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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+66 = 216
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 y2 = 2 – 2 = 0 The END
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 Fall '08
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 Linear Equations, Equations, Quadratic equation, Elementary algebra, Quintic equation, Boddu S Rao

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