{"id":14709,"date":"2020-09-09T12:57:03","date_gmt":"2020-09-09T07:27:03","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=14709"},"modified":"2021-07-02T15:01:54","modified_gmt":"2021-07-02T09:31:54","slug":"kseeb-solutions-for-class-7-maths-chapter-4-ex-4-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-7-maths-chapter-4-ex-4-2\/","title":{"rendered":"KSEEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2"},"content":{"rendered":"

Students can Download Chapter 4 Simple Equations Ex 4.2, 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 4 Simple Equations Ex 4.2<\/h2>\n

Question 1.
\nGiven first the step you will use to separate the variable and then solve the equation:
\na) x – 1 = 0
\nSolution:
\nThe given equation = x – 1 = 0
\nAdd 1 to both the sides
\n\"KSEEB
\nChecking
\nx – 1 = 0
\nput the value of x = 1
\n1 – 1 = 0
\n\u2234 LHS = RHS (checked)<\/p>\n

b) x + 1 = 0
\nThe given equation = x + 1 = 0
\nSubtract 1 from both the sides
\n\"KSEEB
\nVerification
\nx+ 1 = 0
\nsubstitute the value of x = -1 then x + 1 = 0
\n(-1) + 1 = 0
\n0 = 0
\nLHS = RHS (verified)<\/p>\n

c) x – 1 = 5
\nThe given equation is x – 1 = 5
\nAdd 1 to both sides
\n\"KSEEB
\nSubstitute the value in the equation x = 6.
\nx – 1 = 5
\n6 – 1 = 5
\n5 = 5
\n\u2234 LHS = RHS (verified)<\/p>\n

\"KSEEB<\/p>\n

d) x + 6 = 2
\nThe given equation is x + 6 = 2
\nSubtract 6 from both the sides
\n\"KSEEB
\nSubstitute the value of x = -4 in the given equation,
\nx + 6 = 2
\n-4 + 6 = 2
\n2 = 2
\n\u2234 LHS = RHS (verified)<\/p>\n

e) y – 4 = -7
\nThe given equation is y – 4 = -7
\nAdd 4 to both the sides
\n\"KSEEB
\nsubstitute the value of y = -3 in the given equation
\ny – 4 = -7
\n-3 – 4 = -7
\n-7 = -7
\n\u2234 LHS = RHS (verified)<\/p>\n

f) y – 4 = 4
\nSolution:
\nThe given equation is y – 4 = 4
\nAdd 4 to both the sides
\n\"KSEEB
\nsubstitute the value of y = 8 in the given equation
\ny – 4 = 4
\n8 – 4 = 4
\n4 = 4
\n\u2234 LHS = RHS (verified)<\/p>\n

g) y + 4 = 4
\nSolution:
\nThe given equation is y + 4 = 4
\nsubtract 4 from both sides
\n\"KSEEB
\nsubstitute the value of y = 0 in the given equation.
\ny + 4 = 4
\n0 + 4 = 4
\n4 = 4
\n\u2234 LHS = RHS (verified)<\/p>\n

h) y + 4 = -4
\nSolution:
\nThe given equation is y + 4 = -4
\nsubtract 4 from both the sides
\n\"KSEEB
\ny = -8 (verified)
\nsubstitute the value of y = -8 in the given equation
\ny + 4 = -4
\n-8 + 4 = -4
\n-4 = -4
\n\u2234 LHS = RHS (verified)<\/p>\n

Question 2.
\nGive first the step you will use to separate the variable and then solve the equation:<\/p>\n

a) 3l = 42
\nSolution:
\nThe given equation is 3l = 42
\nDivide by 3 both the sides
\n\"KSEEB
\nSubstitute the value of l = 14 in the given equation
\n3l = 42
\n3(14) = 42
\n42 = 42
\n\u2234 LHS = RHS (verified)<\/p>\n

