# AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.3

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 11 Algebraic Expressions Ex 11.3 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 11th Lesson Algebraic Expressions Exercise 11.3

Question 1.
Multiply the binomials:
(i) 2a – 9 and 3a + 4
(ii) x – 2y and 2x – y
(iii) kl + lm and k – l
(iv) m2 – n2 and m + n
Solution:
i) 2a – 9 and 3a + 4
(2a – 9) (3a + 4) = 2a (3a + 4) – 9(3a + 4)
= 6a2 + 8a – 27a – 36
= 6a2 – 19a – 36

ii) x – 2y and 2x – y
(x – 2y) (2x – y) = x(2x – y) – 2y(2x – y)
= 2x2 – xy – 4xy + 2y2
= 2x2 – 5xy + 2y2

iii) kl + lm and k – l
(kl + lm) (k – l) = kl(k – l) + lm(k – l)
= k2l – l2k + klm – l2m

iv) m2 – n2 and m + n
(m2 – n2) (m + n) = m2(m + n) – n2(m + n)
= m3 + m2n – n2m – n3

Question 2.
Find the product:
(i) (x + y)(2x – 5y + 3xy)
(ii) (mn – kl + km) (kl – lm)
(iii) (a – 2b + 3c)(ab2 – a2b)
(iv) (p3 + q3)(p – 5q+6r)
Solution:
i) (x + y) (2x – 5y + 3xy)
= x(2x – 5y + 3xy) + y(2x – 5y + 3xy)
= 2x2 – 5xy + 3x2y + 2xy – 5y2 + 3xy2
= 2x2 – 5y2 – 3xy + 3x2y + 3xy2

ii) (mn – kl + km) (kl – lm)
= kl(mn – kl + km) – lm(mn – kl + km)
= klmn – k2l2 + k2lm – lm2n + kl2m – klm2

iii) (a – 2b + 3c) (ab2 – a2b) = a(ab2 – a2b) – 2b(ab2 – a2b) + 3c(ab2– a2b)
= a2b2 – a3b – 2ab3 + 2a2b2 + 3ab2c – 3a2bc
= 3a2b2 – a3b – 2ab3 + 3ab2c – 3a2bc

iv) (p3 + q3) (p – 5q + 6r) = p3(p – 5q + 6r) + q3(p – 5q + 6r)
= p4 – 5p3q + 6p3r + pq3 – 5q4 + 6rq3
= p4 – 5q4 – 5p3q + 6p3r + pq3 + 6rq3

Question 3.
Simplify the following:
(i) (x-2y) (y – 3x) + (x+y) (x-3y) – (y – 3x) (4x – 5y)
(ii) (m + n) (m2 – mn + n2)
(iii) (a – 2b + 5c) (a – b) – (a – b – c) (2a + 3c) + (6a + b) (2c – 3a – 5b)
(iv) (pq-qr-i-pr) (pq-i-qr) – (pr-i-pq) (p-i-q – r)
Solution:
i) (x – 2y) (y – 3x) + (x + y) (x – 3y) – (y – 3x) (4x – 5y)
= (y – 3x) [x – 2y – (4x – 5y)] + (x + y)(x – 3y)
= (y – 3x) [x – 2y – 4x + 5y] + (x + y) (x – 3y)
= (y – 3x) (3y – 3x) + (x + y) (x – 3y)
= y(3y – 3x) – 3x(3y – 3x) + x(x – 3y) + y(x – 3y)
= 3y2 – 3xy – 9xy + 9x2 + x2 – 3xy + xy – 3y2
= 10x2 – 14xy

ii) (m + n) (m2– mn + n2)
= m(m2 – mn + n2) + n(m2 – mn + n2)
= m3 – m2n + n2m + nm2 – mn2 + n3
= m3 + n3

iii) (a – 2b + 5c) (a – b) – (a – b – c) (2a + 3c) + (6a + b) (2c – 3a – 5b)
= a(a – 2b + 5c) – b(a – 2b + 5c) – 2a(a – b – c) – 3c(a – b – c) + 6a(2c – 3a – 5b) + b(2c – 3a – 5b)
= a2 – 2ab + 5ac – ab + 2b2 – 5bc – 2a2 + 2ab + 2ac – 3ac + 3bc + 3c2 + 12ac – 18a2 – 30ab + 2bc – 3ab – 5b2
= – 19a2 – 3b2 – 34ab + 16ac + 3c2

iv) (pq – qr + pr) (pq + qr) – (pr + pq) (p + q – r)
= pq(pq – qr + pr) + qr(pq – qr + pr) – pr(p + q – r) – pq(p + q – r)
= p2q2 – pq2r + p2qr + pq2r – q2r2 + pqr2 – p2r – pqr + pr2 – p2q – pq2 + pqr
= p2q2 – q2r2 + p2qr + pqr2 – p2r + pr2 – p2q – pq 2

Question 4.
If a, b, care positive real numbers such that $$\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}$$ ,find the value of $$\frac{(a+b)(b+c)(c+a)}{a b c}$$
Solution:
$$\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}$$ = k then
$$\frac{a+b-c}{c}$$ = k ⇒ a + b – c = kc
⇒ a + b = (ck + c) = c(k + 1) …………… (1)
Similarly b + c = a(k + 1) ……………(2)
c + a = b(k + 1) ………………..(3)