{"id":22000,"date":"2022-01-13T17:07:26","date_gmt":"2022-01-13T11:37:26","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=22000"},"modified":"2022-01-28T10:06:33","modified_gmt":"2022-01-28T04:36:33","slug":"ap-board-7th-class-maths-solutions-chapter-3-ex-3-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-7th-class-maths-solutions-chapter-3-ex-3-3\/","title":{"rendered":"AP Board 7th Class Maths Solutions Chapter 3 Simple Equations Ex 3.3"},"content":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf<\/a> Chapter 3 Simple Equations Ex 3.3 Textbook Exercise Questions and Answers.<\/p>\n

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.3<\/h2>\n

Question 1.
\nSolve the following equations and check the result.
\n(i) 5x – 17 = 18
\nAnswer:
\nGiven 5x – 17 = 18
\n\u21d2 5x – 17 + 17 = 18 + 17 (Add 17 on both sides)
\n\u21d2 5x = 35
\n\u21d2 \\(\\frac{5 x}{5}\\) = \\(\\frac{35}{5}\\) (Divide by 5 on both sides)
\n\u21d2 x = 7<\/p>\n

Check:
\nSubstitute x = 7 in 5x – 17 = 18
\nLHS = 5x – 17
\n= 5(7) – 17
\n= 35 – 17 = 18 = RHS
\nHence verified.<\/p>\n

\"AP<\/p>\n

(ii) 29 – 7y = 1
\nAnswer:
\nGiven 29 – 7y = 1
\n\u21d2 29 – 7y – 29 = 1 – 29 (Subtract 29 on both sides)
\n\u21d2 – 7y = -28 .
\n\u21d2 \\(\\frac{-7 y}{-7}\\) = \\(\\frac{-28}{-7}\\) (Divide by – 7 on both sides)
\n\u21d2 y = 4<\/p>\n

Check:
\nSubstitute y = 4 in
\n29 – 7y = 1
\nLHS = 29 – 7y
\n= 29 – 7(4)
\n= 29 – 28 = 1 = RHS
\nHence verified.<\/p>\n

(iii) a – 2.3 = 1.5
\nAnswer:
\nGiven a – 2.3 = 1.5
\n\u21d2 a – 2.3 + 2.3 = 1.5 + 2.3 (Add 2.3 ort both sides)
\n\u21d2 a = 3.8<\/p>\n

Check:
\nSubstitute a = 3.8 in a – 2.3 =1.5
\nLHS = a – 2.3
\n= 3.8 – 2.3
\n= 1.5 = RHS
\nHence verified.<\/p>\n

(iv) b + 3\\(\\frac{1}{2}\\) = \\(\\frac{7}{4}\\)
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

\"AP<\/p>\n

(v) \\(\\frac{7 p}{10}\\) + 9 = 15
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

\"AP<\/p>\n

(vi) 6(q – 5) = 42.
\nAnswer:
\nGiven 6(q – 5) = 42
\n\u21d2 \\(\\frac{6(q-5)}{6}\\) = \\(\\frac{42}{6}\\) (Divide by 6 on both sides)
\n\u21d2 q – 5 = 7
\n\u21d2 q – 5 + 5 = 7 + 5 (Add 5 on both sides)
\n\u21d2 q = 12<\/p>\n

Check:
\nSubstitute q = 12 in
\n6(q – 5) = 42
\nLHS = 6(q – 5)
\n= 6(12 – 5)
\n= 6(7) = 42 = RHS
\nHence verified.<\/p>\n

(vii) – 3(m + 5) + 1 = 13
\nAnswer:
\nGiven – 3(m + 5)4-1 = 13
\n\u21d2 – 3(m + 5)+ 1 – 1 = 13 – 1 (Subtract 1 on both sides)
\n\u21d2 – 3(m + 5) = 12
\n\u21d2 \\(\\frac{-3(m+5)}{-3}\\) = \\(\\frac{12}{-3}\\) (Divide by – 3 on both sides)
\n\u21d2 m + 5 = – 4
\n\u21d2 m + 5 – 5 = – 4 – 5 (Subtract 5 on both sides)
\n\u21d2 m = – 9<\/p>\n

Check:
\nSubstitute m = – 9 in – 3(m + 5) + 1 = 13
\nLHS = – 3(m + 5) + 1
\n= – 3(- 9 + 5) + 1 = (- 3 \u00d7 – 4) + 1
\n= + 12 + 1 = 13 = RHS
\nHence verified.<\/p>\n

\"AP<\/p>\n

(viii) \\(\\frac{n}{2}+\\frac{n}{3}+\\frac{n}{5}\\) = 31
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 2.
\nSolve the following equations and check the result.<\/p>\n