\"KSEEB<\/p>\n

b) \\(\\frac{\\mathbf{b}}{2}\\) = 6
\nSolution:
\nThe given equation is \\(\\frac{\\mathbf{b}}{2}\\) = 6
\nMultiplied by 2 both the sides b
\n\"KSEEB
\nsubstitute the value of b = 12 in the given equation
\n\"KSEEB
\n\u2234 LHS = RHS (verified)<\/p>\n

c)
\n\\(\\frac{\\mathbf{p}}{7}\\) = 4
\nSolution:
\n\"KSEEB
\nMultiplied by 7 to both the sides
\n\"KSEEB
\nSubstitute the value of p = 28 in the given equation
\n\"KSEEB
\n\u2234 LHS = RHS (verified)<\/p>\n

d) 4x = 25
\nSolution:
\nThe given equation is 4 x = 25
\nDivide both sides by 4
\n\"KSEEB
\nsubstitue the value of x = \\(\\frac{25}{4}\\) in the given equation
\n4x = 25
\n\"KSEEB
\n\u2234 LHS = RHS (verified)<\/p>\n

e) 8y = 36
\nSolution:
\nThe given equation is 8y = 36
\nDivide both sides by 8
\n\"KSEEB
\nSubstitute the equation by y = \\(\\frac{9}{2}\\) in the given equation
\n8y = 36
\n\"KSEEB
\n\u2234 LHS = RHS<\/p>\n

f)
\n\"KSEEB
\nSolution:
\n\"KSEEB
\nSubstitute the equation by z = \\(\\frac{15}{4}\\) in the given equation
\n\"KSEEB
\n\u2234 LHS = RHS (verified)<\/p>\n

g)
\n\"KSEEB
\nSolution:
\n\"KSEEB
\nSubstitute the value of a = \\(\\frac{7}{3}\\) in the given equation
\n\"KSEEB
\n\u2234 LHS = RHS (verified)<\/p>\n

h) 20t = -10
\nSolution:
\nThe given equation is 20 t = -10
\nDivide by 20 to both sides
\n\"KSEEB
\n\"KSEEB
\nLHS = RHS (verified)<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nGiven the steps you will use to separate the variable and then solve the equation:
\na) 3n – 2 = 46
\nSolution:
\nThe given equation is 3n – 2 = 46
\nAdd 2 to both sides 3n – 2 + 2 = 46+ 2
\n3n = 48
\nDivide the equation by 3
\n\"KSEEB
\n\u2234 n = 16 This is required solution.
\nsubstitute the value of n = 16 in the given equation
\n3n – 2 = 46
\n3(16) – 2 = 46
\n48 – 2 = 46
\n46 = 46
\n\u2234 LHS = RHS (verified)<\/p>\n

b) 5m + 7 = 17
\nThe given equation is 5m + 7 = 17
\nSubtract 7 from both sides
\n\"KSEEB
\n5m = 10
\nDivide the equation by 5 both the sides
\n\"KSEEB
\nSubstitute the value of m = 2 in the given equation
\n5m + 7 = 17
\n5(2) + 7 =17
\n10 + 7 = 17
\n\u2234 LHS = RHS (verified)<\/p>\n

c) \\(\\frac{20 p}{3}\\) = 40
\nSolution:
\n\"KSEEB
\nSubstitute the value of p = 6 in the given equation. .
\n\"KSEEB
\n40 = 40
\n\u2234 LHS = RHS (verified)<\/p>\n

d)
\n\"KSEEB
\nSolution:
\n\"KSEEB
\nMultiply by 10 to both the sides
\n\"KSEEB
\nDivide by 3 to both sides
\n\"KSEEB
\nSubstitute the value of p = 20 in the given equation
\n\"KSEEB<\/p>\n

Question 4.
\nSolve the following equations :
\na) 10p = 100
\nSolution:
\nThe given equation is 10p = 100
\nDivide by 10 both the sides
\n\"KSEEB
\nSubstitute the value of p in the given equation
\n10p = 100
\n10 x 10 = 100
\n100 = 100
\n\u2234 LHS = RHS (verified)<\/p>\n

