Telangana SCERT Solution Class VII (7) Math Chapter 10 Algebraic Expressions
Get Telangana state board class 7 Math chapter no. 10 ‘Algebraic Expressions’ all exercise problem and solution.
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Exercise – 1
1.) Find the rule which gives the number of matchsticks required to make the following patterns-
(i) A pattern of letter ‘H’
Answer:
Here we have to find rule for pattern of letter ‘H’
Let, the number of matchsticks = p
We required 3 matchsticks.
Rule for pattern of letter ‘H’ = 3p
(ii) A pattern of letter ‘V’
Answer:
Here we have to find rule for pattern of letter ‘V’
Let, the number of matchsticks = m
We required 2 matchsticks.
Rule for pattern of letter ‘V’ = 2m
2.) Given below is a pattern made from coloured tiles and white tiles
(i) Draw the next two figures in the pattern above.
Answer: Here we have to draw next 2 figures.
(ii) Fill the table given below and express the pattern in the form of an algebraic expression.
| Figure Number | 1 | 2 | 3 | 4 | 5 |
| No. of coloured tiles | 4 |
Answer:
Here we have to express the pattern in the form of an algebraic expression.
Here No. of coloured tiles = 4
Let, the number of coloured tiles = n
Rule for No. of coloured tiles = 4n here n=1, 2, 3,………..
The pattern in the form of an algebraic expression = 4n
When n = 1
4n = 4 x 1 = 4
When n = 2
4n = 4 x 2 = 8
When n = 3
4n = 4 x 3 = 12
When n = 4
4n = 4 x 4 = 16
When n = 5
4n = 4 x 5 = 20
| Figure Number | 1 | 2 | 3 | 4 | 5 |
| No. of coloured tiles | 4 | 8 | 12 | 16 | 20 |
(iii) Fill the table given below and express the pattern in the form of an algebraic expression.
| Figure Number | 1 | 2 | 3 | 4 | 5 |
| No. of total tiles | 5 |
Answer:
Here we have to express the pattern in the form of an algebraic expression.
Here No. of coloured tiles = 4
No. of total tiles = 5
Let, the number of coloured tiles = n
Rule for No. of coloured tiles = 4n here n=1, 2, 3…
The pattern in the form of an algebraic expression for No. of total tiles = 4n + 1
When n = 1
4n + 1 = 4 x 1 + 1= 5
When n = 2
4n + 1 = 4 x 2+ 1= 9
When n = 3
4n + 1= 4 x 3+ 1= 13
When n = 4
4n+ 1= 4 x 4 + 1= 17
When n = 5
4n + 1= 4 x 5 + 1= 21
|
Figure Number |
1 | 2 | 3 | 4 | 5 |
| No. of total tiles | 5 | 9 | 13 | 17 |
21 |
3.) Write the following statements using variables, constants and arithmetic operations.
(i) 6 more than p
Answer:
We have to write given statementusing variables, constants and arithmetic operations.
6 more than p = p + 6
(ii) ‘x’ is reduced by 4
Answer:
We have to write given statementusing variables, constants and arithmetic operations.
‘x’ is reduced by 4 = x −4
(iii) 8 subtracted from y
Answer:
We have to write given statementusing variables, constants and arithmetic operations.
8 subtracted from y = y −8
(iv) q multiplied by ‘-5’
Answer:
We have to write given statementusing variables, constants and arithmetic operations.
q Multiplied by ‘-5’ = −5q
(v) y divided by 4
Answer:
We have to write given statementusing variables, constants and arithmetic operations.
y divided by 4 = y/4
(vi) One-fourth of the product of ‘p’ and ‘q’
Answer:
We have to write given statementusing variables, constants and arithmetic operations.
One-fourth of the product of ‘p’ and ‘q’ = pq/4
(vii) 5 added to the three times of ‘z’
Answer:
We have to write given statementusing variables, constants and arithmetic operations.
