{"id":27744,"date":"2021-07-01T17:40:00","date_gmt":"2021-07-01T12:10:00","guid":{"rendered":"https:\/\/mcqbooks.com\/?p=27744"},"modified":"2022-03-06T18:49:55","modified_gmt":"2022-03-06T13:19:55","slug":"ncert-solutions-for-class-8-maths-chapter-9-ex-9-3","status":"publish","type":"post","link":"https:\/\/mcqbooks.com\/ncert-solutions-for-class-8-maths-chapter-9-ex-9-3\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 9 Algebraic Expressions and Identities Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3<\/h2>\n

Question 1.
\nCarry out the multiplication of the expressions in each of the following pairs.
\n(i) 4p, q + r
\n(ii) ab, a – b
\n(iii) a + b, 7a2<\/sup>b2<\/sup>
\n(iv) a2<\/sup> – 9, 4a
\n(v) pq + qr + rp, 0
\nSolution:
\n(i) (4p) \u00d7 (q + r)
\n= (4p \u00d7 q) + (4p \u00d7 r)
\n= 4pq + 4pr<\/p>\n

(ii) (ab) \u00d7 (a – b)
\n= (ab \u00d7 a) – (ab \u00d7 b)
\n= a2<\/sup>b – ab2<\/sup><\/p>\n

(iii) (a + b) (7a2<\/sup>b2<\/sup>)
\n= (a \u00d7 7a2<\/sup>b2<\/sup>) + (b \u00d7 7a2<\/sup>b2<\/sup>)
\n= 7a3<\/sup>b2<\/sup> + 7a2<\/sup>b3<\/sup><\/p>\n

\"NCERT<\/p>\n

(iv) (a2<\/sup> – 9) \u00d7 4a
\n= (a2<\/sup> \u00d7 4a) – (9 \u00d7 4a)
\n= 4a3<\/sup> – 36a<\/p>\n

(v) (pq + qr + rp) \u00d7 0
\n= (pq \u00d7 0) + (qr \u00d7 0) + (rp \u00d7 0)
\n= 0 + 0 + 0
\n= 0<\/p>\n

Question 2.
\nComplete the table.
\n\"NCERT
\nSolution:
\n(i) a \u00d7 (b + c + d)
\n= (a \u00d7 b) + (a \u00d7 c) + (a \u00d7 d)
\n= ab + ac + ad<\/p>\n

(ii) (x + y – 5) (5xy)
\n= (x \u00d7 5xy) + (y \u00d7 5xy) – (5 \u00d7 5xy)
\n= 5x2<\/sup>y + 5xy2<\/sup> – 25xy<\/p>\n

(iii) p \u00d7 (6p2<\/sup> – 7p + 5)
\n= (p \u00d7 6p2<\/sup>) + p \u00d7 (-7p) + p \u00d7 5
\n= 6p3<\/sup> + (-7p2<\/sup>) + 5p
\n= 6p3<\/sup> – 7p2<\/sup> + 5p<\/p>\n

(iv) 4p2<\/sup>q2<\/sup> \u00d7 (p2<\/sup> – q2<\/sup>)
\n= (4p2<\/sup>q2<\/sup> \u00d7 p2<\/sup>) + (4p2<\/sup>q2<\/sup> \u00d7 -q2<\/sup>)
\n= 4p4<\/sup>q2<\/sup> + (-4p2<\/sup>q4<\/sup>)
\n= 4p4<\/sup>q2<\/sup> – 4p2<\/sup>q4<\/sup><\/p>\n

(v) (a + b + c) \u00d7 abc
\n= (a \u00d7 abc) + (b \u00d7 abc) + (c \u00d7 abc)
\n= a2<\/sup>bc + ab2<\/sup>c + abc2<\/sup><\/p>\n

\"NCERT<\/p>\n

Question 3.
\nFind the product.
\n(i) (a2<\/sup>) \u00d7 (2a22<\/sup>) \u00d7 (4a26<\/sup>)
\n(ii) \\(\\left(\\frac{2}{3} x y\\right) \\times\\left(\\frac{-9}{10} x^{2} y^{2}\\right)\\)
\n(iii) \\(\\left(-\\frac{10}{3} \\mathrm{pq}^{3}\\right) \\times\\left(\\frac{6}{5} \\mathrm{p}^{3} \\mathrm{q}\\right)\\)
\n(iv) x \u00d7 x2<\/sup> \u00d7 x3<\/sup> \u00d7 x4<\/sup>
\nSolution:
\n(i) (a2<\/sup>) \u00d7 (2a22<\/sup>) \u00d7 (4a26<\/sup>)
\n= (1 \u00d7 2 \u00d7 4) \u00d7 (a2<\/sup> \u00d7 a22<\/sup> \u00d7 a26<\/sup>)
\n= 8 \u00d7 (a2+22+26<\/sup>)
\n= 8a50<\/sup>
\n\"NCERT
\n(iv) x \u00d7 x2<\/sup> \u00d7 x3<\/sup> \u00d7 x4<\/sup>
\n= x1+2+3+4<\/sup>
\n= x10<\/sup><\/p>\n

Question 4.
\n(a) Simplify 3x(4x – 5) + 3 and find its values for
\n(i) x = 3
\n(ii) x = \\(\\frac{1}{2}\\)
\n(b) Simplify a (a2<\/sup> + a + 1) + 5 and find its value for
\n(i) a = 0
\n(ii) a = 1
\n(iii) a = -1.
\nSolution:
\n(a) 3x (4x – 5) + 3
\n= (3x \u00d7 4x) + (3x \u00d7 -5) + 3
\n= 12x2<\/sup> – 15x + 3
\n(i) when x = 3
\n12x2<\/sup> – 15x + 3
\n= 12 \u00d7 (3)2<\/sup> – 15(3) + 3
\n= 12 \u00d7 9 – 45 + 3
\n= 108 – 45 + 3
\n= 111 – 45
\n= 66<\/p>\n