(i) 3(p – 7) – 4 = 5
\nAnswer:
\nGiven 3(p – 7) – 4 = 5
\n\u21d2 3(p – 7) – 4 + 4 = 5 + 4 (Add 4 on both sides)
\n\u21d2 3(p – 7) = 9
\n\u21d2 \\(\\frac{3(p-7)}{3}\\) = \\(\\frac{9}{3}\\) (Divide by 3 on both sides)
\n\u21d2 p – 7 = 3
\n\u21d2 p – 7 + 7 = 3 + 7 (Add 7 on both sides)
\n\u21d2 p = 10<\/p>\n

Check:
\nSubstitute p = 10 in
\n3(p – 7) – 4 = 5
\nLHS = 3(p – 7) – 4.
\n= 3(10 – 7) – 4.
\n= 3 \u00d7 3 – 4
\n= 9 – 4 = 5 = RHS
\nHence verified.<\/p>\n

(ii) 5(q – 3) – 3(q – 2) = 0
\nAnswer:
\nGiven 5(q – 3) – 3(q – 2) = 0
\n\u21d2 5q – 15 – 3q + 6 = 0 (Distributive property)
\n\u21d2 2q – 9 = 0 .
\n\u21d2 2q – 9 + 9 = 0 + 9 (Add 9 on both sides)
\n\u21d2 2q = 9
\n\u21d2 \\(\\frac{2 \\mathrm{q}}{2}\\) = \\(\\frac{9}{2}\\) (Divide by 2 on both sides)
\n\u21d2 q = \\(\\frac{9}{2}\\)<\/p>\n

Check : Substitute q = \\(\\frac{9}{2}\\) in
\n5(q – 3) – 3(q – 2) = 0
\nLHS = 5(q – 3) – 3(q – 2)
\n= 5\\(\\left(\\frac{9}{2}-3\\right)\\) – 3\\(\\left(\\frac{9}{2}-2\\right)\\)
\n= 5\\(\\left(\\frac{9-6}{2}\\right)\\) – 3\\(\\left(\\frac{9-4}{2}\\right)\\)
\n= 5 \u00d7 \\(\\frac{3}{2}\\) – 3 \u00d7 \\(\\frac{5}{2}\\)
\n= \\(\\frac{15}{2}-\\frac{15}{2}\\) = 0 = RHS
\nHence verified.<\/p>\n

\"AP<\/p>\n

(iii) 4x – 0.3x – 1.2 = 0.6
\nAnswer:
\nGiven 0.4x – 0.3x – 1.2 = 0.6
\n\u21d2 0.1 x – 1.2 = 0.6
\n\u21d2 0.1 x – 1.2 + 1.2 = 0.6 + 1.2 (Add 1.2 on both sides)
\n\u21d2 0.1 x = 1.8
\n\u21d2 \\(\\frac{1 \\mathrm{x}}{10}\\) = \\(\\frac{18}{10}\\)
\n\"AP
\n(Multiply by 10 on both sides)
\n\u21d2 x = 18<\/p>\n

Check:
\nSubstitute x = 18 in
\n0.4x – 0.3x – 1.2 = 0.6
\nLHS = 0.4x – 0.3x – 1.2
\n= 0.4(18) – 0.3(18) – 1.2
\n= 7.2 – 5.4 – 1.2
\n= 7.2 – 6.6 = 0.6 = RHS
\nHence verified.<\/p>\n

(iv) 4(3y + 4) = 7.6
\nAnswer:
\nGiven 4(3y + 4) = 7.6.
\n\u21d2 \\(\\frac{4(3 y+4)}{4}\\) = \\(\\frac{7.6}{4}\\) (Divide by 4 on both sides)
\n\u21d2 3y + 4 = 1.9
\n\u21d23y + 4 – 4= 1.9 – 4 (Subtract 4 on both sides)
\n\u21d2 3y = – 2.1
\n\u21d2 \\(\\frac{3 y}{3}\\) = – \\(\\frac{2.1}{3}\\) (Divide by 3 on both sides)
\n\u21d2 y = – 0.7<\/p>\n

Check:
\nSubstitute y = – 0.7 in
\n4(3y + 4) = 7.6
\nLHS = 4(3y + 4)
\n= 4[3 (- 0.7) + 4]
\n= 4[- 2.1 + 4]
\n= 4 \u00d7 1.9 = 7.6 = RHS
\nHence verified.<\/p>\n