5 added to the three times of ‘z’ = 3z+5
(viii) x multiplied by 5 and added to ’10’
Answer:
We have to write given statementusing variables, constants and arithmetic operations.
x multiplied by 5 and added to ’10’ = 10+5x
(ix) 5 subtracted from two times of ‘y’
Answer:
We have to write given statementusing variables, constants and arithmetic operations.
5 subtracted from two times of ‘y’ = 2y −5
(x) y multiplied by 10 and added to 13
Answer:
We have to write given statementusing variables, constants and arithmetic operations.
y multiplied by 10 and added to 13 = 10y + 13
4.) Write the following expressions in statements.
(i) x + 3
Answer:
Here, we have to give algebraic expression and we have to write this in statements.
x + 3 = 3 is added to x
(ii) y – 7
Answer:
Here, we have to give algebraic expression and we have to write this in statements.
y – 7 = 7 is subtracted from ‘y’
(iii) 10l
Answer:
Here, we have to give algebraic expression and we have to write this in statements.
10l = l is multiplied by 10
(iv) x/5
Answer:
Here, we have to give algebraic expression and we have to write this in statements.
x/5 = x is divided by 5
(v) 3m + 11
Answer:
Here, we have to give algebraic expression and we have to write this in statements.
3m + 11 = m is multiplied by 3 and added to 11
(vi) 2y – 5
Answer:
Here, we have to give algebraic expression and we have to write this in statements.
2y – 5 = y is multiplied by 2 and subtracted 5
5.) Some situations are given below. State the number in situations is a variable or constant?
Example: Our age – its value keeps on changing so it is an example of a variable quantity.
(i) The number of days in the month of January
Answer:The number of days in the month of January is 31 which is constant.
The number in this situation is constant.
(ii) The temperature of a day
Answer:The temperature of a day is variable.
The number in this situation is variable.
(iii) Length of your classroom
Answer:
Length of your classroom is constant.
The number in this situation is constant.
(iv) Height of the growing plant
Answer:
Height of the growing plant is variable.
The number in this situation is variable.
Exercise 2
1.) Identify and write the like terms in each of the following groups.
(i) a2, b2, –2a2, c2, 4a
Answer:
We know,
Like terms are terms which contain the same variables with the same exponents.
a2, b2, –2a2, c2, 4a from this like terms are (a2,–2a2)
(ii) 3a, 4xy, –yz, 2zy
Answer:
We know,
Like terms are terms which contain the same variables with the same exponents.
3a, 4xy, –yz, 2zy from this like terms are (–yz, 2zy)
(iii) –2xy2, x2y, 5y2x, x2z
Answer:
We know,
Like terms are terms which contain the same variables with the same exponents.
–2xy2, x2y, 5y2x, x2z from this like terms are (–2xy2, 5y2x)
(iv) 7p, 8pq, –5pq, –2p, 3p
Answer:
We know,
Like terms are terms which contain the same variables with the same exponents.
7p, 8pq, –5pq, –2p, 3p from this like terms are (7p, –2p, 3p) and (8pq, –5pq)
2.) State whether the expression is a numerical expression or an algebraic expression
(i) x + 1
Answer:
We know,
If every term of an expression is a constant term, then the expression is called numerical expression.
If an expression has at least one algebraic term, then the expression is called an algebraic expression.
x + 1 contains variable… This is algebraic expression.
(ii) 3m2
Answer:
We know,
If every term of an expression is a constant term, then the expression is called numerical expression.
If an expression has at least one algebraic term, then the expression is called an algebraic expression.
3m2 contains variable… This is algebraic expression.
(iii) –30 + 16
Answer:
We know,
If every term of an expression is a constant term, then the expression is called numerical expression.
If an expression has at least one algebraic term, then the expression is called an algebraic expression.
–30 + 16 contains all constant….. This is numerical expression.
(iv) 4p2–5q2
Answer:
We know,
If every term of an expression is a constant term, then the expression is called numerical expression.
