Maharashtra Board Class 7 Math Solution Chapter 8 – Algebraic Expressions and Operations on them
Balbharati Maharashtra Board Class 7 Math Solution Chapter 8: Algebraic Expressions and Operations on them. Marathi or English Medium Students of Class 7 get here Algebraic Expressions and Operations on them full Exercise Solution.
Std |
Maharashtra Class 7 |
Subject |
Math Solution |
Chapter |
Algebraic Expressions and Operations on them |
Practice Set 32
¤ Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.
(i) 7x
(ii) 5y – 7z
(iii) 3x^{3}-5x^{2} – 11
(iv) 1 – 8a – 7a^{2} – 7a^{3}
(v) 5m – 3
(vi) a
(vii) 4
(viii) 3y^{2} – 7y + 5
ANS:
Expressions containing only one term is called as Monomial.
Expressions containing two term is called as Binomial.
Expressions containing three term is called as Trinomial.
Expressions containing more than three term is called as polynomial.
Now, we are classify given algebraic expressions as monomials, binomials, trinomials or polynomials.
Monomial:
(i) 7x
(vi) a
(vii) 4
Binomial:
(ii) 5y – 7z
(v) 5m – 3
Trinomial:
(iii) 3x^{3}-5x^{2} – 11
(viii) 3y^{2} – 7y + 5
Polynomial:
(iv) 1 – 8a – 7a^{2} – 7a^{3}
Practice Set 33
¤ Add.
(i) 9p + 16q; 13p + 2q
ANS:
Here we are adding 9p + 16q and 13p + 2q these two algebraic expressions.
We adding p variable terms with p variable and q variable terms with q variable.
9p + 16q
+ 13p + 02q
________________________________
22p + 18q
Addition of 9p + 16q; 13p + 2q is 22p + 18q
(ii) 2a + 6b + 8c; 16a + 13c + 18b
ANS:
Here we are adding 2a + 6b + 8c and 16a + 13c + 18b these two algebraic expressions.
We adding a variable terms with a variable, b variable terms with b variable and c variable terms with c variable.
We rearranging 2^{nd} expression
16a + 18b + 13c
2a + 6b + 8c
+16a + 18b + 13c
——————————–
18a + 24b + 21c
Addition of 2a + 6b + 8c; 16a + 13c + 18b is 18a + 24b + 21c
(iii) 13x^{2}-12y^{2}; 6x^{2}-8y^{2}
ANS:
Here we are adding 13x^{2}-12y^{2 }and 6x^{2}-8y^{2} these two algebraic expressions.
We adding x^{2} variable terms with x^{2} variable and y^{2} variable terms with y^{2} variable.
When there is two negative sign and addition operation is there we have to add both and give negative sign.
13x^{2}-12y^{2}
+ 6x^{2}– 8y^{2}
^{———————————————————}
19 x^{2} – 20 y^{2}
Addition of 13x^{2}-12y^{2}; 6x^{2}-8y^{2} is 19 x^{2} – 20 y^{2}
(iv) 17a^{2} b^{2} + 16c; 28c-28a^{2} b^{2}
ANS:
Here we are adding 17a^{2} b^{2} + 16c and 28c-28a^{2} b^{2} these two algebraic expressions.
We adding a^{2} b^{2} variable terms with a^{2} b^{2} variable and c variable terms with c variable.
We rearranging 2^{nd} expression
-28a^{2} b^{2} + 28c
When there is one negative sign and one positive sign and addition operation is there we have to subtract smaller number from larger number and give sign of larger number.
17a^{2} b^{2} + 16c
+ -28a^{2} b^{2} + 28c
——————————-
-11 a^{2} b^{2} + 44c
Addition of 17a^{2} b^{2} + 16c; 28c-28a^{2} b^{2 }is -11 a^{2} b^{2} + 44c
(v) 3y^{2}-10y + 16; 2y-7
ANS:
Here we are adding 3y^{2}-10y + 16 and 2y-7 these two algebraic expressions.
We adding y variable terms with y variable and constant term with constant term.
We writing 2^{nd} expression as 0y^{2}+ 2y – 7
When there is one negative sign and one positive sign and addition operation is there we have to subtract smaller number from larger number and give sign of larger number.
3y^{2}-10y + 16
+ 0y^{2}+ 2y – 7
———————————-
3y^{2} – 8y + 9
Addition of 3y^{2}-10y + 16; 2y-7 is 3y^{2} – 8y + 9.
(vi) -3y^{2} + 10y-16; 7y^{2} + 8
ANS:
Here we are adding -3y^{2} + 10y-16 and 7y^{2} + 8 these two algebraic expressions.
We adding y^{2} variable terms with y^{2} variable and constant term with constant term.
