Solutions to Algebra Questions and Problems for Grade 7
Solutions to these Algebra Questions and Problems are presented.
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Evaluate each of the expressions for the given value(s) of the variable(s).
- 12 x 3 + 5 x 2 + 4 x - 6 for x = -1
- 2 a 2 + 3b 3 - 10 for a = 2 and b = -2
- (- 2 x - 1) / (x + 3) for x = 2
- 2 + 2 |x - 4| for x = - 4
Substitute each variable by its given numerical value and simplify.- 12 (-1) 3 + 5 (-1) 2 + 4 (-1) - 6 = 12 (-1) + 5 (1) - 4 - 6 = - 12 + 5 - 4 - 6 = - 17
- 2 (2) 2 + 3 (- 2) 3 - 10 = 2(4) + 3(-8) - 10 = 8 - 24 - 10 = - 26
- (- 2 (2) - 1) / ((2) + 3) = (- 4 - 1) / (5) = - 5 / 5 = - 1
- 2 + 2 |(- 4) - 4| = 2 + 2 |- 8 | = 2 + 2(8) = 2 + 16 = 18
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Expand and simplify each of the expressions below.
- - 2( x - 8) + 3 (x - 7)
- 2 (a + 1) + 5b + 3(a + b) + 3
- a (b + 3) + b (a - 2) + 2 a - 5 b + 8
- (1 / 2) (4x + 4) + (1 / 3)(6x + 12)
- 4 ( - x + 2 - 3(x - 2) )
Use the distribution rule a (b + c) = a × b + a × c = a (b) + a (c) to expand product terms and group like terms.- - 2( x - 8) + 3 (x - 7) = - 2(x) - 2(-8) + 3(x) + 3(-7)
= -2 x + 16 + 3 x - 21 = (- 2x + 3x) +(16 - 21) = x - 5 - 2 (a + 1) + 5b + 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 = (2 a + 3 a) + ( 5 b + 3 b) + (2 + 3) = 5 a + 8 b + 5 - a (b + 3) + b (a - 2) + 2 a - 5 b + 8 = a (b) + a (3) + b(a) + b(-2) + 2 a - 5 b + 8
= a b + 3 a + b a - 2 b + 2 a - 5 b + 8 = ( a b + b a) +(3 a + 2 a) + (-2 b - 5 b) + 8
= 2 a b + 5 a - 7 b + 8 - (1 / 2) (4x + 4) + (1 / 3)(6x + 12) = (1/2)(4 x) + (1/2)( 4 ) +(1/3)(6x) + (1/3)(12)
= 4 x / 2 + 4 / 2 + 6 x / 3 + 12 / 3 = 2 x + 2 + 2 x + 4
= (2 x + 2 x) + (2 + 4) = 4 x + 6 - 4 ( - x + 2 - 3(x - 2) ) = 4 ( - x) + 4 (2) + 4 (-3) (x - 2)
= - 4 x + 8 - 12 (x - 2) = - 4 x + 8 - 12 x + 24 = (- 4 x - 12 x) +(8 + 24) = - 16 x + 32
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Simplify each of the expressions below.
- x / y + 4 / y
- (2 x / 4 ) × (1 / 2)
- (3 x / 5 ) ÷ (x / 5)
We use the rules of multiplication, division and addition to simplify the expressions with numerators and denominators given below.- x / y + 4 / y = (x + 4) / y
- (2 x / 4 ) × (1 / 2) = (2 x × 1) / (4 × 2) = 2 x / 8 = x / 4
- (3 x / 5 ) ÷ (x / 5) = (3 x / 5) × (5 / x) = (3 x × 5) / (5 × x) = 1 5 x / 5 x = (15 / 5) (x / x) = 3 (1) = 3
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Simplify each of the expressions below.
- 3 x 2 × 5 x 3
- [ (2 y) 4 9 x 3] ÷ [ 4 y 4 (3 x) 2 ]
We use the rules of multiplication and division of expressions with exponents.- 3 x 2 × 5 x 3 = (3×5) (x 2 × x 3) = 1 5 x 2 + 3 = 1 5 x 5
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[ (2 y) 4 9 x 3] ÷ [ 4 y 4 (3 x) 2 ] = [ (2) 4 y 4 9 x 3 ] ÷ [ 4 y 4 (3 )2 x 2 ]
= [ 16 y 4 9 x 3 ] ÷ [ 4 y 4 9 x 2 ]
= [ (16 ×9 ) ÷ (4×9)] [ (y 4 x 3) ÷ ( y 4 x 2) ] = 4 y 4 - 4 x3 - 2 = 4 x
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Factor fully each of the expressions below.