If an expression has at least one algebraic term, then the expression is called an algebraic expression.
4p2–5q2 contains variable… This is algebraic expression.
(v) 96
Answer:
We know,
If every term of an expression is a constant term, then the expression is called numerical expression.
If an expression has at least one algebraic term, then the expression is called an algebraic expression.
96contains all constant….. This is numerical expression.
(vi) x2–5yz
Answer:
We know,
If every term of an expression is a constant term, then the expression is called numerical expression.
If an expression has at least one algebraic term, then the expression is called an algebraic expression.
x2–5yz contains variable… This is algebraic expression.
(vii) 215x2yz
Answer:
We know,
If every term of an expression is a constant term, then the expression is called numerical expression.
If an expression has at least one algebraic term, then the expression is called an algebraic expression.
215x2yz contains variable… This is algebraic expression.
(viii) 95 ÷ 5 × 2
Answer:
We know,
If every term of an expression is a constant term, then the expression is called numerical expression.
If an expression has at least one algebraic term, then the expression is called an algebraic expression.
95 ÷ 5 × 2 contains all constant….. This is numerical expression.
(ix) 2 + m + n
Answer:
We know,
If every term of an expression is a constant term, then the expression is called numerical expression.
If an expression has at least one algebraic term, then the expression is called an algebraic expression.
2 + m + n contains variable… This is algebraic expression.
(x) 310 + 15 + 62
Answer:
We know,
If every term of an expression is a constant term, then the expression is called numerical expression.
If an expression has at least one algebraic term, then the expression is called an algebraic expression.
310 + 15 + 62 contains all constant….. This is numerical expression.
(xi) 11a2+6b2–5
Answer:
We know,
If every term of an expression is a constant term, then the expression is called numerical expression.
If an expression has at least one algebraic term, then the expression is called an algebraic expression.
11a2+6b2–5 contains variable… This is algebraic expression.
3.) Which of the following multinomials is monomial, binomial or trinomial?
(i) y2
Answer:
Here we have to classify given expressions as monomial, binomial or trinomial.
We know,
Expressions contains only 1 term is called as “monomial”
Expressions contains 2 terms is called as “binomial”
Expressions contains 3 terms is called as “Trinomial”
y2contains only 1 term. This is “monomial”
(ii) 4y – 7z
Answer:
Here we have to classify given expressions as monomial, binomial or trinomial.
We know,
Expressions contains only 1 term is called as “monomial”
Expressions contains 2 terms is called as “binomial”
Expressions contains 3 terms is called as “Trinomial”
4y – 7z contains 2 terms. This is “binomial”
(iii) 1 + x + x2
Answer:
Here we have to classify given expressions as monomial, binomial or trinomial.
We know,
Expressions contains only 1 term is called as “monomial”
Expressions contains 2 terms is called as “binomial”
Expressions contains 3 terms is called as “Trinomial”
1 + x + x2 contains 3 terms this is “Trinomial”
(iv) 7mn
Answer:
Here we have to classify given expressions as monomial, binomial or trinomial.
We know,
Expressions contains only 1 term is called as “monomial”
Expressions contains 2 terms is called as “binomial”
Expressions contains 3 terms is called as “Trinomial”
7mn contains only 1 term. This is “monomial”
(v) a2 + b2
Answer:
Here we have to classify given expressions as monomial, binomial or trinomial.
We know,
Expressions contains only 1 term is called as “monomial”
Expressions contains 2 terms is called as “binomial”
Expressions contains 3 terms is called as “Trinomial”
a2 + b2 contains 2 terms. This is “binomial”
(vi) 100xyz
Answer:
Here we have to classify given expressions as monomial, binomial or trinomial.
We know,
Expressions contains only 1 term is called as “monomial”
Expressions contains 2 terms is called as “binomial”
Expressions contains 3 terms is called as “Trinomial”
100xyz contains only 1 term. This is “monomial”
(vii) ax + 9
Answer:
Here we have to classify given expressions as monomial, binomial or trinomial.
