{"id":42722,"date":"2023-04-03T12:00:17","date_gmt":"2023-04-03T06:30:17","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=42722"},"modified":"2023-04-04T10:30:48","modified_gmt":"2023-04-04T05:00:48","slug":"ml-aggarwal-class-7-solutions-for-icse-maths-chapter-9-ex-9-3","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-7-solutions-for-icse-maths-chapter-9-ex-9-3\/","title":{"rendered":"ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 9 Linear Equations and Inequalities Ex 9.3"},"content":{"rendered":"

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 9 Linear Equations and Inequalities Ex 9.3<\/h2>\n

Question 1.
\nIf the replacement set is (- 5, – 3, – 1, 0, 1, 3, 4), find the solution set of:
\n(i) x < -2
\n(ii) x > 1
\n(iii) x \u2265 -1
\n(iv) -5 < x < 3
\n(v) -3 \u2264 x < 4
\n(vi) 0 \u2264 x < 7.
\nSolution:
\nReplacement set = {-5, -3, -1, 0, 1, 3, 4}
\n(i) Solution set of x < – 2 = {-5, -3}
\n(ii) Solution set of x > 1 = {3, 4}
\n(iii) Solution set of x \u2265 -1 = {- 1, 0, 1, 3, 4}
\n(iv) Solution set of -5 < x < 3 = {-3, -1, 0, 1}
\n(v) Solution set of -3 \u2264 x < 4 = {-3, -1,0, 1, 3}
\n(vi) Solution set of 0 \u2264 x < 7 = {0, 1, 3, 4}<\/p>\n

Question 2.
\nRepresent the following inequations graphically:
\n(i) x \u2264 3, x \u2208 N
\n(ii) x < 4, x \u2208 W
\n(iii) -2 \u2264 x < 4, x \u2208 I
\n(iv) -3 \u2264 x \u2264 2, x \u2208 I
\nSolution:
\n(i) Given x \u2264 3, x \u2208 N
\nThe solution set = {1, 2, 3}
\nThe solution set is shown by thick dots on the number line.
\n\"ML
\n(ii) x < 4, x \u2208 W
\nThe solution set = {0, 1, 2, 3}
\nThe Solution set is shown by thick dots on the number line.
\n\"ML
\n(iii) -2 \u2264 x < 4, x \u2208 I
\nThe solution set = {-2, -1, 0, 1, 2, 3}
\nThe graph of the solution set is shown by thick dots on the number line.
\n\"ML
\n(iv) -3 \u2264 x \u2264 2, x \u2208 I
\nThe solution set = {-3, -2, -1, 0, 1, 2}
\nThe graph of the solution set is shown by thick dots on the number line.
\n\"ML<\/p>\n

Question 3.
\nSolve the following inequations.
\n(i) 4 – x > -2, x \u2208 N
\n(ii) 3x + 1 \u2264 8, x \u2208 W
\nAlso represent their solutions on the number line.
\nSolution:
\n(i) Given, 4 – x > -2
\nSubtract 4 from both sides
\n\u21d2 -4 + 4 – x > -2 – 4 – x > -6
\n\u21d2 x < 6 (Reverse the symbols)
\nAs x \u2208 N, the solution set = {1, 2, 3, 4, 5}
\nThe graph of the solution set
\n\"ML
\n(ii) Given 3x + 1 \u2264 8.
\nSubtracting -1 from both sides,
\n3x + 1 – 1 \u2264 8 – 1
\n3x \u2264 7
\nDividing both sides by 3
\n\u21d2 x \u2264 \\(\\frac { 7 }{ 3 }\\)
\nAs x = W, the solution set = {0, 1, 2}
\nThe graph of the solution set
\n\"ML<\/p>\n

Question 4.
\nSolve 3 – 4x < x – 12, x \u2208 {-1, 0, 1, 2, 3, 4, 5, 6, 7}.
\nSolution:
\nGiven 3 – 4x < x – 12
\nSubtracting 3 from both sides
\n\u21d2 -3 + 3 – 4x < x – 12 – 3
\n\u21d2 -4x < x – 15
\nSubtracting x from both sides
\n\u21d2 -4x – x < x – x – 15
\n\u21d2 -5x < -15 \u21d2 x > 3
\n(Dividing by – 5 and reverse the symbols)
\nAs x \u2208 {-1, 0, 1, 2, 3, 4, 5, 6, 7}
\nThe solution set = {4, 5, 6, 7}<\/p>\n

Question 5.
\nSolve -7 < 4x + 1 \u2264 23, x \u2208 I.
\nSolution:
\nGiven, -7 < 4x + 1 \u2264 23.
\nWe take -7 < 4x + 1 \u2264 23.
\nSubtracting -1 from all sides,
\n-7 – 1 < 4x + 1 – 1 \u2264 23 – 1
\n-8 < 4x \u2264 22
\n\\(\\frac { -8 }{ 4 }\\) < \\(\\frac { 4x }{ 4 }\\) \u2264 \\(\\frac { 22 }{ 4 }\\) (Dividing by 4)
\n-2 < x \u2264 5.5
\nAs x \u2208 I, the solution set = {-1, 0, 1, 2, 3, 4, 5}.<\/p>\n

ML Aggarwal Class 7 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 9 Linear Equations and Inequalities Ex 9.3 Question 1. If the replacement set is (- 5, – 3, – 1, 0, 1, 3, 4), find the solution set of: (i) x < -2 (ii) x > 1 (iii) x \u2265 -1 (iv) -5 < x < …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/42722"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=42722"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/42722\/revisions"}],"predecessor-version":[{"id":158984,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/42722\/revisions\/158984"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=42722"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=42722"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=42722"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}