Algebraic Expression and Operations
Based on Maharashtra Board Class 7th Mathematics Chapter 8-Notes, Solutions, Test, Videos, PDF
Notes
| An algebraic expression is a combination of constants and variables connected by the signs +, -, Ć,Ā Ć·,Ā |
\(5x+4y\),\( 4n+3\),\( 5(p+q)\), are algebraic expressions.
In the above examples \(x; y; n; p; q\) are variables.
| Symbols which do not have a fixed value and take various values are called variables. |
| The quantities with fixed numerical values are called constants. |
Terms in an algebraic expression are separated by + or - signs.
\(5x-4y+7z\) is an algebraic expression.
\(5x, 4y\) and \(7z\) are its terms
Like Terms -Ā The terms having the same variables with the same powers are called like terms.
Ex. (1)- \(6x \), \(\frac{-4}{7}x\),Ā \(3x\)Ā
Ā Ā Ā (2)- \(3x^2y\) , \(-2x^2y\),Ā Ā \(\frac{3}{8}x^2y\)
Unlike Terms -Ā The terms not having the same variables are called unlike terms
Ex. (1)- \(6xy \), \(\frac{-4}{7}y\),Ā \(3xz\)Ā
Ā Ā Ā (2)- \(3x^2y\) , \(-2z^2y\),Ā Ā \(\frac{3}{8}x^2z\)
Monomial : An algebraic expression containing only one term is called a monomial.
Ex:Ā \(3x^2y\),Ā Ā \(3xy\),Ā \(xy\)
Binomial :Ā An algebraic expression containing two term is called a binomial.
Ex.: \(3x^2y+7x\),Ā Ā \(4xy+7x\),Ā \(xy+mn\)
Trinomial :Ā An algebraic expression containing three term is called a trinomial.
Ex: \(ax^2 + bx + c\),Ā \(x^2 + y^3 + c\)
Polynomial :Ā An algebraic expression containing more than three term is called a Polynomial
Ex.: \(ax^2 + bx + cy+2Z\),Ā \(pq^2 + nm + zy+2\)
Like terms are added as we would add up things of the same thing
Ex - 2 Mango + 4 Mango = 6 Mango .......... (2+4) Mango
Addition of Monomial Expressions
Add -(i)Ā \(-3x, 4x\) and \(7x\)
\(-3x+4x+7x\)
= \((-3+4+7)x\)
= \(8x\)
Add- (ii) \(6x^2, -5x^2\) and \(-4x^2\)
\(6x^2-5x^2-4x^2\)
= \((6-5-4)x^2\)
= \(-3x^2\)
Addition of Binomial Expressions
Add (i) \(2x-3y\); \(5x+10y\)
- Horizontal Arrangement
\(2x-3y+5x+10y\)
= \(2x+5x-3y+10y\)Ā Ā Ā Ā Ā .......... Arrange like terms togetherĀ
= \(7x+7y\)
- VerticalĀ Arrangement
.Ā Ā Ā \(2x-3y\)
.Ā Ā +
.Ā Ā Ā \(5x+10y\)
.Ā Ā --------------
.Ā Ā Ā \(7x+7y\)
Rule : We know that to subtract an integer from another integer is to add its opposite integer to other integer, this same rule is applicable for subtraction of algebraic expressions.Ā
Ex (i) Subtract \(7x\) from \(12x\)Ā
\(12x - 7x\)
= \((12 + (-7))x\)
= \(5x\)
(ii) Subtract \(-7x\) from \(12x\)Ā
= \(12x - (-7x)\)
= \((12 + (+7))x\)
= \((12 +7)x\)
= \(19x\)
(iii) Subtract \(10xy - 8pq\) from \(15pq - 7xy\)Ā
Horizontal arrangement
\((15pq - 7xy) - (10xy - 8pq)\)
=Ā \((15pq - 7xy) + (-10xy + 8pq)\)Ā Ā ....... The opposite ofĀ Ā \(10xy - 8pq\)
=Ā Ā \(15pq + 8pq - 7xy -10xy \)Ā
= \(23pq -17xy\)
Vertical arrangement
.Ā Ā Ā \(15pq - 7xy\)Ā
. \(-\)
.Ā Ā \(- 8pq + 10xy \)
.Ā Ā Ā \(+ \)Ā Ā Ā Ā Ā \( - \)Ā Ā Ā Ā Ā Change the signĀ
.Ā Ā -------------------------
.Ā Ā Ā \(23pq -17xy\)Ā Ā
Change the sign of every term in the expression to be subtractedĀ Ā .