We know,
Expressions contains only 1 term is called as “monomial”
Expressions contains 2 terms is called as “binomial”
Expressions contains 3 terms is called as “Trinomial”
ax + 9 contains 2 terms. This is “binomial”
(viii) p2 – 3pq + r
Answer:
Here we have to classify given expressions as monomial, binomial or trinomial.
We know,
Expressions contains only 1 term is called as “monomial”
Expressions contains 2 terms is called as “binomial”
Expressions contains 3 terms is called as “Trinomial”
p2 – 3pq + r contains 3 terms this is “Trinomial”
(ix) 3y2 – x2 y2 + 4x
Answer:
Here we have to classify given expressions as monomial, binomial or trinomial.
We know,
Expressions contains only 1 term is called as “monomial”
Expressions contains 2 terms is called as “binomial”
Expressions contains 3 terms is called as “Trinomial”
3y2 – x2 y2 + 4x contains 3 terms this is “Trinomial”
(x) 7x2 – 2xy + 9y2 – 11
Answer:
Here we have to classify given expressions as monomial, binomial or trinomial.
We know,
Expressions contains only 1 term is called as “monomial”
Expressions contains 2 terms is called as “binomial”
Expressions contains 3 terms is called as “Trinomial”
7x2 – 2xy + 9y2 – 11 contains more than 3 terms this is “multinomial”
4.) What is the degree of each of the monomials.
(i) 7y
Answer:
Here we have to find degree of each of the monomials.
We know,
The sum of all exponents of the variables present in a monomial is called the degree of monomial
7y = degree of this monomial is 1.
(ii) –xy2
Answer:
Here we have to find degree of each of the monomials.
We know,
The sum of all exponents of the variables present in a monomial is called the degree of monomial
–xy2= degree of this monomial is 3.
(iii) xy2z2
Answer:
Here we have to find degree of each of the monomials.
We know,
The sum of all exponents of the variables present in a monomial is called the degree of monomial
xy2z2= degree of this monomial is 5.
(iv) –11y2z2
Answer:
Here we have to find degree of each of the monomials.
We know,
The sum of all exponents of the variables present in a monomial is called the degree of monomial
–11y2z2 = degree of this monomial is 4.
(v) 3mn
Answer:
Here we have to find degree of each of the monomials.
We know,
The sum of all exponents of the variables present in a monomial is called the degree of monomial
3mn = degree of this monomial is 2.
(vi) –5pq2
Answer:
Here we have to find degree of each of the monomials.
We know,
The sum of all exponents of the variables present in a monomial is called the degree of monomial
–5pq2= degree of this monomial is 3.
5.) Find the degree of each algebraic expression.
(i) 3x–15
Answer:
Here we have to find degree of each algebraic expression.
We know,
The highest of the degrees of all the terms of an expression is called the degree of the expression.
3x–15 = highest of the degree = 1
Degree of the expression is 1
(ii) xy + yz
Answer:
Here we have to find degree of each algebraic expression.
We know,
The highest of the degrees of all the terms of an expression is called the degree of the expression.
xy + yz highest of the degree = 2
Degree of the expression is 2.
(iii) 2y2z + 9yz –7z – 11x2y2
Answer:
Here we have to find degree of each algebraic expression.
We know,
The highest of the degrees of all the terms of an expression is called the degree of the expression.
2y2z + 9yz –7z – 11x2y2 highest of the degree = 4
Degree of the expression is 4.
(iv) 2y2z + 10yz
Answer:
Here we have to find degree of each algebraic expression.
We know,
The highest of the degrees of all the terms of an expression is called the degree of the expression.
2y2z + 10yz highest of the degree = 3
Degree of the expression is 3.
(v) pq + p2q – p2q2
Answer:
Here we have to find degree of each algebraic expression.
