Chapter 3 Linear Equations

1. Chapter 3 Linear Equations

2. 3.1 Linear Equation in One Variable A Linear Equation in one variable is an equation that can be written in the form ax + b = 0  Where a = 0 A linear function can be written as f(x) = ax + b Examples of Linear equation. 2x –1 = 0, -5x = 10 + x, and 3x + 8 = 5

3. Properties of Equality Addition Property of Equality If a, b, c are real numbers, then a = b is equivalent to a+ c = b + c. Multiplication Property of Equality If a, b, c are real numbers with c = 0, then a = b is equivalent to ac = bc.

4. Example 8 Solving a Linear equation graphicallyusing graphing calculator [ -6, 6, 1] by [-4, 4, 1]

5. Standard form of a line ( pg 158 - 159 ) An equation for a line is in standard form when it is written as ax + by = c, where a, b, c are constants With a, b and c are constants with a and b not both 0 • To find x-intercept of a line, let y = 0 in the equation and solve for x • To find y-intercept of a line, let x = 0 in the equation and solve for x

6. Example 115 ( Pg -163 ) [ 1984, 1991, 1] by [0, 350, 50] in 1987 In 1987 CD and LP record sales were both 107 million

7. 3.2- Introduction to problem solving Step1: Read the problems carefully to be sure that you understand it. ( You may need to read the problem more than once. ) Assign a variable to what you are being asked to find. If necessary, write other quantities in terms of this variable. Step 2: Write an equation that relates the quantities described in the problem. You may need to sketch a diagram, make a table, or refer to known formulas. Step 3: Solve the equation and determine the solution. Step 4: Look back and check your answer. Does it seem reasonable? Did you find the required information ?

8. 3.2 Solving a problem The sum of three consecutive even integers is 96. Find the three numbers. Solution Step 1: n is the smallest of the three integers n + 2 : next consecutive even integer n + 4 : larges of the three even integers Step 2: Write an equation that relates these unknown quantities. As the sum of these three even integers is 108 The equation is n + (n + 2) + (n + 4) = 96 Step 3 : Solve the equation in Step 2 n + (n + 2) + (n + 4) = 96 ( n + n +n) + (2 +4) = 96 3n + 6 = 96 3n + 6 = 96 3n = 90 n = 30 smallest of the three numbers is 30, so the three numbers are 30, 32, 34 Step 4: Check your answer. The sum of these three even integers is 30 + 32 + 34= 96 The answer checks.

9. Mixing acid ( Ex 12 pg 172) 20 % 2 liters 60% x liters + = 50% x + 2 liters Step 1x: liters of 60% sulphuric acid x + 2: Liters of 50% sulphuric acid Step 2 Concentration Solution Amount Pure Acid 0.20 (20%) 2 0.20(2) 0.60 (60%) x 0.60x 0.50(50%) x + 2 0.50(x + 2) Equation 0.20(2) + 0.60x = 0.50(x + 2) (pure acid in 20% sol.) (pure acid in 60% sol.) (pure acid in 50% sol.) Step 3 Solve for x 0.20(2) + 0.60x = 0.50(x + 2) 2 (2) + 6x = 5(x + 2) Multiply by 10 4 + 6x = 5x + 10 (Distributive Property) Subtract 5x and 4 from each side x = 6 Six liters of the 60% acid solution should be added to the 2 liters of 20% acid solution. Step 4 If 6 liters of 60% acid solution are added to 2 liters of 20% solution, then there will be 8 liters of acid solution containing 0.60(6) + 0.20(2) = 4 liters of pure acid. Check The mixture represents a 4/8 = 0.50 or 50% mixture

10. Ex 50 (Pg 175) Anti freeze mixture A radiator holds 4 gallons of fluid x represents the amount of antifreeze that is drained and replaced The remaining amount is 4 – x 20 % of + 70 % of solution of = 50% of solution 4 – x gallons x gallons 4 gallons 0.20(4-x) + 0.70x = 0.50(4) 0.8 – 0.2x + 0.7x = 2 0.5x = 1.2 x = 2.4 The amount of antifreeze to be drained and replaces is 2.4 gallons

11. Geometric Formulas h a c Perimeter of triangle P = a+b+c unit Area of triangle A = ½ bh sq.unit Area of Rectangle =LW sq.unit Perimeter P = 2(L + W) unit Area of Parallelogram A = bh sq.unit Area & Volume of cylinder A = 2 rh sq.unit V =r2h cu.unit Area & Volume of a cube A = 6a2 sq.unit V = a3 cu.unit b W L h b h r a a a

12. 3.3 Linear Inequality in One Variable A linear inequality in one variable is an inequality that can be written in the form ax + b > 0, where a = 0. ( The symbol > may be replaced with >, <, or < ) There are similarities among linear functions, equations, and inequalities. A linear function is given by f(x) = ax + b, a linear equation by ax + b = 0, and a linear inequality by ax + b > 0.

13. 3.3 Properties of Inequalities Let a, b, c be real numbers.  • a < b and a+c < b+c are equivalent ( The same number may be added to or subtracted from both sides of an inequality.) • If c > 0, then a < b and ac < bc are equivalent. (Both sides of an inequality may be multiplied or divided by the same positive number) • If c< 0, then a < b and ac > bc are equivalent. Each side of an inequality may be multiplied or divided by the same negative number provided the inequality symbol is reversed.

