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)