algebraic expressions 7th grade worksheets | worksheet 1
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1. . Identify the algebraic expression among the following.a) b) 2 3x−
c)1
32
x =
d)2x y=
2. Classify the following algebraic expressions into monomials,binomials, and trinomials.
a) 23x yz
b) x y z+ −
c) 2 2x y−
3. Simplify the following algebraic expressions by combining liketerms.a) 3 5 7x x x+ −
b) 1
7 32
x x− −
4. Mia’s piggy bank has only dimes and quarters. Write anexpression for the amount of money (in dollars) in her piggybank if x is the number of dimes and y is the number of quarters.
2 3 0x+ =
Algebraic Expressions 7th Grade Worksheets
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5. Find the sum. a) (3.2 ) ( 10 2.5)x x− + − +
b)
1 15 7
3 2m m
− + +
6. Find the algebraic expression that corresponds to the perimeter of the following rectangle.
7. Find the difference. a)( 3 8) (5 12)g g− + − +
b)
35 2
4 4
51x x
+ − − +
8) Simplify the following using distributive property.
7 9(3 6) 2 5
2 2y y
− − + −
9) Subtract 25 4x x− + from the sum of 22 5x x− and 27 2x − .
10) What should be added to 2 23x xy y+ + to get
2 23 3x y− ? 11) What should be subtracted from 3 5 10a b− + in order to get
7 2 5a b− + − ? 12) Factor out the GCF. a) 16 20x y− b) 14 21 7m n+ − 13) The area of a rectangular shaped swimming pool is 3 9x−
square units. Find the possible dimensions of the rectangle as algebraic expressions.
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14) Factor out the coefficient of the variable term.
a) 5 5
8 4k +
b) 7
142
m+
15) A square shaped door mat has a perimeter of 8 12x− units. Find an expression for the length of the mat.
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1) Option b 2) a) Monomial b) Trinomial c) Binomial
3) a) Monomial b) Trinomial c) Binomial
4) 0.10x + 0.25y
5) a) 5.7-11x b) (5/6)m+2
6) 12m+8
7) a) -8g-4 b) 2x-7
8) -39y/2 +31 9) 28 7x x− −
10)
24y xy− −
11) 10a-7b+15 12)a)4(4x-5y)
13) 3 units and (x-3) units
14)a) 5(k+2)/8 b) 7(m+4)/2
15) 2x-3
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1. Identify the algebraic expression among the following. a) 2 3 0x+ = b) 2 3x−
c)1
32
x =
d)2x y= Solution: An algebraic expression is a combination of variables and constants without any “=” symbol. Hence the option is b). Note: a), c), and d) are called the equations NOT the algebraic expressions. 2. Classify the following algebraic expressions into monomials, binomials, and trinomials.
a) 23x yz
b) x y z+ −
c) 2 2x y−
Solution: The algebraic expressions are called monomials if they have one term, binomials if they have two terms, and trinomials if they have three terms. Hence:
a) 23x yz is a monomial.
b) x y z+ − is a trinomial.
c) 2 2x y− is a binomial.
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3. Simplify the following algebraic expressions by combining like terms. a) 3 5 7x x x+ −
b) 1
7 32
x x− −
Solution: a) All terms in the given expression are like terms. We know that 3 5 7 1+ − = . Hence, 3 5 7 (3 5 7) 1 (or) x x x x x x+ − = + − = . b)
1 1 57 3 3 7 7
2 2 2x x x x x
− − − = − − = −
4. Mia’s piggy bank has only dimes and quarters. Write an expression for the amount of money (in dollars) in her piggy bank if x is the number of dimes and y is the number of quarters.
Solution: The number of dimes is x and each dime is of worth $0.10. So the total worth of dimes is 0.10 x . The number of quarters is y and each quarter is of worth $0.25. So the total worth of quarters is 0,25 x .