We writing 2^{nd} expression as 7y^{2} + 0y + 8
When there is one negative sign and one positive sign and addition operation is there we have to subtract smaller number from larger number and give sign of larger number.
-3y^{2} + 10y-16
+ 7y^{2} + 0y + 8
————————————
4y^{2}+ 10y – 8
Addition of -3y^{2} + 10y-16; 7y^{2} + 8 is 4y^{2}+ 10y – 8
Practice Set 34
¤ Subtract the second expression from the first.
(i) (4xy – 9z) ; (3xy – 16z)
ANS:
Here we are subtracting (3xy – 16z) from (4xy – 9z)
We subtract xy term from xy term and z term from z term.
(4xy – 9z)
– (3xy – 16z)
———————
When negative sign is in front of any bracket sign changes of the terms inside the bracket.
– (3xy – 16z) becomes – 3xy + 16z
When there is one negative sign and one positive sign and subtraction operation is there we have to subtract smaller number from larger number and give sign of larger number.
4xy – 9z
– 3xy + 16z
——————————
xy + 7z.
Subtraction of (4xy – 9z) ; (3xy – 16z) is xy + 7z.
(ii) (5x + 4y + 7z) ; (x + 2y + 3z)
ANS:
Here we are subtracting (x + 2y + 3z) from (5x + 4y + 7z).
We subtract x term from x term, y term from y term and z term from z term.
(5x + 4y + 7z)
-(x + 2y + 3z)
———————————-
When negative sign is in front of any bracket sign changes of the terms inside the bracket.
-(x + 2y + 3z) becomes – x – 2y – 3z.
5x + 4y + 7z
– x – 2y – 3z
—————————————-
4x + 2y + 4z
Subtraction of (5x + 4y + 7z) ; (x + 2y + 3z) is 4x + 2y + 4z.
(iii) (14x^{2} + 8xy + 3y^{2} ) ; (26x^{2} – 8xy – 17y^{2} )
ANS:
Here we are subtracting (26x^{2} – 8xy – 17y^{2} ) from (14x^{2} + 8xy + 3y^{2} )
We subtract x^{2} term from x^{2} term, xy term from xy term and y^{2} term from y^{2} term.
(14x^{2} + 8xy + 3y^{2} )
-(26x^{2} – 8xy – 17y^{2} )
—————————————
When negative sign is in front of any bracket sign changes of the terms inside the bracket
-(26x^{2} – 8xy – 17y^{2} ) becomes -26x^{2} + 8xy + 17y^{2}
14x^{2} + 8xy + 3y^{2}
-26x^{2} + 8xy + 17y^{2}
^{————————————————————————}
-12 x^{2} +16xy + 20 y^{2}
Subtraction of (14x^{2} + 8xy + 3y^{2}) ; (26x^{2} – 8xy – 17y^{2} ) is -12 x^{2} +16xy + 20 y^{2}
(iv) (6x^{2} + 7xy + 16y^{2}) ;( 16x^{2} – 17xy)
ANS:
Here we are subtracting ( 16x^{2} – 17xy) from (6x^{2} + 7xy + 16y^{2})
We subtract x^{2} term from x^{2} term, xy term from xy term and y^{2} term from y^{2} term.
We rearranging 2^{nd} expression ( 16x^{2} – 17xy + 0y^{2})
(6x^{2} + 7xy + 16y^{2})
-( 16x^{2} – 17xy + 0y^{2})
———————————-
When negative sign is in front of any bracket sign changes of the terms inside the bracket
-( 16x^{2} – 17xy + 0y^{2}) becomes -16x^{2} + 17xy – 0y^{2}
6x^{2} + 7xy + 16y^{2}
-16x^{2} + 17xy – 0y^{2}
^{—————————————————————}
-10x^{2 }+ 24xy + 16y^{2}
Subtraction of (6x^{2} + 7xy + 16y^{2}) ;( 16x^{2} – 17xy) is -10x^{2 }+ 24xy + 16y^{2}
(v) (4x + 16z) ; (19y – 14z + 16x)
ANS:
Here we are subtracting (19y – 14z + 16x) from (4x + 16z).
We subtract x term from x term, y term from y term and z term from z term.
We rearranging 1^{st} expression (4x + oy + 16z)
Also we rearranging 2^{nd} expression (19y – 14z + 16x) as 16x + 19y – 14z
(4x + oy + 16z)
-(16x + 19y – 14z)
——————————-
When negative sign is in front of any bracket sign changes of the terms inside the bracket
-(16x + 19y – 14z) becomes -16x – 19y + 14z
4x + oy + 16z
-16x – 19y + 14z
——————————-
-12x – 19y + 30z
Subtraction of (4x + 16z) ; (19y – 14z + 16x) is -12x – 19y + 30z
Practice Set 35
1.) Multiply.