Multiplying Monomial by another monomialĀ
- When multiplying two monomials,first multiply the coefficients along with the sign, then multiply the variables.
Ex. 1.Ā Ā Multiply \(4x\)Ā by \(7y\)
\(4x \times 7y\)Ā Ā .... write the multiplication signĀ
= \( 4 \times x \times 7 \times y\)
=Ā \( (4 \times 7) (x \times y)\)Ā Ā ........(write coefficients in one bracket and variables in another bracket)
= \(28xy\)
Ex. 2.Ā Ā Multiply \(-8x^2\)Ā by \(7y^2\)
\(8x^2 \times 7y^2\)Ā Ā .... write the multiplication signĀ
= \( 8 \times x^2 \times 7 \times y^2\)
=Ā \( (8 \times 7) (x^2 \times y^2)\)Ā Ā ........(write coefficients in one bracket and variables in another bracket)
= \(56x^2y^2\)
Multiplying Binomial by another monomialĀ
Ex. 1.Ā Ā Multiply \((3x + 2)\)Ā by \(2x\)
\((3x + 2) \times (2x)\)
\(3x \times 2x + 2 \times 2x\)
\(6x^2 + 4x\)
Ex. 2.Ā Ā Multiply \(6Z\) by \((5x - 7y)\)
\(6z \times (5x - 7y)\)
\(6z \times 5x + 6z \times -7y\)
\(30zx - 42xy\)
Multiplying Binomial by another BinomialĀ
Ex. 1.Ā Ā Multiply \((4x + 7y)\)Ā by \(3x + 2y\)
\((4x + 7y) \times (3x + 2y)\)
= \(4x( 3x + 2y) + 7y(3x+ 2y)\)
= \(4x \times 3x + 4x \times 2y + 7y \times 3x +Ā 7y \times 2y \)
= \(12x^2 + 8xy + 21xy + 14y^2\)
= \(12x^2 + 29xyĀ + 14y^2\)Ā Ā ...... addition of like terms
( Note student can solve example by vertical arrangement method also )
An equation is an equality containing a variable /Ā variables.
An equation containing only oneĀ variable with index 1 are known as simple equations in one variable.
Ex . Solve the following equations.
(i) \(x + 7 = 4\)
\(x + 7 = 4\)
\(x + 7 - 7= 4 - 7\)
\( \dot .. xĀ =Ā -3\)
Ans :\(xĀ =Ā -3\)
(ii) \(4p = 12\)
\(4p = 12\)
\( \frac{4p}{4} = {12\over 4}\)
\( \dot .. pĀ =Ā 3\)
Ans :Ā \(pĀ =Ā 3\)
(iii) \(m - 7 = 4\)
\(m - 7 = 4\)
\(m - 7 + 7= 4 + 7\)
\( \dot .. mĀ =Ā 11\)
Ans :Ā \( mĀ =Ā 11\)
(iv) \( \frac{t}{3} = 9\)
\( \frac{t}{3} = 9\)
\( \frac{t}{3} \times 3 = 9 \times 3\)
\( \dot .. tĀ =Ā 27\)
Ans :Ā \( tĀ =Ā 27\)
- When there are variables on both sides of an equation, eliminate those from the right hand side of equation.
- If a term is transposed from one side to other side of the '=' sign of an equation that term's sign must be changed.
Solve : \(7a + 2 = 26 + 3a\)
\(7a + 2 = 26 + 3a\)
\(7a + 2 - 3a = 26 \)
\(4a + 2Ā = 26 \)
\(4aĀ = 26 - 2 \)
\(4aĀ = 24 \)
\(4a \div 4 = 24 \div 4 \)
\(aĀ = 6 \)
Ans:Ā \(aĀ = 6 \)
Solve : The sum of two consecutive natural number is 151. Find the numbers.