- 9 x - 3
- 24 x + 18 y
- b x + d x
Find a common factor then use the rule of distribution a (b + c) = a b + a c from right to left a b + a c = a (b + c) to factor the given expression.-
9 x - 3
= 3 (3 x) + 3(1) 3 is a common factor
= 3(3x + 1) use distribution from right to left to factor -
24 x + 18 y
= 6(4 x) + 6 (3 y) 6 is a common factor
= 6(4 x + 3 y) factor -
b x + d x
= x(b) + x(d) x is a common factor
= x(b + d) factor
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Solve each of the equations below and check your answer.
- 2 x + 5 = 11
- 3 x = 6 / 5
- 3 (2 x + 2) + 2 = 20
Add, subtract, multiply by or divide by the same number both sides of the equation till it is solved.-
2 x + 5 = 11
2 x + 5 - 5 = 11 - 5 subtract 5 from both sides
2x = 6 simplify
2x / 2 = 6 / 2 divide both sides by 2
x = 3 simplify -
3 x = 6 / 5
5(3 x) = 5(6 / 5) multiply both sides by 5
15 x = 6 simplify
15 x / 15 = 6 / 15 divide both sides by 15
x = 6 / 15 simplify
x = 2 / 5 reduce fraction
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3 (2 x + 2) + 2 = 20
3(2x) + 3(2) + 2 = 20 expand
6 x + 6 + 2 = 20
6 x + 8 = 20 simplify
6x + 8 - 8 = 20 - 8 subtract 8 from both sides
6x = 12 simplify
6 x / 6 = 12 / 6 divide both sides by 6
x = 2 simplify
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Rewrite the expressions 3 × a × a × a - 5 × b × b using exponential.
Solution
Use exponents to rewrite
a × a × a = a 3 and b × b = b 2
Substitute in the given expression
3 × a × a × a - 5 × b × b = 3 a 3 - 5 b 2
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A rectangle has a length given by 2 x + 3 units, where x is a variable. The width of the rectangle is given by x + 1 units. Find the value of x if the perimeter of the rectangle is equal to 32.
Solution
Perimeter P of a rectangle is given by
P = 2 × length + 2 × width
Substitute length by 2 x + 3 and width by x + 1
P = 2 (2x + 3) + 2(x + 1)
Perimeter P is given as 32, hence
2 (2x + 3) + 2(x + 1) = 32
Solve the above equation
4 x + 6 + 2 x + 2 = 32 expand
6 x + 8 = 32 group like terms
6 x + 8 - 8 = 32 - 8 Subtract 8 from both sides
6 x = 24 simplify
6 x / 6 = 24 / 6 divide both sides by 6
x = 4
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A rectangle has a length given by 2x - 1 units, where x is a variable. The width of the rectangle is equal to 3 units. Find the value of x if the area of the rectangle is equal to 27.
Solution
Area A of a rectangle is given by
P = width × length
Substitute width by 3 and length by 2 x - 1
P = 3 (2x - 1)
Area A is given as 27, hence
3 (2x - 1) = 27
Solve the above equation
6 x - 3 = 27 expand
6 x - 3 + 3 = 27 + 3 add 3 to both sides
6 x = 30 simplify
6 x / 6 = 30 / 6 divide both sides by 6
x = 5
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45% of the students in a school are male? Find the ratio of the number of female to the total number of male students in this school.
Solution
If 45% of students are male then
100% - 45% = 55% of students are female
Ratio R of females to males is
R = 55% / 45% = 55/45
Reduce fraction
R = 11/9 or 11:9
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A car travels at the speed x + 30 kilometers in one hour, where x is an unknown. Find x if this car covers 300 kilometers in 3 hours?
Solution
Distance = time × speed, hence
300 = 3(x + 30)
Expand
300 = 3 x + 90
Subtract 90 from both sides of the equation
300 - 90 = 3 x + 90 - 90
210 = 3 x
210/ 3 = 3 x / 3
70 = x
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Solve the proportion : 4 / 5 = a / 16
Solution
Multiply both sides of the equation by the product of the two denominators: 5×16
5×16(4 / 5) = 5×16(a / 16)
Simplify
16×4= 5×a (which is also called cross product)
Divide both sides by 5
16×4 / 5 = 5 a / 5
Simplify
64 / 5 = a
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Find a if the ordered pair (2 , a + 2) is a solution to the equation 2 x + 2 y = 10?
Solution
Substitute the x (=2) and y (= a + 2) values of the ordered pair into the given equation
2 (2) + 2 (a + 2) = 10
Expand term in the above equation
4 + 2 a + 4 = 10
Simplify
2 a + 8 = 10
Solve for a
2 a + 8 - 8 = 10 - 8
2a = 2
a = 1
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Find the greatest common factor of the numbers 25 and 45.