(i) 16xy × 18xy
ANS:
Here we have to multiply 16xy × 18xy.
When there is multiplication of monomial with monomial we have to first do multiplication of coefficients of expressions and then we have to do multiplication of variables.
Coefficient of 1^{st} monomial is 16.
Coefficient of 2^{st} monomial is 18.
Variable of 1^{st} monomial is xy.
Variable of 2^{nd} monomial is xy.
16xy × 18xy
16 x 18 x (xy) x (xy)
= 288x^{2}y^{2}
Multiplication of 16xy × 18xy is 288x^{2}y^{2}
(ii) 23xy^{2} × 4yz^{2}
ANS:
Here we have to multiply 23xy^{2} × 4yz^{2}
When there is multiplication of monomial with monomial we have to first do multiplication of coefficients of expressions and then we have to do multiplication of variables.
Coefficient of 1^{st} monomial is 23.
Coefficient of 2^{st} monomial is 4.
Variable of 1^{st} monomial is xy^{2}.
Variable of 2^{nd} monomial is yz^{2}.
23xy^{2} × 4yz^{2}
= 23 x 4 x (xy^{2}) x (yz^{2})
= 92xy^{3} z^{2}
Multiplication of 23xy^{2} × 4yz^{2 }is 92xy^{3} z^{2}
(iii) (12a + 17b) × 4c
ANS:
Here we have to multiply (12a + 17b) × 4c
When there is multiplication of binomial with monomial we have to do multiplication of monomial and each term of binomial expressions.
(12a + 17b) × 4c
= (4c x 12a) + ( 4c x 17b)
= (48ac + 68bc)
Multiplication of (12a + 17b) × 4c is (48ac + 68bc).
(iv) (4x + 5y) × (9x + 7y)
ANS:
Here we have to multiply (4x + 5y) × (9x + 7y)
When there is multiplication of binomial with binomial we have to do multiplication of each term of first binomial with the 2^{nd} binomial expressions.
(4x + 5y) × (9x + 7y)
= 4x X (9x + 7y) + 5y x (9x + 7y)
= 36x^{2} + 28xy + 45 xy + 35y^{2}
= 36x^{2 }+ 73 x^{2}y^{2} + 35y^{2}
Multiplication of (4x + 5y) × (9x + 7y) is 36x^{2 }+ 73 x^{2}y^{2} + 35y^{2}.
2.) A rectangle is (8x + 5) cm long and (5x + 3) cm broad. Find its area.
ANS:
Given that
Length of rectangle is (8x + 5) cm
Breadth of rectangle is (5x + 3) cm
We know,
The area of rectangle = Length x Breadth
The area of rectangle = (8x + 5) cm x (5x + 3) cm
When there is multiplication of binomial with binomial we have to do multiplication of each term of first binomial with the 2^{nd} binomial expressions.
(8x + 5) x (5x + 3)
= 8x x (5x + 3) + 5 x (5x + 3)
= 40x^{2}+ 24x + 25x + 15
= 40x^{2}+ 49x + 15
The area of rectangle is 40x^{2}+ 49x + 15
Practice Set 36
1.) Simplify (3x-11y) – (17x + 13y) and choose the right answer.
(i) 7x-12y (ii) -14x -54y (iii) -3 (5x + 4y) (iv) -2 (7x + 12y)
ANS:
Here we are subtracting (17x + 13y) from (3x-11y)
We subtract x term from x term and y term from y term.
(3x-11y)
– (17x + 13y)
———————
When negative sign is in front of any bracket sign changes of the terms inside the bracket.
– (17x + 13y) becomes -17x – 13y
3x-11y
-17x – 13y
——————————
-14x – 24y
We take -2 common we get,
-2 (7x + 12y)
Simplification of (3x – 11y) – (17x + 13y) is -2 (7x + 12y)
(iv) -2 (7x + 12y)
2.) The product of (23 x^{2} y^{3} z) and (-15x^{3}yz^{2}) is……………….. .
(i) -345 x^{5 }y^{4 }z^{3} (ii) 345 x^{2} y^{3} z^{5 }(iii) 145 x^{3} y^{2} z (iv) 170 x^{3} y^{2} z^{3 }
ANS:
Here we have to multiply (23 x^{2} y^{3} z) and (-15x^{3}yz^{2}).
When there is multiplication of monomial with monomial we have to first do multiplication of coefficients of expressions and then we have to do multiplication of variables.
Coefficient of 1^{st} monomial is 23.
Coefficient of 2^{st} monomial is – 15.
