AP State Syllabus AP Board 7th Class Maths Solutions Chapter 10 Algebraic Expressions InText Questions and Answers.
AP State Syllabus 7th Class Maths Solutions 10th Lesson Algebraic Expressions InText Questions
Question 1.
In the expressions given below identify all the terms. (Page No. 194)
i) 5x2 + 3y + 7
ii) 5x2y + 3
iii) 3x2y
iv) 5x – 7
v) 5x + 8 – 2(-y)
vi) 7x2 – 2x
Solution:
i) 5x2 + 3y + 7 is a trinomial
ii) 5x2y + 3 is a binomial
iii) 3x2y is a monomial
iv) 5x – 7 is a binomial
v) 5x + 8 – 2 (-y) is a trinomial
vi) 7x2 – 2x is a binomial
Question 2.
Write the following expressions in statements. (Page No. 195)
12x, 12, 25x, -25, 25y, 1, x, 12y, y, 25xy, 5x2y, 7xy2, 2xy, 3xy2, 4x2y.
Solution:
Like terms Groups → {12x, 25x, x}
→ {-25, 12, 1}
→ {25y, 12y, 1}
→ {25xy, 2xy}
→ {5x2y,4x2y}
→ {7xy2, 3xy2}
Try This
Question 1.
i) What is the numerical coefficient of ‘x’ ? (Page No. 195)
Solution:
The numerical coefficient of ‘x’ is 1.
ii) What is the numerical coefficient of -‘y’ ?
Solution:
The numerical coefficient of -y is -1.
iii) What is the literal coefficient of ‘-3z’ ?
Solution:
z.
iv) Is a numerical coefficient a constant ?
Solution:
Yes.
v) Is a literal coefficient always a variable ?
Solution:
Yes.
Question 2.
Write 3 algebraic expressions with 3 terms each. (Page No. 196)
Solution:
i) ax2 + bx + c
ii) px + qy + rz
iii) x2 + y2 + z2
Do This
Question 1.
State true or false and give reasons for your answer. (Page No. 195)
i) 7x2 and 2x are unlike terms.
ii) pq2 and – 4pq2 are like terms.
iii) xy, – 12x2y and 5xy2 are like terms.
Solution:
i) 7x2 and 2x are unlike terms is true. Since the power of the variable x is not same in both the terms.
ii) pq2 and – 4pq2 are like terms is true. Since both the terms are having same variables and same exponents.
iii) xy, -12x2y and 5xy2 are like terms is false. Since all the terms are not contains same exponents.
Question 2.
How many terms are there in each of the following expressions ?
i) x + y
ii) 11x – 3y – 5,
iii) 6 x2 + 5x – 4
iv) x2z + 3
v) 5x2y
vi) x + 3 + y
vii) x – \(\frac{11}{3}\)
viii) \(\frac{3 x}{7 y}\)
ix) 2z – y
x) 3x + 5 (Page No. 196)
Solution:
One term – (v) 5x2y, viii) \(\frac{3 x}{7 y}\)
Two terms – (i) x + y , (iv) x2z + 3, (vii) x – \(\frac{11}{3}\), (ix) 2z – y (x) 3x + 5
Three terms – (ii) 11x – 3y – 5, (iii) 6x2 + 5x – 4, (vi) x + 3 + y
Question 3.
Give two examples for each type of algebraic expression. ( Page No. 197)
Solution:
Monomial : i) 5x2y, ii) \(\frac{3}{2}\) xyz
Binomial :i) ax + by, ii) 2z – 5
Trinomial : i) ax + by + cz, ii) p2 + q2 + r2
Polynomial: i) 5x4 – 2x2 + x – 1, ii) 6 + 5x – 4x2 + 3y3 – 2z4
Question 4.
Identify the expressions given below as monomial, binomial, trinomial, and multinomial. (Page No. 197)
i) 5x2 + y + 6
ii) 3xy
iii) 5x2y + 6x
iv) a + 4x – xy + xyz
Solution:
i) 5x2 + y + 6 → is a trinomial.
ii) 3xy → is a monomial.
iii) 5x2y + 6x → is a binomial.
iv) a + 4x – xy + xyz → is a multinomial.
Question 5.
Find the sum of the like terms. (Page No. 200)
i) 5x, 7x
ii) 7x2y, -6x2y
iii) 2m, 11m
iv) 18ab, 5ab, 12ab
v) 3x2, -7x2, 8x2
vi) 4m2, 3m2, -6m2, m2
Solution:
i) 5x + 7x = 12x
ii) 7x2y + (-6x2y) = (7 – 6) x2y = x2y
iii) 2m + 11m = (2 + 11)m = 13m
iv) 18ab + 5ab + 12ab = (18 + 5 + 12) ab
= 35ab
v) 3x + (-7x) + 8x = (3 – 7 + 8) x2
= (11 – 7) x2
= 4x2
vi) 4m” + 3m2 +(-6m2) + m2 = (4 + 3 – 6 + 1) m2
= (8 – 6) m2
= 2m2
vii) 18pq + (-15 pq) + 3pq = (18 -15 + 3) pq
= (21 – 15) pq
= 6pq
Question 6.