Solution : Let one natural number be \(x\)
Then the next number is \((x + 1)\)
The sum of two numbers is \(x + (x + 1)\)
\(x + (x + 1) = 151\)Ā Ā ... given
\( \dot .. x + x + 1 = 151\)
\( \dot .. 2x + 1 = 151\)
\( \dot .. 2x = 151 - 1\)
\( \dot .. 2x = 150\)
\( \dot .. 2x \div 2 = 150 \div 2\)
\( \dot .. x = 75\) and \(x + 1 = 176\)
Ans: The required numbers are 75 & 76
Solutions
Add :Ā
(i) \(9p + 16q ; 13p + 2q\)
= \((9p + 16q) + (13p + 2q)\) = \(9pĀ + 13pĀ +Ā Ā 16qĀ Ā + 2q\)Ā ……… Arrange like terms = \(22pĀ +Ā 18q\)Ā Ans:Ā Ā \(22pĀ +Ā 18q\)Ā
(ii) \(2a + 6b + 8c ; 16a + 13c + 18b\)
=Ā Ā \((2a + 6b + 8c) + (16a + 13c + 18b)\) =Ā Ā \(2a + 16a + 6b + 18b + 8c + 13c\)Ā ……… Arrange like terms =Ā Ā \(18a + 24b + 21c\)Ā Ans :Ā Ā \(18a + 24b + 21c\)Ā
(iii) \(13x^2 - 12y^2 ; 6x^2 - 8y^2 \)
=Ā Ā \((13x^2 – 12y^2) + (6x^2 – 8y^2)\) =Ā Ā \(13x^2 + 6x^2- 12y^2 – 8y^2\)Ā ……… Arrange like terms =Ā Ā \(19x^2 – 20y^2\)Ā Ans :\(19x^2 – 20y^2\)Ā
(iv) \(17a^2b^2 + 16c ; 28c - 28a^2b^2 \)
=Ā Ā \((17a^2b^2 + 16c) + (28c – 28a^2b^2)\) = \(17a^2b^2 + 16c + 28c – 28a^2b^2\) =Ā Ā \(17a^2b^2 – 28a^2b^2 + 16c + 28c \)Ā ……… Arrange like terms =Ā Ā \(-11a^2b^2 + 44c\)Ā Ans : Ā \(-11a^2b^2 + 44c\)Ā
(v) \(3y^2 - 10y + 16 ; 2y - 7 \)
=Ā Ā \((3y^2 – 10y + 16) + (2y – 7)\) = \(3y^2 – 10y + 16 + 2y – 7\) =Ā Ā \(3y^2 – 10yĀ + 2y + 16Ā – 7 \)Ā ……… Arrange like terms =Ā Ā \(3y^2 – 8y + 9\)Ā Ans :Ā \(3y^2 – 8y + 9\)Ā
(vi) \(-3y^2 + 10y -16 ; 7y^2 + 8\)
=Ā Ā \((-3y^2 + 10y -16) + (7y^2 + 8)\) = \(-3y^2 + 10y -16 + 7y^2 + 8\) =Ā Ā \(-3y^2 + 7y^2 + 10y – 16 + 8 \)Ā ……… Arrange like terms =Ā Ā \(4y^2 + 10y -8\)Ā Ans :Ā \(4y^2 + 10y -8\)Ā
Subtract the second expression from first one :Ā
(i) \((4xy - 9z) ; (3xy - 16z)\)
= \((4xy – 9z) – (3xy – 16z)\) =Ā \(4xy – 9z – 3xy + 16z\)Ā Ā =Ā \(4xy – 3xy – 9zĀ + 16z\)Ā Ā ……… Arrange like terms = \(xy + 7z\)Ā Ans:Ā Ā \(xy + 7z\)
(ii) \((5x + 4y + 7z) ; (x + 2y + 3z)\)
= \((5x + 4y + 7z) – (x + 2y + 3z)\) = \(5x + 4y + 7z – x – 2y – 3z\) =Ā \(5x – x + 4y – 2y + 7z – 3z\) Ā ……… Arrange like terms = \(4x + 2y + 4z\)Ā Ans:Ā Ā \(4x + 2y + 4z\)Ā
(iii) \((14x^2 + 8xy + 3y^2) ; (26x^2 - 8xy -17y^2)\)