We know,
The highest of the degrees of all the terms of an expression is called the degree of the expression.
pq + p2q – p2q2 highest of the degree = 4
Degree of the expression is 4.
(vi) ax2+ bx + c
Answer:
Here we have to find degree of each algebraic expression.
We know,
The highest of the degrees of all the terms of an expression is called the degree of the expression.
ax2 + bx + c highest of the degree = 2
Degree of the expression is 2.
6.) Write any two Algebraic expressions with the same degree
Answer:
We have to write any two Algebraic expressions with the same degree.
4xy2 + 8 and 8y2z + y
3xyz + 7 and 2x2y + z
Exercise – 3
1.) Find the length of the line segment PR in the following figure in terms of ‘a’
Answer:
Here we have to find the length of the line segment PR in the given figure in terms of ‘a’
Here,
PR = PQ + QR
But PQ = 3a and QR = 2a
PR = 3a + 2a
PR = 5a
2.) (i) Find the perimeter of the following triangle.
Answer:
Here we have to find the perimeter of the given triangle.
We know,
The perimeter of the triangle = Sum of all three sides
The perimeter of the triangle = 6x + 2x + 5x
The perimeter of the triangle = 13x
(ii) Find the perimeter of the following rectangle.
Answer:
Here we have to find the perimeter of the given rectangle.
We know,
The perimeter of the rectangle = 2 x (Length + Breadth)
The perimeter of the rectangle = 2 x (3x + 2x)
The perimeter of the rectangle = 2 x 5x
The perimeter of the rectangle = 10x
3.) Subtract the second term from first term.
(i) 8x, 5x
Answer:
Here, we have to subtract the second term from first term.
First term = 8x
Second term = 5x
8x – 5x = 3x or 8x
– 5x
———
3x
(ii) 5p, 11p
Answer:
Here, we have to subtract the second term from first term.
First term = 5p
Second term = 11p
5p – 11p = -6p or 5p
-11p
——————–
-6p
(iii) 13m2, 2m2
Answer:
Here, we have to subtract the second term from first term.
First term = 13m2
Second term = 2m2
13m2– 2m2 = 11 m2
4.) . Find the value of following monomials, if x = 1.
(i) –x
Answer:
Here we have to find value of –x given x = 1.
–x = – 1
(ii) 4x
Answer:
Here we have to find value of 4x given x = 1.
4x = 4 x 1 = 4
(iii) –2x2
Answer:
Here we have to find value of –2x2 given x = 1.
–2x2 = -2 x (1)2 = -2
5.) Simplify and find the value of 4x + x -2x2 + x – 1 when x = –1
Answer:
Here we have to find value of 4x + x -2x2 + x – 1 when x = –1
We put x = -1 in 4x + x -2x2 + x – 1
4x + x -2x2 + x – 1 = 4 x (-1) + (-1) – 2 (-1)2 + (-1) – 1
4x + x -2x2 + x – 1 = -4 – 1 – 2 -1 – 1
4x + x -2x2 + x – 1 = -9
6.) Write the expression 5x2 – 4 – 3x2 + 6x + 8 + 5x – 13 in its simplified form. Find its value
When x = –2
Answer:
We have to write 5x2 – 4 – 3x2 + 6x + 8 + 5x – 13 in its simplified form.
5x2 – 4 – 3x2 + 6x + 8 + 5x – 13 = (5x2– 3x2)+ (6x + 5x) + (8– 4)– 13
5x2 – 4 – 3x2 + 6x + 8 + 5x – 13 = 2x2 + 11x -9
2x2 + 11x – 9 this is simplified form.
