# AP Board 7th Class Maths Solutions Chapter 10 Algebraic Expressions InText Questions

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) x+ 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