14. Examples of linear inequalities are 2x + 1< 0, 1-x > 6, and 5x + 1 < 3 – 2x A solution to an inequality is a value of the variable that makes the statement true. The set of all solutions is called the solution set. Two inequalities are equivalent if they have the same solution set. Inequalities frequently have infinitely many solutions. For example, the solution set to the inequality x- 5> 0 includes all real numbers greater than 5, which can be written as x > 5. Using set builder notation, we can write the solution set as { x x > 5 }. Meaning This expression is read as “ the set of all real numbers x such that x is greater than 5. “

15. Ex -4 Graphical Solutions (Pg 182) 350 300 250 200 150 100 50 Distance (miles) y1 y2 y2 y1 x = 2 0 1 2 3 4 Time ( hours ) Distances of two cars When x = 2 , y1 = y2, ie car 1 and car 2 both are 150 miles From Chicago y1 < y2 when x< 2 car 1 is closer to Chicago than car 2 y1 is below the graph of y2 y1 > y2 when x > 2 Car 1 is farther from Chicago than Car 2 Y1 above the graph of y2

16. …Continued Ex 5 (Pg 183) Solving an inequality graphically Solve 5 – 3x < x – 3 y1 = 5 – 3x and y2 = x – 3 Intersect at the point (2, -1) 5 4 3 2 1 X = 2 y1 y2 -4 -3 -2 -1 0 1 2 3 4 y1 = y2 when x = 2 -1 -2 -3 (2, -1) y1 < y2 when x > 2 , y1 is below the graph of y2 Combining the above result y1 < y2 when x > 2 Thus 5 – 3x < x – 3 is satisfied when x > 2. The solution set is {x / x> 2 }

17. Using Graphing Calculator Y1 = 5 – 3x Y2 = x - 3 Hit window Hit Graph Hit Y Enter inequality Enter [ -5, 5, 1] by [-5, 5, 1]

18. Ex 92 ( Pg 178) Sales of CD and LP recordsUsing Graphing Calculator Hit Y , enter equations Enter table set Enter window Hit Table Hit graph 1987 or after CD sales were greater than or equal to LP records

19. 3.4 Compound inequalities A compound inequality consists of two inequalities joined by the words and or In compound inequality contains word and, a solution must satisfy both two inequalities. 2x > - 5and2x < 3 2(1) > -5 and 2(1) < 3 1 is a solution True True In compound inequality contains word or , a solution must satisfy atleast one of the two inequalities. 5 + 2 > 3 or 5 – 1 < -5  5 is a solution x + 2 > 3 or x – 1 < -5 True False Example 1 Example 2

20. Symbolic Solutions and Number Lines ] x < 6 x> - 4 x < 6 and x > - 4 -8 -6 -4 -2 0 2 4 6 8 ( -8 -6 -4 -2 0 2 4 6 8 ( ] -8 -6 -4 -2 0 2 4 6 8 - 4 < x < 6

21. Three-part inequality Sometimes compound inequality containing the word and can be combined into a three part inequality. For example, rather than writing x > 4 and x < 10 We could write the three-part inequality 4 < x < 10 ( ] Numberline -1 0 1 2 3 4 5 6 7 8 9 10 4< x < 10

22. Compound Inequality Example Solve x + 2 < -1 or x + 2 > 1 x < -3 or x > -1 ( subtract 2 ) The solution set for the compound inequality results from taking the union of the first two number lines. We can write the solution, using Set builder notation, as { x x < - 3} U { x x > - 1 } or {x x < - 3 or x > - 1} x < - 3- x > - 1- x < - 3 or x > -1 - )  - 4 -3 -2 -1 0 1 2 3 4  ( - 4 -3 -2 -1 0 1 2 3 4 ) (  - 4 -3 -2 -1 0 1 2 3 4

23. Interval Notation Inequality Interval Notation Number line Graph - 1 < x < 3 ( - 1, 3) - 3 < x < 2 ( - 3, 2] - 2 < x < 2 [ - 2, 2 ] x < - 1 or x > 2 ( -  , - 1) U (2,  ) - x > - 1 ( - 1,  ) x < 2 ( - , 2 ]- ( ) -4 -3 -2 -1 0 1 2 3 4 ( ] -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4 [ ] ) ( + ( + ]

24. Example Solve graphically and numerically. Write your answer in interval notationx + 1 < -1 or x + 1 > 1 Y1 = -1, Y2 = x + 1, Y3 = 1 Enter y1, y2, y3 Enter window [ -5, 5, 1] by [ -5, 5, 1]  +1 0 Hit Graph - 2 -1 -  x + 1 < - 1 or x + 1 > 1 Solution in interval notation is ( - , - 2) U (0,  )

25. Example 6 (Page 195) Solving a compound inequality numerically and graphically Using Technology • Tution at private colleges and universities from 1980 to 1997 • Can be modelled by f(x) = 575(x – 1980) + 3600 • Estimate when average tution was between $8200 and $10,500. Hit Y and enter equation Hit Window and enter Hit 2nd and Table [ 1980, 1997, 1] by [3000, 12000, 3000 ] Hit 2nd and calc and go to Intersect and enter 4 times to get intersection

26. School Enrollment • 70 • 60 • 55 • 50 Enrollment (millions) • 1980 1990 2000 • Year