Now,
When x = –2
2x2 + 11x – 9 = 2(-2)2 + 11(-2) – 9
2x2 + 11x – 9 = 8 -22 – 9
2x2 + 11x – 9 = -23
7.) If x = 1 ; y = 2 find the values of the following expressions
(i) 4x–3y + 5
Answer:
We have to find value of 4x–3y + 5 when x = 1; y = 2
4x–3y + 5 = 4(1)–3(2) + 5
4x–3y + 5 = 4 – 6 + 5
4x–3y + 5 = -2 + 5
4x–3y + 5 = 3
(ii) x2 + y2
Answer:
We have to find value ofx2 + y2 when x = 1; y = 2
x2 + y2 = (1)2 + (2)2
x2 + y2 = 1 + 4
x2 + y2 = 5
(iii) xy + 3y – 9
Answer:
We have to find value ofxy + 3y – 9 when x = 1; y = 2
xy + 3y – 9 = 1 x 2 + 3 x 2 – 9
xy + 3y – 9 = 2 + 6 – 9
xy + 3y – 9 = 8 – 9
xy + 3y – 9 = -1
8.) Area of a rectangle is given by A = l × b. If l = 9cm, b = 6cm, find its area?
Answer:
Given that,
l = 9cm, b = 6cm
We know,
Area of a rectangle = l × b
Area of a rectangle = 9cm x 6cm
Area of a rectangle = 54 cm2
Simple interest is given by I = PTR/100 If P = 900, T = 2 years; and R = 5%, find the simple interest.
Answer:
Given that,
P = 900, T = 2 years; and R = 5%,
We have to find the simple interest.
We know,
Simple interest = PTR/100
Simple interest = 900 x 5% x 2 / 100
Simple interest = Rs.90
Exercise – 4
1.) Add the following algebraic expressions using both horizontal and vertical methods. Did you get the same answer with both methods.
(i) x2 – 2xy + 3y2; 5y2 + 3xy – 6x2
Answer:
Here, we have to add algebraic expressions using both horizontal and vertical methods.
Horizontal Method:
We 1st arrange expression in standard form.
x2 – 2xy + 3y2 = x2 – 2xy + 3y2
5y2 + 3xy – 6x2 = – 6x2+ 3xy + 5y2
Now, we add x2 – 2xy + 3y2and – 6x2+ 3xy + 5y2
x2 – 2xy + 3y2 + (– 6x2+ 3xy + 5y2)
Now, we add like terms.
(x2 – 6x2) + (– 2xy + 3xy) + (3y2+ 5y2)
= -5x2 + xy + 8y2
Now,
Vertical method:
We 1st arrange expression in standard form.
x2 – 2xy + 3y2 = x2 – 2xy + 3y2
5y2 + 3xy – 6x2 = – 6x2+ 3xy + 5y2
Now, we add x2 – 2xy + 3y2and – 6x2+ 3xy + 5y2
|
x2 – 2xy + 3y2 |
|
– 6x2+ 3xy + 5y2 |
|
-5x2 + xy + 8y2 |
By Both method we get same answers.
(ii) 4a2 + 5b2 + 6ab; 3ab; 6a2 -2b2;4b2 – 5ab
Answer:
Here, we have to add algebraic expressions using both horizontal and vertical methods.
Horizontal Method:
We 1st arrange expression in standard form.
4a2 + 5b2 + 6ab = 4a2 + 5b2 + 6ab
3ab = 3ab
6a2 -2b2= 6a2 -2b2
4b2 – 5ab =4b2 – 5ab
Now, we add all terms.
4a2 + 5b2 + 6ab + 3ab + 6a2 -2b2 + 4b2 – 5ab
We add like terms.
(4a2 + 6a2) + (5b2 + 4b2-2b2) + (6ab + 3ab – 5ab)
= 10a2 + 7b2 + 4ab
Now,
Vertical method:
We 1st arrange expression in standard form.
4a2 + 5b2 + 6ab; 3ab; 6a2 -2b2;4b2 – 5ab
|
4a2 |
+ 5b2 | + 6ab |
|
+ 3ab |
||
|
+ 6a2 |
– 2b2 | |
| + 4b2 |
– 5ab |
|
|
10a2 |
+ 7b2 |
– 5ab |
By Both method we get same answers.