= \((14x^2 + 8xy + 3y^2) – (26x^2 – 8xy -17y^2)\) = \(14x^2 + 8xy + 3y^2 – 26x^2 + 8xy +17y^2\) =Ā \(14x^2 – 26x^2 + 8xy + 8xy + 3y^2 + 17y^2\) Ā ……… Arrange like terms = \(-12x^2 + 16xy + 20y^2\)Ā Ans:Ā \(-12x^2 + 16xy + 20y^2\)Ā Ā
(iv) \((6x^2 + 7xy + 16y^2) ; (16x^2 - 17xy)\)
= \((6x^2 + 7xy + 16y^2) – (16x^2 – 17xy)\) = \(6x^2 + 7xy + 16y^2 – 16x^2 + 17xy\) =Ā \(6x^2 – 16x^2 + 7xy + 17xy + 16y^2 \) Ā ……… Arrange like terms = \(-10x^2 + 24xy + 16y^2\)Ā Ans:Ā \(-10x^2 + 24xy + 16y^2\)Ā Ā Ā
(v) \((4x + 16z) ; (19y - 14z + 16x)\)
= \((4x + 16z) – (19y – 14z + 16x)\) = \(4x + 16z – 19y + 14z – 16x\) =Ā \(4x – 16x + 16z + 14z – 19y\) Ā ……… Arrange like terms = \(-12x + 30z – 19y\)Ā Ans: \(-12x + 30z – 19y\)Ā Ā Ā
Multiply :
(i) \(16xy \times 18xy\)
=Ā \(16 \times x \times y \times 18 \times x \times y\) \(16 \times 18 \times x \timesĀ x \times y \times y\) = \(288x^2y^2\) Ans :Ā \(288x^2y^2\)
(ii) \(23xy^2 \times 4yz^2\)
=Ā \(23 \times x \times y^2 \times 4 \times y \times z^2\) \(23 \times 4 \times x \times y^2 \times y \times z^2\) = \(92xy^3z^2\) Ans :\(92xy^3z^2\)Ā
(iii) \((12a + 17b) \times 4c\)
=Ā \(12a \times 4c + 17b \times 4c\) = \(12 \times a \times 4 \times c + 17 \times b \times 4 \times c\) =Ā \(12 \times 4 \times a \times c + 17 \times 4 \times b \times c\) =Ā \(48ac + 68bc\) Ans :\(48ac + 68bc\)Ā
(iv) \((4x + 5y) \times (9x + 7y)\)
=Ā \(4x(9x + 7y) + 5y(9x + 7y)\) = \(4x \times 9xĀ + 4x \times 7y + 5y \times 9xĀ + 5y \times 7y\) = \(36x^2 + 28xy + 45xy + 35y^2\) = \(36x^2 + 73xy + 35y^2\) Ans : \(36x^2 + 73xy + 35y^2\)Ā
Solve :
(i) The length of a rectangle is \((8x + 5)\) cm and its breadth is \((5x + 3)\) cm. Find its area
The area of rectangle = length x breadth =Ā \((8x + 5) \timesĀ (5x + 3)\) =Ā \((8x(5x + 3) + 5(5x + 3)\) = \(8x \times 5xĀ + 8x \times 3 + 5 \times 5xĀ + 5 \times 3\) = \(40x^2 + 24x + 25x + 15\) = \(40x^2 + 49x + 15\) Ans : The area of rectangle is \((40x^2 + 49x + 15)\) \(cm^2\)Ā
1.Simplify :
(i) \((3x - 11y) - (17x + 13y)\)
\((3x – 11y) – (17x + 13y)\) =Ā \(3x – 11y – 17x – 13y\) =Ā \(3x – 17x – 11y – 13y\) = \(-14x – 24y\) = \(-2(7x + 12y)\) Ans :Ā \(-2(7x + 12y)\)
2. Find the product of \((23x^2y^3z)\) & \((-15x^3yz^2)\)
