Algebra PPT Equations edited

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Let's play now

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Find a number for the symbols Hint: Each symbol has a different value. They are all even numbers no bigger than 10. The values given are sum of the symbols in the boxes horizontally and vertically

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24 14 14 16 18 18 18 18 16 24 14 14 16 18 18 18 18 16 24 14 14 24 14 14

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Now, can you write equations showing how you found the values

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Solving Linear Equations

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Review of linear equations. Construct and solve simple linear equations with variables on both sides. Represent problems mathematically, making correct use of symbols. LESSON OBJECTIVES

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Solve x + 1 = 7 1 x Picture a set of weighing scales The scales are perfectly balanced Removing the same weights from BOTH sides will keep the scales balanced 1 1 1 1 1 1 1 This leaves us with: x = 6 EXAMPLE 1

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Solve 3 x + 5 = 14 1 x x x Picture a set of weighing scales The scales are perfectly balanced Removing the same weights from BOTH sides will keep the scales balanced 1 1 1 1 1 This leaves us with: 3 x = 9 Which means that: x = 3 1 1 1 1 1 1 1 1 1 1 1 1 1 EXAMPLE 2

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Solve 5 x + 1 = 2 x + 7 1 x x x x x x x Picture a set of weighing scales The scales are perfectly balanced Removing the same weights from BOTH sides will keep the scales balanced 1 1 1 1 1 1 1 To check if this is correct: put x = 2 5 x + 1 = 5  2 + 1 = 10 + 1 = 11 2 x + 7 = 2  2 + 7 = 4 + 7 = 11 This leaves us with: 3 x = 6 Which means that: x = 2 3 x + 1= 7 Now we have: EXAMPLE 3

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Solve 3 x + 10 = 5 x + 2 x x x Picture a set of weighing scales The scales are perfectly balanced Removing the same weights from BOTH sides will keep the scales balanced This leaves us with: 1 1 1 x x x x 1 1 1 1 1 x 1 1 1 1 8 = 2 x Which means that: 4 = x EXAMPLE 4

WORKSHEET:

WORKSHEET Solve the following equations. 1) 4 x + 2 = 3 x + 12 2) 4 x  2 = 3 x + 12 3) 3 x + 1 =2 x + 9 4) 3 x  1 = 2 x + 9 5) 4 x + 4 = 2 x + 8 6) 3 x  4 = x + 8 7) 5 x + 5 = 2 x + 20 8) 3 x  15 = 2 x + 7 9) 4 x + 6 = 2 x - 6 10) 4 x + 10 = x + 40

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4 x + 2 = 3 x + 12 subtract 3 x from both sides: x + 2 = 12 subtract 2 from both sides: x = 10 SOLUTION

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4 x  2 = 3 x + 12 subtract 3 x from both sides: x  2 = 12 add 2 to both sides: x = 14 SOLUTION

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3 x + 1 = 2 x + 9 subtract 2 x from both sides: x + 1 = 9 subtract 1 from both sides: x = 8 SOLUTION

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3 x  1 = 2 x + 9 subtract 2 x from both sides: x  1 = 9 add 1 to both sides: x = 10 SOLUTION

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4 x + 4 = 2 x + 8 subtract 2 x from both sides: 2 x + 4 = 8 subtract 4 from both sides: 2 x = 4 divide by 2: x = 2 SOLUTION

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3 x  4 = x + 8 subtract x from both sides: 2 x  4 = 8 add 4 to both sides: 2 x = 12 divide by 2: x = 6 SOLUTION

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5 x + 5 = 2 x + 20 subtract 2 x from both sides: 3 x + 5 = 20 subtract 5 from both sides: 3x = 15 divide by 3: x = 5 SOLUTION

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3 x 1 5 = 2 x + 7 subtract 2x from both sides : x  15 = 7 Add 15 to both sides: x = 22 SOLUTION

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4 x + 6 = 2 x - 6 subtract 2 x from both sides: 2 x + 6 = - 6 subtract 6 to both sides: 2 x = - 12 divide by 2: x = - 6 SOLUTION

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4 x + 10 = x + 40 subtract x from both sides: 3 x + 10 = 40 subtract 10 from both sides: 3 x = 30 divide by 3: x = 10 SOLUTION

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_ ___ ______ __ _ ___________ ________ _______ __ _____ _____ ___ _____ ___ ___ _________ __ __ ___ __ ___ ______ ___ ____ ___ ____

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x 3x x 3x Lets say the width = x cm So the length = 3x cm

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Area = length x Width Perimeter = x +3 x + x +3 x That means : 8 x = 24 = 24 So : x = 3 Hence: width = 3cm And Length = 3 x 3 = 9cm Area = 27cm 2 = 9 x 3 = 27 x 3x x 3x

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__ ____ __ ____ _ _________ ______

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You should now be able to form and identify simple linear equations You should be able to solve linear equations with variables on both sides Represent problems mathematically, making correct use of symbols SUMMARY

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