(v) 20 – (2r – 5) = 25
\nAnswer:
\nGiven 20 – (2r – 5) = 25
\n\"AP
\n(Subtract 20 on both sides)
\n\u21d2 – (2r – 5) = 5
\n\u21d2 – (2r -5) \u00d7 – 1 = 5 \u00d7 – 1
\n(Multiply by – 1 on both sides)
\n\u21d2 2r – 5 = 5
\n\u21d2 2r – 5 + 5= – 5 + 5 (Add 5 on both sides)
\n\u21d2 2r = 0
\n\u21d2 \\(\\frac{2 \\mathrm{r}}{2}\\) = \\(\\frac{0}{2}\\) (Divide by 2 on both sides)
\n\u21d2 r = 0<\/p>\n

Check:
\nSubstitute r = 0 in
\n20 – (2r – 5) = 25
\nLHS = 20 – (2r – 5)
\n= 20 – [2(0) – 5]
\n= 20 – [0 – 5]
\n= 20 – (- 5)
\n= 20 + 5 = 25 = RHS
\nHence verified.<\/p>\n

\"AP<\/p>\n

(vi) 3(5 – t) – 2(t – 2) = – 1
\nAnswer:
\nGiven 3(5 – t) – 2(t – 2) = – 1
\n\u21d2 15 – 3t – 2t + 4 .= – 1 (Distributive property)
\n\u21d2 19 – 5t = – 1
\n\"AP
\n(Subtract 19 on both sides)
\n\u21d2 – 5t = – 20
\n\u21d2 \\(\\frac{-5 t}{-5}\\) = \\(\\frac{-20}{-5}\\) (Divide by – 5 on both sides)
\n\u21d2 t = 4<\/p>\n

Check:
\nSubstitute t = 4 in
\n3(5 -1) – 2(t – 2) = – 1
\nLHS = 3(5 – t) – 2(t – 2)
\n= 3(5 – 4) – 2(4 – 2)
\n= 3(1) – 2(2)
\n= 3 – 4 = – 1 = RHS
\nHence verified.<\/p>\n

(vii) 3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16
\nAnswer:
\nGiven 3(2k + 1) – 2(k T 5) – 5(5 – 2k) = 16
\n\u21d2 6k + 3-2k + 10-25 + 10k = 16 (Distributive property)
\n\u21d2 14k – 12 = 16
\n\u21d2 14k – 12 + 12 – 16 + 12 (Add 12 on both sides)
\n\u21d2 14k = 28
\n\u21d2 \\(\\frac{14 \\mathrm{k}}{14}\\) = \\(\\frac{28}{14}\\) (Divide by 14 on both sides)
\n\u21d2 k = 2<\/p>\n

Check:
\nSubstitute k = 2 in
\n3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16
\nLHS = 3(2k + 1) – 2(k – 5) – 5(5 – 2k)
\n= 3[2 \u00d7 2 + 1] – 2[2 – 5] – 5[5 – 2 \u00d7 2]
\n= 3[4 + 1] – 2(- 3) – 5(5 – 4)
\n= 15 + 6 – 5 – 16 = RHS
\nHence verified.<\/p>\n

\"AP<\/p>\n

(viii) \\(\\frac{3 \\mathrm{~m}}{4}\\) – 5m – \\(\\frac{3}{4}\\)
\nAnswer:
\n\"AP\u21d2 – 17m – 3 = 48
\n\u21d2 – 17m – 3 + 3 =48 + 3 (Add 3 on both sides)
\n\u21d2 – 17m = 51
\n\u21d2 \\(\\frac{-17 \\mathrm{~m}}{-17}\\) = \\(\\frac{51}{-17}\\) (Divide by – 17 on both sides)
\n\u21d2 m = – 3<\/p>\n

\"AP<\/p>\n

(ix) \\(\\frac{4 n}{5}+\\frac{n}{4}-\\frac{n}{2}=\\frac{11}{10}\\)
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

\"AP<\/p>\n

(x) \\(\\frac{x}{2}-\\frac{4}{5}+\\frac{x}{5}+\\frac{3 x}{10}=\\frac{1}{5}\\)
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

\"AP<\/p>\n

Question 3.
\nWrite any three equivalent equations having the solution x = 2.
\nAnswer:
\n\"AP<\/p>\n

Question 4.
\nWrite any three equivalent equations having the solution a = – 5.
\nAnswer:
\n\"AP<\/p>\n","protected":false},"excerpt":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.3 Textbook Exercise Questions and Answers. AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.3 Question 1. Solve the following equations and check the result. (i) 5x – 17 = 18 Answer: Given 5x – 17 = 18 \u21d2 5x …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[9],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/22000"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=22000"}],"version-history":[{"count":5,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/22000\/revisions"}],"predecessor-version":[{"id":23880,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/22000\/revisions\/23880"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=22000"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=22000"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=22000"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}