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Solving Absolute Value Equations in Temperature Analysis of Steel Production

This article explores how to solve absolute value equations to determine the temperature range in steel production. Specifically, it discusses the furnace temperature for liquid iron production, calculated to be 1700°C ± 5°C. The absolute value equation is formulated and solved, revealing that the highest temperature is 1705°C and the lowest is 1695°C. The article also examines scenarios involving equations with no solutions and one solution, emphasizing the importance of checking solutions to ensure accuracy.

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Solving Absolute Value Equations in Temperature Analysis of Steel Production

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  1. SOLVING ABSOLUTE VALUE EQUATIONS Cristina Mocanu Math Department Al Ain

  2. In steel industry liquid iron is obtained in huge furnaces from minerals and coal. This process takes place at very high temperature, within 5 degrees at 1700 oC. Write and solve an absolute value equation to find the least and the greatest possible values of the furnace temperature. Cristina Mocanu Math Department Al Ain

  3. The equation contains absolute value and can be written as below: central value range l x - 1700 l = 5 What are the numbers whose absolute value is 5? Cristina Mocanu Math Department Al Ain

  4. 5 -5 or Case 1 a = b x – 1700 = 5 x = 5 + 1700 x = 1705 Case 2 a = - b x – 1700 = - 5 x = - 5 + 1700 x = 1695 Cristina Mocanu Math Department Al Ain

  5. Check: l x - 1700 l = 5 l 1705 - 1700 l = 5 l 5 l = 5 5 = 5 true l x - 1700 l = 5 l 1695 - 1700 l = 5 l - 5 l = 5 5 = 5 true Cristina Mocanu Math Department Al Ain

  6. The solutions are 1705 and 1695. The highest temperature is 1705 0C and the least is 1695 0C. On the number line, you can see that both solutions are 5 units away from 1700. 5 units 5 units 1695 1700 1705 Cristina Mocanu Math Department Al Ain

  7. EQUATIONS WITH NO SOLUTION Solve the equation: │2x + 5│- 3 = - 7 │2x + 5│= - 7 + 3 │2x + 5│= - 4 This sentence is never true. The solution set is Ø Cristina Mocanu Math Department Al Ain

  8. EQUATIONS WITH ONE SOLUTION │2x - 5│= 4x + 3 Case 2: a = - b 2x -5 = -(4x + 3) 2x - 5 = -4x - 3 6x - 5 = -3 6x = 2 x = 1/3 • Case 1: • a = b • 2x -5 = 4x + 3 • 5 = 2x + 3 • -8 = 2x • -4 = x Cristina Mocanu Math Department Al Ain

  9. Check: Case 2: │2(1/3) -5│ = 4(1/3) + 3 │2/3 - 5│ = 4/3 + 3 │-13/3│ = 13/3 13/3 = 13/3 true Case 1: │2(-4) -5│ = 4(-4) + 3 │-13│ = -16+ 3 13 = -13 false Cristina Mocanu Math Department Al Ain

  10. The only solution is 1/3. The number -4 is called extraneous solution. Don’t forget to check the values and decide at the end the solution set! Cristina Mocanu Math Department Al Ain

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