Week 2, Day 2 Solving linear equations and application problems

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1 Week 2, Day 2 Solving linear equations and application problems September 8th, 2014 Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

2 Linear equations in one variable A linear equation in one variable, x say, is an equation that equates two expressions that are sums and differences of multiples of x and constants. The set of values of x that satisfy the equation is called the solution set. There are three possibilities for solution sets The most common situation: there is exactly one solution The equation is an identity 5x + 7 = 32 solution set = {5} 7x + 9 = 7x + 9 solution set = R The equation has no solutions 2x + 4 = 2x + 7 solution set = Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

3 Solving linear equations in one variable To solve linear equations First clear any fractions : Multiply the equation by a common denominator (a common multiple of the denominators in the equation). Distribute any parentheses. Combine like items Isolate the variable ; add or subtract multiples of the variable and constants from the appropriate sides of the equation so as to leave only multiples of x on one side of the equation and constants on the other If the equation isn t either an identity or a contradiction, divide by the coefficient of x to obtain x = the solution Check the answer you got satisfies the equation Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

4 Example 1 Solve the linear equation 6(x + 2) 3x = 3(x + 1) + 9 Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

5 Example 1 Solve the linear equation 6(x + 2) 3x = 3(x + 1) + 9 First we use the distributive property 6x x = 3x Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

6 Example 1 Solve the linear equation First we use the distributive property 6(x + 2) 3x = 3(x + 1) + 9 6x x = 3x Combining like terms on each side of the equals sign 3x + 12 = 3x + 12 This is true for any value of x so the solution set is the set of all real numbers Solution set = R Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

7 Example 2 Solve the equation 1 2 (5x 4) 5 3 (x 2) = 2 (3x + 1) 5 Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

8 Example 2 Solve the equation First clear the fractions. 1 2 (5x 4) 5 3 (x 2) = 2 (3x + 1) 5 Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

9 Example 2 Solve the equation 1 2 (5x 4) 5 3 (x 2) = 2 (3x + 1) 5 First clear the fractions. A common denominator is 30. Multiply the equation by this gives 15(5x 4) 50(x 2) = 12(3x + 1) Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

10 Example 2 Solve the equation 1 2 (5x 4) 5 3 (x 2) = 2 (3x + 1) 5 First clear the fractions. A common denominator is 30. Multiply the equation by this gives 15(5x 4) 50(x 2) = 12(3x + 1) Now distribute 75x 60 50x = 36x + 12 Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

11 Example 2 Solve the equation 1 2 (5x 4) 5 3 (x 2) = 2 (3x + 1) 5 First clear the fractions. A common denominator is 30. Multiply the equation by this gives Now distribute Combining like terms gives 15(5x 4) 50(x 2) = 12(3x + 1) 75x 60 50x = 36x x + 40 = 36x + 12 Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

12 Subtract 25x and 12 from both sides of the equation 28 = 11x Dividing by the coefficient of x, i.e. 11, gives the solution x = = Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

13 Subtract 25x and 12 from both sides of the equation 28 = 11x Dividing by the coefficient of x, i.e. 11, gives the solution x = = Now substitute this into the original equation ( ) ( ) = 2 5 ( ) 5 ( ) = ( ) ( ) 6 = 2 ( ) ( 3 28 ) ( ) = Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

14 General problem solving strategy 1 Read the problem carefully 2 Define the problem i.e identify what you are required to find 3 Assign the variables 4 Translate the problem into an equation 5 Solve the equation 6 Check your answer 7 Answer the original question Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

15 A uniform motion problem Landon can climb a certain hill at a rate 2.5 mph slower than his rate descending the hill. It takes him 2 hours to climb the hill and 45 minutes to come down the hill. What was his rate coming down the hill? Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

16 A uniform motion problem Landon can climb a certain hill at a rate 2.5 mph slower than his rate descending the hill. It takes him 2 hours to climb the hill and 45 minutes to come down the hill. What was his rate coming down the hill? To find Landons rate of descent we need to find the distance to the top of the hill. Let the distance to the top of the hill be D Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

