{"id":16041,"date":"2020-09-10T17:33:11","date_gmt":"2020-09-10T12:03:11","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=16041"},"modified":"2021-07-02T15:01:52","modified_gmt":"2021-07-02T09:31:52","slug":"kseeb-solutions-for-class-7-maths-chapter-12-ex-12-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-7-maths-chapter-12-ex-12-3\/","title":{"rendered":"KSEEB Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3"},"content":{"rendered":"

Students can Download Chapter 12 Algebraic Expressions Ex 12.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths<\/a>, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3<\/h2>\n

Question 1.
\nIf m = 2, find the value of:
\ni) m – 2
\n2 – 2 = 0<\/p>\n

ii) 3m – 5
\n3 \u00d7 2 – 5 = 6 – 5 = 1<\/p>\n

iii) 9 – 5m
\n9 – 5 \u00d7 2 \u00f7 9 – 10 = -1<\/p>\n

\"KSEEB<\/p>\n

iv) 3m2<\/sup> – 2m – 7
\n= 3 \u00d7 22<\/sup> – 2 \u00d7 2 – 7
\n= 3 \u00d7 4 – 4 – 7
\n= 12 – 11 = 1<\/p>\n

v)
\n\"KSEEB
\nSolution:
\n\"KSEEB<\/p>\n

Question 2.
\nIf p = -2, find the value of :
\nSolution:
\ni) 4p + 7
\n= 4x – 2 + 7
\n= -8 + 7 = -1<\/p>\n

ii) -3p2<\/sup> + 4p + 7
\n= -3 (-2)2<\/sup> + 4 (-2) + 7
\n= 3 \u00d7 4 – 8 + 7
\n= 12 – 8 + 7
\n= -13<\/p>\n

\"KSEEB<\/p>\n

iii) -2p3<\/sup> – 3p2<\/sup> – 4p + 7
\n= -2 (-2)3<\/sup> – 3(2)2<\/sup> + 4 (-2) + 7
\n= -2(-8) – 3(4) – 8 + 7
\n= 16 – 12 – 8 + 7
\n= 16 + 7 – 12 – 8
\n= 23 – 20 = 3<\/p>\n

Question 3.
\nFind the value of the following expressions, when x = -1:<\/p>\n

i) 2x – 7
\n= 2(-1) – 7
\n= -2 – 7 = -9<\/p>\n

ii) -x + 2
\n= -1 + 2 = 1<\/p>\n

iii) x2<\/sup> + 2x + 1
\n= 1 – 2 + 1 = 0<\/p>\n

iv) 2x2<\/sup> – x – 2
\n= 2(-1)2<\/sup> – (-1) – 2
\n= 2 \u00d7 1 + 1 – 2
\n= 2 + 1 – 2 = 1<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nIf a = 2, b = -2, find the value of:
\ni) a2<\/sup> + b2<\/sup>
\n= 22<\/sup> + (-2)2<\/sup>
\n= 4 + 4 = 8<\/p>\n

ii) a2<\/sup> + ab + b2<\/sup>
\n= 22<\/sup> + 2 \u00d7 -2 + (-2)
\n= 4 – 4 + 4 = 4<\/p>\n

iii) a2<\/sup> – b2<\/sup>
\n= (2)2<\/sup> – (-2)2<\/sup> = 4 – 4 = 0<\/p>\n

Question 5.
\nWhen a = 0, b = -1, find the value of the given expressions :
\ni) 2a + 2b
\n= 2 \u00d7 0 + 2 \u00d7 -1
\n= o – 2 = -2<\/p>\n

ii) 2a2<\/sup> + 2b2<\/sup> + 1
\n= 2(0)2<\/sup> + (-1)2<\/sup> + 1
\n= 0 + 1 + 1 = 2<\/p>\n

iii) 2a2<\/sup>b + 2ab2<\/sup> + ab
\n= 2 \u00d7 02<\/sup> \u00d7 -1 + 2 \u00d7 0 \u00d7 (-1)2<\/sup> + 0 \u00d7 -1
\n= 0 + 0 + 0 = 0<\/p>\n

