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) m^{2} – n^{2} and m + n

Solution:

i) 2a – 9 and 3a + 4

(2a – 9) (3a + 4) = 2a (3a + 4) – 9(3a + 4)

= 6a^{2} + 8a – 27a – 36

= 6a^{2} – 19a – 36

ii) x – 2y and 2x – y

(x – 2y) (2x – y) = x(2x – y) – 2y(2x – y)

= 2x^{2} – xy – 4xy + 2y^{2}

= 2x^{2} – 5xy + 2y^{2}

iii) kl + lm and k – l

(kl + lm) (k – l) = kl(k – l) + lm(k – l)

= k^{2}l – l^{2}k + klm – l^{2}m

iv) m^{2} – n^{2} and m + n

(m^{2} – n^{2}) (m + n) = m^{2}(m + n) – n^{2}(m + n)

= m^{3} + m^{2}n – n^{2}m – n^{3}

Question 2.

Find the product:

(i) (x + y)(2x – 5y + 3xy)

(ii) (mn – kl + km) (kl – lm)

(iii) (a – 2b + 3c)(ab^{2} – a^{2}b)

(iv) (p^{3} + q^{3})(p – 5q+6r)

Solution:

i) (x + y) (2x – 5y + 3xy)

= x(2x – 5y + 3xy) + y(2x – 5y + 3xy)

= 2x^{2} – 5xy + 3x^{2}y + 2xy – 5y^{2} + 3xy^{2}

= 2x^{2} – 5y^{2} – 3xy + 3x^{2}y + 3xy^{2}

ii) (mn – kl + km) (kl – lm)

= kl(mn – kl + km) – lm(mn – kl + km)

= klmn – k^{2}l^{2} + k^{2}lm – lm^{2}n + kl^{2}m – klm^{2}

iii) (a – 2b + 3c) (ab^{2} – a^{2}b) = a(ab^{2} – a^{2}b) – 2b(ab^{2} – a^{2}b) + 3c(ab^{2}– a^{2}b)

= a^{2}b^{2} – a^{3}b – 2ab^{3} + 2a^{2}b^{2} + 3ab^{2}c – 3a^{2}bc

= 3a^{2}b^{2} – a^{3}b – 2ab^{3} + 3ab^{2}c – 3a^{2}bc

iv) (p^{3} + q^{3}) (p – 5q + 6r) = p^{3}(p – 5q + 6r) + q^{3}(p – 5q + 6r)

= p^{4} – 5p^{3}q + 6p^{3}r + pq^{3} – 5q^{4} + 6rq^{3}

= p^{4} – 5q^{4} – 5p^{3}q + 6p^{3}r + pq^{3} + 6rq^{3}

Question 3.

Simplify the following:

(i) (x-2y) (y – 3x) + (x+y) (x-3y) – (y – 3x) (4x – 5y)

(ii) (m + n) (m^{2} – mn + n^{2})

(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)

= 3y^{2} – 3xy – 9xy + 9x^{2} + x^{2} – 3xy + xy – 3y^{2}

= 10x^{2} – 14xy

ii) (m + n) (m^{2}– mn + n^{2})

= m(m^{2} – mn + n^{2}) + n(m^{2} – mn + n^{2})

= m^{3} – m^{2}n + n^{2}m + nm^{2} – mn^{2} + n^{3}

= m^{3} + n^{3}

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)

= a^{2} – 2ab + 5ac – ab + 2b^{2} – 5bc – 2a^{2} + 2ab + 2ac – 3ac + 3bc + 3c^{2} + 12ac – 18a^{2} – 30ab + 2bc – 3ab – 5b^{2}

= – 19a^{2} – 3b^{2} – 34ab + 16ac + 3c^{2}

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)

= p^{2}q^{2} – pq^{2}r + p^{2}qr + pq^{2}r – q^{2}r^{2} + pqr^{2} – p^{2}r – pqr + pr^{2} – p^{2}q – pq^{2} + pqr

= p^{2}q^{2} – q^{2}r^{2} + p^{2}qr + pqr^{2} – p^{2}r + pr^{2} – p^{2}q – 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)