17 A uniform motion problem Landon can climb a certain hill at a rate 2.5 mph slower than his rate descending the hill. It takes him 2 hours to climb the hill and 45 minutes to come down the hill. What was his rate coming down the hill? To find Landons rate of descent we need to find the distance to the top of the hill. Let the distance to the top of the hill be D Time(hours) Distance(miles) Rate=Distance/Time Ascending 2 D D 2 Descending 3 4 D D 3 4 = D 4 3 = 4D 3 Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

18 We are told that the rate of ascent is 2.5 mph slower than the rate of descent. Translating this into an equation gives Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

19 We are told that the rate of ascent is 2.5 mph slower than the rate of descent. Translating this into an equation gives 4D = D 2 To clear the equation of fractions we need to multiply the equation by the lowest common denominator, Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

20 We are told that the rate of ascent is 2.5 mph slower than the rate of descent. Translating this into an equation gives 4D = D 2 To clear the equation of fractions we need to multiply the equation by the lowest common denominator,which in this case is 6. ( ) ( ) 4D D = 6 2 Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

21 We are told that the rate of ascent is 2.5 mph slower than the rate of descent. Translating this into an equation gives 4D = D 2 To clear the equation of fractions we need to multiply the equation by the lowest common denominator,which in this case is 6. ( ) ( ) 4D D = 6 2 Distributing 24D 3 15 = 6D 2 Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

22 We are told that the rate of ascent is 2.5 mph slower than the rate of descent. Translating this into an equation gives 4D = D 2 To clear the equation of fractions we need to multiply the equation by the lowest common denominator,which in this case is 6. ( ) ( ) 4D D = 6 2 Distributing 24D 3 15 = 6D 2 Simplifying the fractions gives 8D 15 = 3D Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

23 Adding 15 to both sides of the equation gives 8D = 3D + 15 Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

24 Adding 15 to both sides of the equation gives 8D = 3D + 15 Now subtracting 3D from both sides of the equation 5D = 15 Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

25 Adding 15 to both sides of the equation gives 8D = 3D + 15 Now subtracting 3D from both sides of the equation 5D = 15 Dividing by 5 gives 5D 5 = 15 5 D = 3 miles Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

26 Adding 15 to both sides of the equation gives 8D = 3D + 15 Now subtracting 3D from both sides of the equation Dividing by 5 gives 5D 5 = 15 5 So Landon s rate of descent will be 5D = 15 D = 3 miles Rate of descent = 4D 3 = = 12 3 = 4 mph Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

27 Now checking that the answer is reasonable. The rate of ascent is given by Rate of ascent = D 2 = 3 2 = 1.5 mph and Rate of descent 2.5 = = 1.5 = Rate of ascent so the solution we have found Rate of descent = 4 mph satisfies the question that was asked. Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

28 Group Work 9/8/ Solve the linear equation 2. Solve the linear equation (3x + 6) + 2x + 2 = 5(x 1) (2x 4) + x 8 = 7 (x 3) 2 3. Manuel traveled 3 hours nonstop to Mexico, a total of 112 miles. He took a train part of the way, which averaged 50 mph. He then took a bus for the remaining distance, which averaged 30 mph. How long was Manuel on the bus? Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

29 Group Work Solutions 1 This is a contradiction it has solution set =. 2. The solution set is { 11 3 }. 3. Let T be the time Manuel spent on the bus then Distance on bus = 30T Distance on train = 50(3 T ) Total distance = 112 = 30T + 50(3 T ) This has solution T = = hours. Week 2, Day 2 Solving linear equations and application problems September 8th, / 1

Name: Math 9. Comparing & Ordering Rational Numbers. integers (positive or negative numbers, no decimals) and b 0

Name: Math 9. Comparing & Ordering Rational Numbers. integers (positive or negative numbers, no decimals) and b 0 Page 1 Ch.2- Rational Numbers 2.1 Name: Math 9 What is a rational number? Comparing & Ordering Rational Numbers A number that can be expressed as a b (a fraction), where a and b are integers (positive