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So the algebraic expression that represents the total money in the piggy bank is 0.10 0.25x y+ dollars. 5. Find the sum. a) (3.2 ) ( 10 2.5)x x− + − +
b)
1 15 7
3 2m m
− + +
Solution: a) We will find the sum by combining the like terms. (3.2 ) ( 10 2.5) (3.2 2.5) ( 10 )
5.7 11
x x x x
x
− + − + = + + − −
= − b) We will find the sum by combining the like terms.
1 1 1 15 7 ( 5 7)
3 2 3 2
52
6
m m m m
m
− + + = + + − +
= +
6. Find the algebraic expression that corresponds to the perimeter of the following rectangle.
Solution: The length of the rectangle is 4 5l m= + The width of the rectangle is 2 1w m= − The perimeter is, 2( ) 2(4 5 2 1)
2(6 4)
12 8
l w m m
m
m
+ = + + −
= +
= +
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7. Find the difference. a)( 3 8) (5 12)g g− + − +
b)
35 2
4 4
51x x
+ − − +
Solution: a) We will just distribute the minus sign and perform the usual addition. ( 3 8) (5 12) 3 8 5 12
( 3 5 ) (8 12)
8 4
g g g g
g g
g
− + − + = − + − −
= − − + −
= − − b) Here also, we will just distribute the minus sign and perform the usual addition.
( )
3 5 3 55 12 5 12
4 4 4 4
3 55 12
4 4
2 7
x x x x
x x
x
+ − − + = + + −
= + + −
= − 8) Simplify the following using distributive property.
7 9(3 6) 2 5
2 2y y
− − + −
Solution: The distributive property is ( )a b c ab ac+ = + .
7 9 21(3 6) 2 5 21 10 9
2 2 2
219 (21 10)
2
3931
2
yy y y
yy
y
− − + − = − + + −
= − − + +
= − +
9) Subtract 25 4x x− + from the sum of 22 5x x− and 27 2x − . Solution: Using the given instruction,
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( ) ( ) ( )
( )
2 2 2
2 2
2
2 5 7 2 5 4
9 5 2 5 4
8 7
x x x x x
x x x x
x x
− + − − − +
= − − − + −
= − − 10) What should be added to
2 23x xy y+ + to get 2 23 3x y− ?
Solution:
Here, the sum is 2 23 3x y− and one of the addends is
2 23x xy y+ + . To find the other addend, we have to subtract the given addend from the sum.
( ) ( )2 2 2 2
2 2 2 2
2
3 3 3
3 3 3
4
x y x xy y
x y x xy y
y xy
− − + +
= − − − −
= − − 11) What should be subtracted from 3 5 10a b− + in order to get
7 2 5a b− + − ? Solution: Here the difference is 7 2 5a b− + − and the minuend is 3 5 10a b− + . To get the subtrahend, we subtract difference from the minuend. (3 5 10) ( 7 2 5)
3 5 10 7 2 5
10 7 15
a b a b
a b a b
a b
− + − − + −
= − + + − +
= − + 12) Factor out the GCF. a) 16 20x y− b) 14 21 7m n+ − Solution: a) The GCF of 16 and 20 is 4. So 16 20 4(4 5 )x y x y− = − b) The GCF of 14, 21, and -7 is 7. So 14 21 7 7(2 3 )m n m n+ − = + − 13) The area of a rectangular shaped swimming pool is 3 9x−
square units. Find the possible dimensions of the rectangle as algebraic expressions.
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Solution: The area of the given swimming pool = 3 9 3( 3) Length Widthx x− = − = So the possible dimensions are 3 units and ( 3)x − units. 14) Factor out the coefficient of the variable term.
a) 5 5
8 4k +
b) 7
142
m+
Solution:
a) Here, the variable is k and its coefficient is 5
8 .
( )5 5 5
28 4 8
k k+ = +
b) Here, the variable is m and its coefficient is 7
2 . 7 7
14 ( 4)2 2
m m+ = +
15) A square shaped door mat has a perimeter of 8 12x− units. Find an expression for the length of the mat. Solution: The perimeter of the mat = 8 12 4(2 3)x x− = − . If we divide it by 4, we get its length.
So the length of the mat = 4(2 3)
(2 3)4
xx
−= −
units.