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Solving_Equations(Prepared_by_Boddu_S_Rao)Opp

Solving_Equations(Prepared_by_Boddu_S_Rao)Opp - 100’s of...

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Unformatted text preview: 100’s of free ppt’s from www.pptpoint.com library Solving Equations Solving Equations Overview Boddu S Rao A linear or First degree Equation A linear or First degree Equation • Eg. 2y+1 = 13 1.Solving linear equations by using inverse operations using x – 7 = -3 + 7 = +7 (adding 7 on each side) +7 -------------X =4 3x = 21 3x = 21 3x / 3 = 21 /3 ( dividing each by 3) X = 7 Solving Equations with parentheses parentheses Solve 3(b+2) +2b = 21 3(b+2) +2b = 21 3(b+2) 3b+6 +2b = 21 5b+6 = 21 5b+6-6 = 21-6 5b = 15 5b/5 = 15/ 5 Therefore b = 3 Solving equation contains fractions m/3 – m/4 = 2 m/3 – m/4 = 2 • 12(m/3) – 12(m/4) = 12(2) [ L.C.M. of 3 &4 is 12) • 4m – 3m = 24 • So, m= 24 Solving Equations by cross multiplying If x / 3 = (x + 4)/5 , what is the value of x? If x / 3 = (x + 4)/5 , what is the value of x? 5(x) = 3(x+4) 5x = 3x +12 2x = 12 x = 12 / 2 Therefore x= 6 Using A root of an equation to answer a question answer If 3y – 2 = 4 , what is the value If of y – 2 ? of If 3y – 2 = 4 , what is the value of y–2? If 3y – 2 = 4 ,then 3y = 6 If Therefore y = 2 The value of y-2 = 2 – 2 = 0 The END The END Thank you for viewing …. Thank you for viewing …. Presented by Boddu S Rao ...
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