\"KSEEB<\/p>\n

b) 10p + 10 = 100
\nSolution:
\nThe given equation is 10p + 10 = 100
\nsubtract 10 from both the sides
\n\"KSEEB
\nDivide by 10 both the sides
\n\"KSEEB
\nSubstitute the value of p = 9 in the given equation
\n10p + 10 = 100
\n10 \u00d7 9 + 10 = 100
\n90 + 10 = 100
\n100 = 100
\nLHS = RHS (verified)<\/p>\n

c) \\(\\frac{\\mathbf{p}}{4}\\) = 5
\nSolution:
\n\"KSEEB
\n5 = 5
\n\u2234 LHS = RHS (verified)<\/p>\n

d) \\(\\frac{-p}{3}\\) = 5
\nThe given equation is \\(\\frac{-p}{3}\\) = 5
\nThe equation is multiplied by -3 both the sides
\n\"KSEEB
\nSubstitute the value of p = -15 in the given equation
\n\"KSEEB
\n\u2234 LHS = RHS (verified)<\/p>\n

e)
\n\"KSEEB
\nSolution:
\n\"KSEEB
\nSubstitute the value p = 8 in the given equation
\n\"KSEEB<\/p>\n

f) 3s = -9
\nThe given equation is 3s = -9
\nDivide the equation by 3 both the sides
\n\"KSEEB
\nSubstitute the value of s = -3 in the given equation
\n3s = -9
\n3(-3)=-9
\n-9 = -9
\n\u2234 LHS = RHS (verified)<\/p>\n

g) 3s + 12 = 0
\nSolution:
\nThe given equation is 3s + 12 = 0
\nsubtract 12 from both the sides
\n\"KSEEB
\nDivide by 3 both the sides
\n\"KSEEB
\nSubstitute the value of s = -4 in the given equation
\n3s = 12 = 0
\n3(-4) = 12 = 0
\n-12 + 12 = 0
\n0 = 0
\n\u2234 LHS = RHS<\/p>\n

h) 3s = 0
\nSolution:
\nThe given equation is 3s = 0
\nDivide by 3 both the sides
\n\"KSEEB
\nSubstitute the value of s = 0 in the given equation
\n3s = 0
\n3(0) = 0
\n0 = 0
\n\u2234 LHS = RHS (verified)<\/p>\n

i) 2q = 6
\nSolution:
\nThe given equation is 2q = 6
\nDivide the equation by 2 both the sides
\n\"KSEEB
\nsubstitute the value of q = 3 in the given equation
\n2q = 6
\n2(3) = 6 6 = 6
\n\u2234 LHS = RHS (verified)<\/p>\n

\"KSEEB<\/p>\n

j) 2q – 6 = 0
\nSolution:
\nThe given equation is 2q – 6 = 0
\nAdd 6 to both the sides
\n\"KSEEB
\nSubstitute the value of q = 3 in the given equation
\n2q – 6 = 0
\n2(3) – 6 = 0
\n6 – 6 = 0
\n0 = 0
\nLHS = RHS (verified)<\/p>\n

k) 2q + 6 = 0
\nSolution:
\nThe given equation is 2q + 6 = 0
\nSubtract 6 from both the sides
\n\"KSEEB
\nDivide by 2 both the sides by 2
\n\"KSEEB
\nSubstitute the value of q = -3 in the given equation
\n2q + 6 = 0
\n2(-3) + 6 = 0
\n-6 + 6 = 0
\n0 = 0
\n\u2234 LHS = RHS (verified)<\/p>\n

l) 2q + 6 = 12
\nThe given equation is 2q + 6 = 12
\nSubtract 6 from both the sides
\n\"KSEEB
\nDivide by 2 both the sides
\n\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 4 Simple Equations Ex 4.2, 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 4 Simple Equations Ex 4.2 Question 1. Given first the step …<\/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\/14709"}],"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=14709"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/14709\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=14709"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=14709"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=14709"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}