iv) a2<\/sup> + ab + 2
\n= (0)2<\/sup> + 0 \u00d7 -1 + 2
\n= 0 + 0 + 2 = 2<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nSimplify the expressions and find the value if x is equal to 2
\nSolution:
\ni) x + 7 + 4 (x \u2013 5)
\n= x + 7 + 4x \u2013 20 (re-arranging the terms)
\n= 5x \u2013 13 = 5(2) \u2013 12
\n= 10 \u2013 13 = -3<\/p>\n

ii) 3(x + 2) + 5x \u2013 7
\n= 3x + 6 + 5x \u2013 1
\n= 8x \u2013 1 = 8(2) \u2013 1
\n= 16 \u2013 1 = 15<\/p>\n

iii) 6x + 5 + 5(x \u2013 2)
\n= 6x+ 5x \u2013 10
\n= 11x \u2013 10 = 11(2) \u2013 10
\n= 22 \u2013 10 = 12<\/p>\n

iv) 4(2x \u2013 1) + 3x + 11
\n= 8x \u2013 4 + 3x + 11
\n= 8x + 3x \u2013 4 + 11 (re-arranging the terms)
\n= 11x + 7 = 11(2) + 7 = 22 + 7 = 29<\/p>\n

Question 7.
\nSimplify these expressions and find their values if x = 3, a = -1, b = -2.
\ni) 3x – 5 – x + 9
\n= 3x – x + 9 – 5
\n= 2x + 4 = 2(3) + 4 = 6 + 4 = 10<\/p>\n

ii) 2 – 8x + 4x + 4
\n= 2 + 4 – 4x
\n= -4x + 6
\n= -4 \u00d7 3 + 6
\n= -12 + 6 = -6<\/p>\n

\"KSEEB<\/p>\n

iii) 3a + 5 – 8a + 1
\n= 3a – 8a + 5 + 1
\n= -5a + 6
\n= -5(-1) + 6
\n= 5 + 6 = 11<\/p>\n

iv) 10 – 3b – 4 – 5b
\n= -3b – 5b + 10 – 4
\n= -8b + 6
\n= -8 (-2) + 6
\n= 16 + 16 = 22<\/p>\n

v) 2a – 2b – 4 – 5 + a
\n= 2a – 2b – 9
\n= 3a – 2b – 9
\n= 3(-1) – 2 (-2) – 9
\n= -3 + 4 – 9
\n= -12 + 4 = -8<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\ni) If z = 10, find the value of z3<\/sup> – 3(z – 10)
\n= (10)3<\/sup> – 3 (10 – 10)
\n= 1000 – 3(0)
\n= 1000 – 0 = 1000<\/p>\n

ii) If p = -10, find the value of p2<\/sup> – 2p – 100
\n= (10)2<\/sup> – 2(-10) – 100
\n= 100 + 20 – 100
\n= 120 – 100
\n= 20<\/p>\n

Question 9.
\nWhat should be the value of a if the value of 2x2<\/sup> + x – a equals to 5, when x = 0 ?
\nSolution:
\n= 2x2<\/sup> + x – a
\n= 2(0)2<\/sup> + 0 – a
\n= 5
\n= 0 + 0 – a = 5
\n\u2234 a = -5<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nSimplify the expression and find its value when a = 5 and b = -3, 2(a2<\/sup> + ab) + 3 – ab
\n= 2a2<\/sup> + 2ab + 3 – ab
\n= 2(5)2<\/sup> + 2 \u00d7 5 \u00d7 -3 + 3 – 5 \u00d7 -3
\n= 2 \u00d7 25 – 30 + 3 + 15
\n= 50 – 30 + 3 + 15
\n= 50 + 3 + 15 – 30
\n= 68 – 30 = 38<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 12 Algebraic Expressions Ex 12.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 Question 1. If m = 2, …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[84],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/16041"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=16041"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/16041\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=16041"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=16041"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=16041"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}