C) Factors of 50: 1, 2, 5, 10, 25, 50. Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. The GCF is 10.
Write the number "Six hundred seventy-two million two hundred fifty-nine" using the digits 0, 1, 2, ...9.
Break it down by periods:
Combine them: 672,000,259
Which of the following expressions are equivalent?
A) \(2(x + 3) - 2\)
B) \(2x + 4\)
C) \(3(x + 3) - x - 5\)
Simplify expressions A and C to compare them to B:
A) \(2(x + 3) - 2 = 2x + 6 - 2 = 2x + 4\)
C) \(3(x + 3) - x - 5 = 3x + 9 - x - 5 = 2x + 4\)
All three expressions (A, B, and C) simplify to \(2x + 4\), so they are all equivalent.
Find the lowest common multiple (LCM) for each pair of the given numbers.
A) 2 and 3
B) 7 and 14
C) 25 and 15
A) Multiples of 2: 2, 4, 6... Multiples of 3: 3, 6... The LCM is 6.
B) Multiples of 7: 7, 14, 21... Since 14 is a multiple of 7, the LCM is 14.
C) Multiples of 25: 25, 50, 75, 100... Multiples of 15: 15, 30, 45, 60, 75... The LCM is 75.
Add and/or subtract and simplify.
A) \(\dfrac{1}{3} + \dfrac{2}{3}\)
B) \(\dfrac{2}{5} - \dfrac{1}{7}\)
A) The denominators are the same, so add the numerators:
$$\dfrac{1 + 2}{3} = \dfrac{3}{3} = 1$$
B) Find a common denominator (35):
$$\dfrac{2 \times 7}{5 \times 7} - \dfrac{1 \times 5}{7 \times 5} = \dfrac{14}{35} - \dfrac{5}{35} = \dfrac{9}{35}$$
What is \(\dfrac{2}{3}\) of 21?
Multiply the fraction by the whole number:
$$\dfrac{2}{3} \times 21 = \dfrac{2 \times 21}{3} = \dfrac{42}{3} = 14$$
Alternative method: Divide 21 by 3 to get 7, then multiply by 2 to get 14.
What is 40% of \(\dfrac{1}{4}\)?
Convert 40% to a fraction or decimal. 40% = \(\dfrac{40}{100} = \dfrac{2}{5}\).
Now multiply by \(\dfrac{1}{4}\):
$$\dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{2}{20}$$
Simplify the fraction:
$$\dfrac{1}{10}$$ (or \(0.1\))
What is 20% of 50%?
Convert the percentages to decimals or fractions:
20% = 0.20
50% = 0.50
Multiply them:
$$0.20 \times 0.50 = 0.10$$
Convert back to a percentage: 0.10 = 10%.
How many seconds are there in one hour?
There are 60 minutes in an hour, and 60 seconds in a minute. Multiply to find the total seconds:
$$60 \times 60 = 3,600 \text{ seconds}$$
How many minutes are there in the month of January?
January has 31 days. There are 24 hours in a day, and 60 minutes in an hour.
$$\text{Total minutes} = 31 \times 24 \times 60$$
$$31 \times 1,440 = 44,640 \text{ minutes}$$
What is the location of each of the points: \((-2, 0)\), \((0, 3)\) and \((-2, -3)\) in a coordinate plane?
Order \(\dfrac{7}{5}\), \(\dfrac{12}{10}\), \(\dfrac{21}{20}\) and 111% from the smallest to the largest.
Convert all values to decimals for easy comparison:
Now order the decimals from smallest to largest: \(1.05 < 1.11 < 1.20 < 1.40\).
Replacing with the original values gives:
\(\dfrac{21}{20}\) , 111% , \(\dfrac{12}{10}\) , \(\dfrac{7}{5}\)