(iii) 2x+ 9y -7z; 3y + z + 3x; 2x – 4y – z
Answer:
Here, we have to add algebraic expressions using both horizontal and vertical methods.
Horizontal Method:
We 1st arrange expression in standard form.
2x+ 9y -7z = 2x+ 9y -7z
3y + z + 3x = 3x +3y + z
2x – 4y – z = 2x – 4y – z
Now, we add all terms.
2x + 9y -7z + 3x +3y + z + 2x – 4y – z
We add like terms.
(2x + 3x + 2x) + (9y +3y – 4y) + (-7z + z – z)
= 7x + 8y -7z
Vertical method:
We 1st arrange expression in standard form.
2x+ 9y -7z = 2x+ 9y -7z
3y + z + 3x = 3x +3y + z
2x – 4y – z = 2x – 4y – z
Now, we add all terms.
|
2x |
+ 9y | -7z |
| + 3x | +3y |
+ z |
|
+ 2x |
– 4y | – z |
| 7x | + 8y |
-7z |
By Both method we get same answers.
(iv) 2x2 – 6x + 3; -3x2-x – 4; 1 + 2x – 3x2
Answer:
Here, we have to add algebraic expressions using both horizontal and vertical methods.
Horizontal Method:
We 1st arrange expression in standard form.
2x2 – 6x + 3 = 2x2 – 6x + 3
-3x2– x – 4 = -3x2– x – 4
1 + 2x – 3x2 = – 3x2 + 2x + 1
Now, we add all terms.
2x2 – 6x + 3 + (-3x2– x – 4) + (– 3x2 + 2x + 1)
We add like terms.
(2x2-3x2 -3x2) + (– 6x – x + 2x) + (3 – 4 + 1)
-4x2 -5x
Vertical method:
We 1st arrange expression in standard form.
2x2 – 6x + 3 = 2x2 – 6x + 3
-3x2– x – 4 = -3x2– x – 4
1 + 2x – 3x2 = – 3x2 + 2x + 1
Now, we add all terms.
|
2x2 |
– 6x | + 3 |
| -3x2 | – x |
– 4 |
|
– 3x2 |
+ 2x | + 1 |
| -4x2 |
-5x |
By Both method we get same answers.
2.) Simplify:2x2+ 5x – 1 + 8x + x2 + 7 – 6x + 3 – 3x2
Answer:
Here we have to simplify 2x2+ 5x – 1 + 8x + x2 + 7 – 6x + 3 – 3x2
We add like terms.
2x2+ 5x – 1 + 8x + x2 + 7 – 6x + 3 – 3x2= (2x2+ x2– 3x2) + (5x + 8x – 6x) + (– 1 + 7 + 3)
2x2+ 5x – 1 + 8x + x2 + 7 – 6x + 3 – 3x2 = 7x + 9
5.) Subtract the second expression from the first expression
(i) 2a+b, a–b
Answer:
We have to subtract second expression from the first expression.
First expression = 2a+b
Second expression = a–b
|
2a |
+b |
|
a |
–b |
| (-) |
(+) |
| a | +2b |
Ans. a + 2b
(ii) x+2y+z , –x–y–3z
Answer:
We have to subtract second expression from the first expression.
First expression = x+2y+z
Second expression = –x–y–3z
|
x |
+2y | +z |
| –x | –y |
–3z |
|
(+) |
(+) | (+) |
| 2x | +3y |
+4z |
Ans. 2x+3y+4z
(iii) 3a2–8ab–2b2, 3a2 –4ab+6b2
Answer:
We have to subtract second expression from the first expression.
First expression = 3a2–8ab–2b2
Second expression = 3a2 –4ab+6b2
|
3a2 |
–8ab | –2b2 |
| 3a2 | –4ab |
+6b2 |
|
(-) |
(+) | (-) |
| −4ab |
−8b2 |
Ans. −4ab −8b2
(iv) 4pq–6p2–2q2, 9p2
Answer:
We have to subtract second expression from the first expression.
