Do Now 1/25/12  Take out HW from last night. Mid-Term Review worksheet #1 Mid-Term Review worksheet #1 Mid-Term Review worksheet #2 Mid-Term Review worksheet #2  Copy HW in your planner. Mid-Term tomorrow! Mid-Term tomorrow! Go to bed by 8:00 Go to bed by 8:00 Eat breakfast in the morning Eat breakfast in the morning

Algebra Mid-Term Review Worksheet #1  1) -162  2) 3x² - 5y + 1  3) 10  4) x = b – (c/a)  5) -2  6) -1  7) y = 8  8) x = -6  9) no solution  10) no solution  11) vertical  12) y = -5x – 22  13) -15, 9  14) 2 < z < 5  15) undefined / 0  16) x < 1.6  17) x > 9 or x 9 or x < 9  18) y ≤ 2 and y ≥ -3  19) x > -1 or x -1 or x < -3  20) -1  21) whole, rational, integer, real  22) 75, 80, 85  23) m = -5/2; y-int = 5  24) y – 2 = 7(x + 4)  25) t ≤ -2  26) empty set / no solution; x = 4  27) 2x + y = 0  28) identity/ all real numbers  29) B  30)

Objective  SWBAT review Chapters 4-7 concepts for Mid-Term

Chapter 4 Graphing Linear Equations and Functions

Section 4.1 “Coordinate Plane” y-axis x-axis Origin (0,0) (0,0) Quadrant I (+,+) Quadrant II (-,+) Quadrant III (-,-) Quadrant IV (+,-)

Section 4.2 “Graph Linear Equations” Solve the equation for y. STEP 1 SOLUTION Graph the equation y + 2x = 4. STEP 2 Make a table by choosing a few values for x and then finding values for y. STEP 3 Plot the points. Notice the points appear on a line. Connect the points drawing a line through them.x-2012y86420

Section 4.3 “Graph Using Intercepts” Graph the equation 6x + 7y = 42. 6x + 7y = 42 x =  x- intercept 7 Find the x-intercept 6x + 7(0)=42 6(0) + 7y = 42 y =  y- intercept 6 6x + 7y = 42 Find the y-intercept Plot points. The x-intercept is 7, so plot the point (7, 0). The y- intercept is 6, so plot the point (0, 6). Draw a line through the points.

(0, 6) and (5, –4) m = y 2 – y 1 x 2 – x 1 Let (x 1, y 1 ) = (0, 6) and (x 2, y 2 ) = (5, – 4). – 4 – 6 5 – 0 = Write formula for slope. Substitute. Simplify = – = – 2 Find the slope of the line that passes through the points Section 4.4 “Find Slope and Rate of Change”

Section 4.5 “Graph Using Slope-Intercept Form” SLOPE-INTERCEPT FORM- a linear equation written in the form y = mx + b slopey-intercept y-coordinatex-coordinate

Determine which of the lines are parallel. Find the slope of each line. Line a: m = – 1 – 0 – 1 – 2 – 3 – (–1 ) 0 – 5 = – 1– 1– 1– 1 – 3– 3– 3– 3 13 = Line b: m = – 2– 2– 2– 2 – 5– 5– 5– 5 = 2 5 = Line c: m = – 5 – (–3) – 2 – 4 – 2– 2– 2– 2 – 6– 6– 6– 6 = 1 3 = Line a and line c have the same slope, so they are parallel.

Section 4.7 “Graph Linear Functions” Function Notation- a linear function written in the form y = mx + b where y is written as a function f. f(x) = mx + b slope y-intercept x-coordinate f(x) is another name for y. It means “the value of f at x.” g(x) or h(x) can also be used to name functions This is read as ‘f of x’

Graph a Function x-2012f(x) STEP 1 SOLUTION Graph the Function f(x) = 2x – 3 STEP 2 Make a table by choosing a few values for x and then finding values for y. STEP 3 Plot the points. Notice the points appear on a line. Connect the points drawing a line through them. The domain and range are not restricted therefore, you do not have to identify.

Compare graphs with the graph f(x) = x. Graph the function g(x) = x + 3, then compare it to the parent function f(x) = x. Graph the function g(x) = x + 3, then compare it to the parent function f(x) = x. xf(x) f(x) = x xf(x) g(x) = x + 3 The graphs of g(x) and f(x) have the same slope of 1.

Chapter 5 Writing Linear Equations

Section 5.1 & 5.2 “Write Equations in Slope-Intercept Form” SLOPE-INTERCEPT FORM- a linear equation written in the form y = mx + b slopey-intercept y-coordinatex-coordinate

Section 5.3 “Write Linear Equations in Point-Slope Form” POINT-SLOPE FORM- of a linear equation is written as: slope y-coordinate point 1 x-coordinate point 1 y x run rise

Section 5.4 “Write Linear Equations in Standard Form” (General Form) The STANDARD FORM of a linear equation is represented as Ax + By = C where A, B, and C are real numbers

Section 5.5 “Write Equations of Parallel and Perpendicular Lines” PARALLEL LINES If two nonvertical lines in the same plane have the same slope, then they are parallel. If two nonvertical lines in the same plane have the same slope, then they are parallel. If two nonvertical lines in the same plane are parallel, then they have the same slope. If two nonvertical lines in the same plane are parallel, then they have the same slope.