(ii) When x = \\(\\frac{1}{2}\\)
\n12x2<\/sup> – 15x + 3
\n\"NCERT<\/p>\n

(b) a(a2<\/sup> + a + 1) + 5
\n= (a \u00d7 a2<\/sup>) + (a \u00d7 a) + (a \u00d7 1) + 5
\n= a3<\/sup> + a2<\/sup> + a + 5<\/p>\n

(i) When a = 0
\na3<\/sup> + a2<\/sup> + a + 5
\n= (0)3<\/sup> + (0)2<\/sup> + 0 + 5
\n= 0 + 0 + 0 + 5
\n= 5<\/p>\n

(ii) When a = 1
\na3<\/sup> + a2<\/sup> + a + 5
\n= (13<\/sup>) + (12<\/sup>) + 1 + 5
\n= 1 + 1 + 1 + 5
\n= 8<\/p>\n

\"NCERT<\/p>\n

(iii) When a = -1
\na3<\/sup> + a2<\/sup> + a + 5
\n= (-1)3<\/sup> + (-1)2<\/sup> + (-1) + 5
\n= (-1) + 1 – 1 + 5
\n= -1 + 1 – 1 + 5
\n= 4<\/p>\n

Question 5.
\n(a) Add: p(p – q), q(q – r) and r (r – p)
\n(b) Add: 2x(z – x – y) and 2y(z – y – x)
\n(c) Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)
\n(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c (-a + b + c)
\nSolution:
\n(a) p(p – q) + q(q – r) + r(r – p)
\n= p \u00d7 p – p \u00d7 q + q \u00d7 q + q(-r) + r \u00d7 r + r \u00d7 (-p)
\n= p2<\/sup> – pq + q2<\/sup> – qr + r2<\/sup> – rp
\n= p2<\/sup> + q2<\/sup> + r2<\/sup> – (pq + qr + rp)<\/p>\n

(b) 2x (z – x – y) + 2y(z – y – x)
\n= 2x \u00d7 z + 2x (-x) + 2x (-y) + 2yz + 2y (-y) + 2y(-x)
\n= 2xz – 2x2<\/sup> – 2xy + 2yz – 2y2<\/sup> – 2xy
\n= -2x2<\/sup> – 2y2<\/sup> – 4xy + 2yz + 2xz<\/p>\n

(c) 4l(10n – 3m + 2l) – 3l(l – 4m + 5n)
\n= (4l \u00d7 10n) + (4l \u00d7 -3m) + (4l \u00d7 2l) + (-3l \u00d7 1) + (-3l \u00d7 -4m) + (-3l \u00d7 5n)
\n= 40ln + (-12lm) + 8l2<\/sup> + (-3l2<\/sup>) + 12lm + (-15ln)
\n= 40ln – 12lm + 8l2<\/sup> – 3l2<\/sup> + 12lm – 15ln
\n= 40ln – 15ln – 12lm + 12lm + 8l2<\/sup> – 3l2<\/sup>
\n= 25ln + 5l2<\/sup>
\n= 5l2<\/sup> + 25ln<\/p>\n

\"NCERT<\/p>\n

(d) Simplify the I part we get 3a (a + b + c) – 2b(a – b + c)
\n= (3a \u00d7 a) + (3a \u00d7 b) + (3a \u00d7 c) – [(2b \u00d7 a) + 2b(-b) + (2b) \u00d7 c]
\n= 3a2<\/sup> + 3ab + 3ac – (2ab – 2b2<\/sup> + 2bc)
\n= 3a2<\/sup> + 3ab + 3ac – 2ab + 2b2<\/sup> – 2bc
\n= 3a2<\/sup> + 2b2<\/sup> + 3ab – 2ab + 3ac – 2bc
\n= 3a2<\/sup> + 2b2<\/sup> + ab + 3ac – 2bc
\nSimplify the 2nd part 4c \u00d7 (-a + b + c) = (4c \u00d7 – a) + (4c \u00d7 b) + (4c \u00d7 c)
\nAccording to the given question = -4ac + 4bc + 4c2<\/sup>
\n2nd part – 1st part
\n= -4ac + 4bc + 4c2<\/sup> – (3a2<\/sup> + 2b2<\/sup> + ab + 3ac – 2bc)
\n= -4ac + 4bc + 4c2<\/sup> – 3a2<\/sup> – 2b2<\/sup> – ab – 3ac + 2bc
\n= -4ac – 3ac + 4bc + 2bc + 4c2<\/sup> – 3a2<\/sup> – 2b2<\/sup> – ab
\n= -7ac + 6bc + 4c2<\/sup> – 3a2<\/sup> – 2b2<\/sup> – ab
\n= -3a2<\/sup> – 2b2<\/sup> + 4c2<\/sup> – ab + 6bc – 7ac<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3 Question 1. Carry out the multiplication of the expressions in each of the following pairs. … Read more<\/a><\/p>\n","protected":false},"author":10,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[7],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/posts\/27744"}],"collection":[{"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/users\/10"}],"replies":[{"embeddable":true,"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/comments?post=27744"}],"version-history":[{"count":1,"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/posts\/27744\/revisions"}],"predecessor-version":[{"id":37281,"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/posts\/27744\/revisions\/37281"}],"wp:attachment":[{"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/media?parent=27744"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/categories?post=27744"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mcqbooks.com\/wp-json\/wp\/v2\/tags?post=27744"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}