\((23x^2y^3z) \times (-15x^3yz^2)\) =Ā \((23 \times -15)( x^2y^3z \times x^3yz^2)\) =Ā \(-345x^5y^4z^3\) Ans :Ā \(-345x^5y^4z^3\)
3. Solve the following equations
(i) \(4x + \frac{1}{2} =Ā \frac{9}{2}\)
Solution ;Ā \(4x + \frac{1}{2} =Ā \frac{9}{2}\) \(4x + \frac{1}{2} –Ā \frac{1}{2} Ā =Ā \frac{9}{2} – \frac{1}{2}\)Ā Ā … subtracting \(\frac{1}{2}\) from both side \(\dot .. 4xĀ =Ā \frac{9 – 1}{2}\) \(\dot .. 4xĀ =Ā \frac{8}{2}\) \(\dot .. 4xĀ =Ā 4\) \(\dot .. \frac{4x}{4}Ā =Ā \frac{4}{4}\)Ā Ā ….dividing both side by 4 \(\dot .. xĀ =Ā 1\) Ans :Ā \(xĀ =Ā 1\)
(ii) \(10 = 2y + 5\)
Solution ;Ā \(10 = 2y + 5\) \(\dot .. 2y + 5 = 10\) …. Transposing LHS and RHS \(\dot .. 2y + 5 – 5 = 10 – 5\)Ā …. subtracting 5 from both sides \(\dot .. 2yĀ =Ā 5\) \(\dot .. \frac{2y}{2}Ā =Ā \frac{5}{2}\)Ā …Dividing both side by 2 \(\dot .. yĀ =Ā \frac{5}{2}\) Ans :Ā \(yĀ =Ā \frac{5}{2}\)
(iii) \(5m - 4 = 1\)
Solution ;\(5m – 4 = 1\) \(\dot .. 5m – 4 + 4 = 1 + 4\) …. Adding 4 to both sides \(\dot .. 5m =Ā 5\)Ā Ā \(\dot .. \frac{5m}{5}Ā =Ā \frac{5}{5}\)Ā … dividing both side by 5 \(\dot .. mĀ = 1\) Ans :Ā \(mĀ = 1\)
(iv) \(6x - 1 = 3x + 8\)
Solution ;\(6x – 1 = 3x + 8\) \(\dot .. 6x – 1 + 1 = 3x + 8 + 1\) …. Adding 1 to both sides \(\dot .. 6x =Ā 3x + 9\)Ā Ā \(\dot .. 6x – 3x =Ā 3x + 9 – 3x\) … subtracting \(3x\) from bothe sides \(\dot .. 3xĀ = 9\) \(\dot .. \frac{3x}{3}Ā =Ā \frac{9}{3}\)Ā … dividing both side by 3 \(\dot .. xĀ = 3\) Ans :Ā \(xĀ = 3\)
(v) \(2(x - 4) = 4x + 2\)
Solution ;Ā \(2(x – 4) = 4x + 2\) \(\dot .. 2x – 8 = 4x + 2\)Ā \(\dot .. 2x – 8 + 8 = 4x + 2 + 8\)Ā Ā … adding 8 to both sidesĀ Ā \(\dot .. 2x =Ā 4x + 10\)Ā \(\dot .. 2x – 4x =Ā 4x + 10 – 4x\)Ā … subtracting \(4x\) from bothe sides \(\dot .. -2x = 10 \) \(\dot .. \frac{-2x}{-2}Ā =Ā \frac{10}{-2}\)Ā Ā … dividing both side by -2 \(\dot .. xĀ = -5\) Ans :Ā \(xĀ = -5\)
(vi) \(5(x + 1) = 74\)
Solution ;Ā \(5(x + 1) = 74\) \(\dot .. 5x + 5 = 74\)Ā \(\dot .. 5xĀ + 5 – 5 = 74 – 5\)Ā Ā … subtracting 5 from both sidesĀ Ā \(\dot .. 5x =Ā 69\)Ā \(\dot .. \frac{5x}{5}Ā =Ā \frac{69}{5}\)Ā Ā … dividing both side by 5 \(\dot .. xĀ =Ā \frac{69}{5}\) Ans :Ā \(xĀ =Ā \frac{69}{5}\)