First expression = 4pq–6p2–2q2
Second expression = 9p2
|
4pq |
–6p2 | –2q2 |
|
9p2 |
||
|
|
(-) | |
| 4pq | −15 p2 |
− 2q2 |
Ans. 4pq −15 p2 − 2q2
(v) 7–2x–3x2, 2x2–5x–3
Answer:
We have to subtract second expression from the first expression.
First expression = 7–2x–3x2 = –3x2 – 2x + 7
Second expression = 2x2–5x–3
|
–3x2 |
– 2x | + 7 |
| 2x2 | –5x |
–3 |
|
(-) |
(+) | (+) |
| −5x2 | +3x |
–3 |
Ans. −5x2 +3x+10
(vi) 5x2–3xy–7y2, 3x2–xy–2y2
Answer:
We have to subtract second expression from the first expression.
First expression = 5x2–3xy–7y2
Second expression = 3x2–xy–2y2
|
5x2 |
– 3xy | –7y2 |
| 3x2 | –xy |
–2y2 |
|
(-) |
(+) | (+) |
| 2x2 | − 2xy |
− 5y2 |
Ans. 2x2 − 2xy − 5y2
(vii) 6m3+4m2+7m–3 , 3m3 +4
Answer:
We have to subtract second expression from the first expression.
First expression = 6m3+4m2+7m–3
Second expression = 3m3 +4
|
6m3 |
+4m2 | +7m | –3 |
| 3m3 |
+4 |
||
|
(-) |
(-) | ||
| 3m3 | + 4m2 | + 7m |
−7 |
Ans. 3m3 + 4m2 + 7m −7
9.) The sum of 3 expressions is 8 + 13a + 7a2.Two of them are 2a2 + 3a + 2 and 3a2 –4a + 1. Find the third expression.
Answer:
Given that,
The sum of 3 expressions is 8 + 13a + 7a2
Two of them are 2a2 + 3a + 2 and 3a2 –4a + 1
We have to find 3rd expression.
Let 3rd expression is “Y”
8 + 13a + 7a2 = 2a2 + 3a + 2 + 3a2 –4a + 1 + Y
8 + 13a + 7a2 = (2a2+ 3a2) + (3a–4a) + (2 + 1) + Y
8 + 13a + 7a2 = 5a2 – a + 3 + Y
Y = 8 + 13a + 7a2– 5a2+ a – 3
Y = (7a2– 5a2) + (13a + a) + 5
Y = 2a2 + 14a + 5
3rd expression is 2a2 + 14a + 5
10.) If A = 4x2 + y2 – 6xy;
B=3y2 + 12x2 + 8xy;
C=6x2 + 8y2 + 6xy
Find (i) A + B + C (ii) (A – B) –C
Answer:
Given that,
A = 4x2 + y2 – 6xy;
B=3y2 + 12x2 + 8xy;
C=6x2 + 8y2 + 6xy
We have to find
(i) A + B + C
4x2 + y2 – 6xy + 3y2 + 12x2 + 8xy + 6x2 + 8y2 + 6xy
We add like terms.
(4x2+ 12x2+ 6x2) + (– 6xy+ 8xy+ 6xy) + (y2+ 3y2+ 8y2)
A + B + C = 22x2 + 8xy+ 12y2
(ii) (A – B) –C
[4x2 + y2 – 6xy –(3y2 + 12x2 + 8xy)] – 6x2 + 8y2 + 6xy
We add or subtract like terms.
[(4x2 – 12x2) + (y2 – 3y2) + (– 6xy – 8xy)] – 6x2 + 8y2 + 6xy
(- 8x2 –2y2– 14xy)- 6x2 + 8y2 + 6xy
We add like terms.
((- 8x2– 6x2) + (– 2y2 – 8y2) + (- 14xy – 6xy)
(A – B) –C = − 14x2 − 10y2 − 20xy