Variable of 1^{st} monomial is x^{2} y^{3} z.
Variable of 2^{nd} monomial is x^{3} y z^{2}.
16xy × 18xy
23 x – 15 x (x^{2} y^{3} z) x (x^{3} y z^{2})
= -345 x^{5 }y^{4 }z^{3}
The product of (23 x^{2} y^{3} z) and (-15x^{3}yz^{2}) is-345 x^{5 }y^{4 }z^{3}
(i) -345 x^{5 }y^{4 }z^{3}
3.) Solve the following equations.
(i) 4x + 1/ 2 = 9/ 2
ANS:
We have to solve 4x + 1/ 2 = 9/ 2 this equation.
4x + 1/ 2 = 9/ 2
4x = 9/ 2 – 1/ 2
When denominator is same we doing subtraction of numerator only.
4x = 8 / 2
4x = 4
X = 4/ 4
X = 1
Value of x is 1.
(ii) 10 = 2y + 5
ANS:
We have to solve 10 = 2y + 5this equation.
10 = 2y + 5
2y = 10 – 5
2y = 5
y = 5/2
Value of y is 5/2.
(iii) 5m – 4 = 1
ANS:
We have to solve 5m – 4 = 1this equation.
5m – 4 = 1
5m = 1 + 4
5m = 5
m = 5 / 5
m = 1
Value of m is 1.
(iv) 6x-1 = 3x + 8
ANS:
We have to solve6x-1 = 3x + 8 this equation.
6x-1 = 3x + 8
We are taking variable term one side and constant term on other side.
When we move from one side to another and in between = sign is there then its sign changes.
6x- 3x = 8 + 1
3x = 9
X = 9/3
X= 3
Value of x is 3.
(v) 2 (x-4) = 4x + 2
ANS:
We have to solve 2 (x-4) = 4x + 2 this equation.
2 (x-4) = 4x + 2
Firstly we are multiplying monomial with binomial.
2 (x-4) = 2x – 8
Now,
2x – 8 = 4x + 2
We are taking variable term one side and constant term on other side.
When we move from one side to another and in between = sign is there then its sign changes.
2x – 4x = 2 + 8
-2x = 10
X = 10 / -2
X = -5
Value of x is -5.
(vi) 5 (x + 1) = 74
ANS:
We have to solve 5 (x + 1) = 74 this equation.
5 (x + 1) = 74
Firstly we are multiplying monomial with binomial.
5 (x + 1) = 5x + 5
Now,
5x + 5 = 74
5x = 74 – 5
5x = 69
X = 69 /5
Value of x is 69 /5.
4.) Rakesh’s age is less than Sania’s age by 5 years. The sum of their ages is 27 years. How old are they?
ANS:
Given that,
Rakesh’s age is less than Sania’s age by 5 years.
Let,
We are taking Sania’s age is x years.
Therefore
Rakesh’s age is less than Sania’s age by 5 years.
Rakesh’s age = (x – 5) years.
Now,
The sum of their ages is 27 years
X + (x – 5) = 27
2x – 5 = 27
2x = 27 + 5
2x = 32
X = 32/2
X = 16 years.
Bur x is Sania’s age
Sania’s age is 16 years.
Rakesh’s age is less than Sania’s age by 5 years.
Rakesh’s age = 16 – 5 = 11 years.
Rakesh’s age =11 years.
5.) When planting a forest, the number of jambhul trees planted was greater than the number of ashoka trees by 60. If there are altogether 200 trees of these two types, how many jambhul trees were planted?
ANS:
The number of jambhul trees planted was greater than the number of ashoka trees by 60.
Let,
The number of ashoka trees planted is x.
Therefore
The number of jambhul trees planted = x + 60
There are altogether 200 trees
We are adding both trees.
X + (x + 60) = 200
2x + 60 = 200
2x = 200 – 60
2x = 140
X = 140 / 2
X = 70
But,
X is the number of ashoka trees planted.
The number of ashoka trees planted is 70.
The number of jambhul trees planted was greater than the number of ashoka trees by 60.
= 70 + 60
= 130.
The number of jambhul trees planted= 130.
6 .) Virat made twice as many runs as Rohit. The total of their scores is 2 less than a double century. How many runs did each of them make?
ANS:
Given that,
Virat made twice as many runs as Rohit.
Let,
Rohit score x runs.
Therefore
Viratscore 2x runs.
The total of their scores is 2 less than a double century.
Addition of their scores is 198.
X + 2x = 198
3x = 198
X = 198 /3
X = 66
But,
Rohit score x runs.
Rohit score 66 runs.
Virat made twice as many runs as Rohit.
Virat score 2x runs.
Virat score 2x 66 runs.
Virat score 132 runs.