4. Rakesh's age is 5 year less than sania's age. The sum of their age is 27. How old are they ?
Solution : Let Sania’s age be \(x\) years ThenĀ Rakesh’s age is \(x-5\) years The sum of their ages is 27 years \(\dot .. x + (x – 5)Ā =Ā 27\) \(\dot .. x + x – 5Ā =Ā 27\) \(\dot .. 2x – 5Ā =Ā 27\) \(\dot .. 2x – 5 + 5 =Ā 27 + 5\)Ā … adding 5 to both sides \(\dot .. 2xĀ =Ā 32\) \(\dot .. \frac{2x}{2}Ā = \frac{32}{2}\) ….. dividing both sides by 2 \(\dot .. xĀ =Ā 16\) and \(x – 5 =16 – 5 = 11\)Ā Ā Ans : Sania’s age is 16 years andĀ Rakesh’s age is 11 yearsĀ
5. In a grove, 60 more saplings of Jambul were planted than those of Asoka. If there are altogather 200 saplings. Find the saplings of Jambul.
Solution : Let \(x\) sapling of Asoka were planted. Then \(x + 60 \) sapling of Jambul were planted. Together their number is 200 \(\dot .. x + (x + 60)Ā = 200\) \(\dot .. x + x + 60Ā = 200\)Ā Ā Ā \(\dot .. 2x + 60Ā = 200\) \(\dot .. 2x + 60 – 60Ā = 200 – 60\)Ā ….subtracting 60 from bothe sides \(\dot .. 2xĀ = 140\) \(\dot .. \frac{2x}{2}Ā = \frac{140}{2}\) \(\dot .. xĀ = 70,Ā and,Ā x + 60 = 70 + 60 = 130\) Ans :130 sapling of Jambul were planted.Ā
6. Shubhangi has twice as many 20 rupee notes as she has 50 rupee notes. Altogather, she has Rs. 2700. How many 50 Rupee notes does she have
Solution : Let shubhangi have \(x\) note of 50 rupee Therefore total amount of 50 rupee note is \(50 \times x\) Then she has \(2x \) notes of 20 rupee Therefore total amount of 20 rupee note is \(20 \times 2x\) The total amount with her is Rs. 2700. \(\dot .. 50 \times x + 20 \times 2x Ā = 2700\) \(\dot .. 5x + 40xĀ = 2700\)Ā Ā Ā \(\dot .. 90xĀ = 2700\) \(\dot .. \frac{90x}{90}Ā = \frac{2700}{90}\) …dividing both sides by 90 \(\dot .. xĀ = 30\) Ans : shubhangi has 30 notes of 50 rupee
7. Virat made twice as many runs as Rohit. The total of their scores is 2 less than double century. How many runs did each of them make.
Solution : Let Rohit score \(x\) runs Virat scored \(2x\) runs They together scores 2 less than double century runs means they scored \(200 – 2 = 198\) runsĀ \(\dot .. x + 2x = 198\) \(\dot .. 3x = 198\) \(\dot .. \frac{3x}{3} = \frac{198}{3}\) \(\dot .. x = 66\) andĀ \(2x=132\) Ans :Ā Virat scored 132) runs andĀ Rohit score 66 runsĀ
Videos
Click on below link to view :
Theory :
1- Introduction
2- Terms
3- Factors
4- Coefficient
Exercise Solution Videos :
Click on below link to Download PDF
MSBSHSE-Class 7-Maths-Chapter-8-Algebraic Expression and Operations-Book
Visit Main Page : Class 7th MSBSHSE ā Mathematics (English medium)Ā - view chapter wise online/offline Books, Solutions, Notes, PDF, Test and Videos, based on Maharashtra Board class 7 Mathematics.