Subtract the first term from the second term. (Page No. 200)
i) 2xy, 7xy
ii) 5a2, 10a2
iii) 12y, 3y
iv) 6x2y, 4x2y
v) 6xy, -12xy
Solution:
i) 2xy, 7xy
7xy – 2xy = (7 – 2) xy = 5xy
ii) 5a2, 10a2
10a -5a = (10-5) a2 = 5a2
iii) 12y, 3y
3y – 12y = (3 – 12)y = -9y
iv) 6x2y, 4x2y
4x2y – 6x2y = (4 – 6) x2y = -2x2y
v) 6xy,-12xy
(-12xy) – 6xy = (-12 – 6) xy = -18xy
Question 7.
Simplify the following. (Page No. 201)
i) 3m + 12m -5m
ii) 25yz – 8yz – 6yz
iii) 10m2 – 9m + 7m – 3m2
iv) 9x2 – 6 + 4x + 11 – 6x2 – 2x + 3x2 – 2
v) 3a2 – 4a2b + 7a2 – b2 – ab
vi) 5x2 + 10 + 6x + 4 + 5x + 3x2 + 8
Solution:
i) 3m + 12m -5m =(3+12-5)m
= (15 – 5) m
= 10m
ii) 25yz – 8yz – 6yz = (25 – 8 – 6) yz
= (25 – 14) yz
= 11 yz
iii) 10m2 – 9m + 7m – 3m2 – 5m – 8 = (10m2 – 3m2) + (-9m + 7m – 5m) – 8
= (10 – 3)m2 + (-9 + 7 – 5) m – 8
= 7m2 + (-7m) – 8
= 7m2 – 7m – 8
iv) 9x2 – 6 + 4x + 11- 6x2 – 2x + 3x2 – 2
= (9x2-6x2 + 3x2) + (4x-2x) + (-6 + 11 -2)
= (9 – 6 + 3) x2 + (4 – 2) x + (11 – 8)
= (12 – 6) x2 + 2x + 3
= 6x2 + 2x + 3
v) 3a2 + 4a2b – 7a2 – b2 – ab = (3a2 + 7a2) – 4a2b – b2 – ab
= 10a2 – b2 – 4a2b – ab
vi) 5x2 + 10 + 6x + 4 + 5x + 3X2 + 8 = (5x2 + 3x2) + (6x + 5x) + (10 + 4 + 8)
= 8x2 + 1 lx + 22
Question 8.
Write the following expressions in standard form. (Page No. 202)
Solution:
Expression – Standard form
i) 3x + 18 + 4x2 → 4x2 + 3x + 18
ii) 8 – 3x + 4x → -3x2 + 4x + 8
iii) -2m + 6 – 3m2 → -3m2 – 2m + 6
iv) y3 + 1 + y + 3 → y3 + 3y2 + y + 1
Question 9.
Identify the expressions that are in standard form. (Page No. 202)
i) 9x2 + 6x + 8
ii) 9x2 + 15 + 7x
iii) 9x2 + 7
iv) 9x3 + 15x + 3
v) 15x2 + x3 + 3x
vi) x2y + xy + 3
vii) x3 + x2y2 + 6xy
Solution:
(i), (iii), (iv), (vi) are in standard form.
Question 10.
Write 5 different expressions in standard form. (Page No. 202)
Solution:
i) ax2 + bx + c
ii) ax + b
iii) 4x3 + 5x2 – 6x + 2
iv) 5x4 – 3x3 – 2x – 2
v) px3 + qx2 + r
Try This
Question 1.
Find the value of the expression ‘-9x’ if x = -3. (Page No. 203)
Solution:
The value of -9x when x = -3
-9x = -9 (-3) = + 27
Question 2.
Write an expression’ whose value is equal to -9, when x = -3. (Page No. 203)
Solution:
When x = -3, then the value of an expression 3x is -9.
∴ 3x = 3 (-3)= -9.
Do This
Question 1.
Answer the following expressions (Page No. 206)
i) x – 2y, 3x + 4y
ii) 4m2 – 7n2 + 5 mil, 3m2 + 5n2 – 2mn
iii) 3a – 4b, 5c – 7a + 2b
Solution:
i) x – 2y, 3x + 4y = (x – 2y) + (3x + 4y)
= (x + 3x) + (-2y + 4y) = 4x + 2y
ii) 4m2 – 7n2 + 5mn, 3m2 + 5n2 – 2mn = (4m2 – 7n2 + 5mn) + (3m2 + 5n2 – 2mn)
= (4m2 + 3m2) + (-7n2 + 5n2) + (5mn – 2mn)
= 7 m2 + (-2n2) + 3mn = 7 m2 – 2n2 + 3mn
iii) 3a – 4b, 5c – 7a + 2b = (3a – 4b) + (5c – 7a + 2b) = (3a – 7a) + (-4b + 2b) + 5c = -4a – 2b + 5c