Solution
Factors of 25 and 45 are
25: 1, 5, 25
45: 1, 3, 5 , 9, 15, 45
When we examine the two lists, 25 and 45 have two common factors: 1and 5 and the greatest is 5. Hence greatest common factor of the numbers 25 and 45 is
5
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Write the number " one billion, two hundred thirty four million, seven hundred fifty thousand two " using digits.
Solution
1234750002
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Write the number 393,234,000,034 in words.
Solution
three hundred ninety three billion, two hundred thirty four million thirty four
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Find the lowest common multiple to the numbers 15 and 35.
Solution
Multiples of 15 and 35 are
15: 15 , 30 , 45 , 60 , 75 , 90 , 105 , 120 , 135 , ....
35: 35 , 70 , 105 , 140, ...
When we examine the two lists and15 and 35 have the lowest common multiple equal to 105
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Find x if 2 / 3 of x is 30?
Solution
2 / 3 of x is 30 is mathematically written as
(2 / 3) × x = 30
We now need to solve the above equation. Multiply the two sides of the equations by 3
3 × (2 / 3) × x = 3 × 30
Simplify
2 x = 90
Divide both sides of the equation by 2
2x / 2 = 90 / 20
x = 45
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What is 20% of 1 / 3?
Solution
20% of 1 / 3 is written as
20% × (1 / 3)
= (20 / 100)×(1 / 3) write 20% as a fraction: 20 / 100
= (20×1) / (100 × 3) multiply fractions
= 20 / 300 simplify
= 1 / 15 reduce fraction
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Order 12 / 5 , 250% , 21 / 10 and 2.3 from the smallest to the largest.
Solution
Convert all number in decimal form
12 / 5 = 2.4
250% = 2.5
21 / 10 = 2.1
2.3 = 2.3
Order from smallest to largest
21/10 , 2.3 , 12 / 5 , 250%
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The sum of 3 positive consecutive integers is equal to 96. Find the largest of these numbers.
Solution
Let x be the smallest of these numbers. Since the three integers are consecutive, we can write them as: x , x + 1 and x + 2. Their sum is 96, hence
x + (x + 1) + (x + 2) = 96
x + x + 1 + x + 2 = 96 Expand terms in parentheses
(x + x + x) + (1 + 2) = 96 group like terms
3 x + 3 = 96 simplify
3 x + 3 - 3 = 96 - 3 subtract 3 from both sides of the equation
3 x = 93 simplify
3 x / 3 = 93 / 3 Divide by 3
x = 31 Solve
The largest of these numbers is x + 2, hence
x + 2 = 31 + 2 = 33
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Dany scored 93 in physics, 88 in mathematics, and a score in chemistry that is double his score in geography. The average score of all 4 courses is 79. What were his scores in chemistry and geography?
Solution
Let x be Dany's score in geography. His score in chemistry is double his score in geography and it is equal to
2 x
The average of all four scores is 79. Hence
(93 + 88 + x + 2 x) / 4 = 79
Multiply both sides of the equation by 4
4×(93 + 88 + x + 2 x) / 4 = 4×79
Simplify
93 + 88 + x + 2 x = 316
Group like terms
3 x + 181 = 316
Solve for x
3 x = 135
3 x / 3 = 135 / 3
x = 45
score in geography = x = 45
score in chemistry = 2 x = 2 × 45 = 90
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Linda scored a total of 265 points in mathematics, physics and English. She scored 7 more marks in mathematics than in English and she scored 5 more marks in physics than in mathematics. Find her scores in all three subjects.
Solution
Let x be Linda's score in English. Her score in mathematics is 7 more her score in English, hence her score in mathematics is
x + 7
Linda scored 5 more marks in physics than in mathematics, hence her score in physics is
(x +7) + 5 = x + 12
The total of all three scores is 265, hence
x + (x + 7) + (x + 12) = 265
group like terms
3 x + 19 = 265
Solve for x
3 x = 246
3 x / 3 = 246 / 3
x = 82
score in English = x = 82
score in mathematics = x + 7 = 82 + 7 = 89
score in physics = x + 12 = 82 + 12 = 94
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There are bicycles and cars in a parking lot. There is a total of 300 wheels including 100 small wheels for bicycles. How many cars and how many bicycles are there?
Solution
100 small wheels for bicycles gives
100 / 2 = 50 bicycles
A total of 300 wheels and 100 wheels for bicycles gives
300 - 100 = 200 wheels for cars
200 wheels for cars gives
200 / 4 = 50 cars.
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The difference between two numbers is 17 and their sum is 69. Find the largest of these two numbers.
Solution
Let x be the smallest number, then the largest number is
x + 17 since their difference is 17
The sum of the two numbers is 69, hence
x + (x + 17) = 69
group like terms and solve
2 x + 17 = 69
2 x = 52
2 x / 2= 52 / 2
x = 26
the largest of these two numbers is
x + 17 = 43
Links and References
Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With AnswersHigh 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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