Write an equation of the line that passes through (–2,11) and is parallel to the line y = -x + 5. STEP 1 Identify the slope. The graph of the given equation has a slope of -1. So, the parallel line through (– 2, 11) has a slope of -1. STEP 2 Find the y- intercept. Use the slope and the given point. y = mx + b 11 = -1(–2) + b 9 = b Write slope-intercept form. Substitute -1 for m, - 2 for x, and 11 for y. Solve for b. STEP 3 Write an equation. Use y = mx + b. y = -x + 9 Substitute -1 for m and 9 for b.

Section 5.5 “Write Equations of Parallel and Perpendicular Lines” PERPENDICULAR LINES PERPENDICULAR LINES If two nonvertical lines in the same plane have slopes that are negative reciprocals, then the lines are perpendicular. If two nonvertical lines in the same plane have slopes that are negative reciprocals, then the lines are perpendicular. If two nonvertical lines in the same plane are perpendicular, then their slopes are negative reciprocals If two nonvertical lines in the same plane are perpendicular, then their slopes are negative reciprocals ½ and -2 are negative reciprocals. 3 and -1/3 are negative reciprocals.

Modeling Data When data show a positive or negative correlation, you can model the trend in the data using a LINE OF FIT Using a Line of Fit to Model Data 1). 2). 3). 4). Decide whether the data can be a modeled by a line. (Does it have positive or negative correlation?) Make a scatter plot of the data. Draw a line that appears to fit the data closely. Write an equation using two points on the line. (The points do not have to be actual data pairs, but they do have to be on the line.)

Draw a line of fit for the scatter plot. Write an equation that models the number of ounces of water left in the water bottle as a function of minutes on the treadmill. Minutes on treadmill Ounces of water in water bottle y-axis x-axis35 Write an equation using two points on the line. 12 Minutes on the treadmill Use the points (5,12) and (30,4). 12 – 4 = _8_ 5 – Find the y-intercept. Use (5,12). y = mx + b 12 = 8 (5) + b b = y = 8 8 x

Chapter 6 Solving and Graphing Linear Inequalities

Writing Equations with Inequalities SymbolMeaning Key phrases = Is equal to The same as < Is less than Fewer than ≤ Is less than or equal to At most, no more than > Is great than More than ≥ Is greater than or equal to At least, no less than

 On a number line, the GRAPH OF AN INEQUALITY in one variable is the set of points that represent all solutions of the inequality. Graph x < Graph x ≥ “Less than” and “greater than” are represented with an open circle. “Less than or equal to” and “greater than or equal to” are represented with closed circle.

Multiplying and/or dividing each side of an inequality by a NEGATIVE number only produces an equivalent inequality IF the inequality sign is REVERSED!! Multiplying and/or dividing each side of an inequality by a NEGATIVE number only produces an equivalent inequality IF the inequality sign is REVERSED!! Solve Inequalities When Multiplying and Dividing by a NEGATIVE”

14x + 5 < 7(2x – 3) Write original inequality. 14x + 5 < 14x – 21 Distributive property 5 < – 21 Subtract 14x from each side. There are no solutions because 5 < – 21 is false. Solve: 14x + 5 < 7(2x – 3) Solve: 14x + 5 < 7(2x – 3) **HINT** If an inequality is equivalent to an inequality that is false, such as 5 < -21, then the solution of the inequality has NO SOLUTION.

12x – 1 > 6(2x – 1) 12x – 1 > 6(2x – 1) Write original inequality. Distributive property Subtract 12x from each side. 12x – 1 > 12x – 6 – 1 > – 6 All real numbers are solutions because – 1 > – 6 is true. 12x – 1 > 6(2x – 1) 12x – 1 > 6(2x – 1) **HINT** If an inequality is equivalent to an inequality that is true, such as -1 > -6, then the solutions of the inequality are ALL REAL NUMBERS.

Solve Solve a Compound Inequality with AND –7 < x – 5 < -1. Graph your solution. Separate the compound inequality into two inequalities. Then solve each inequality separately. –7 < x – 5 Write two inequalities. –7 + 5 < x –5 + 5 Add 5 to each side. –2 < x Simplify. The compound inequality can be written as – 2 < x < 4. and x – 5 < -1 x – < x < 4 and and – 3 – 2 – – 3 – 2 – Graph:

Write original inequality. Solve 2x Graph your solution. Solve the two inequalities separately. 2x + 3 < 9 or 3x – 6 > 12 2x + 3 – 3 < 9 – 3 or 3x – > Addition or Subtraction property of inequality 2x < 6 or 3x > 18 Simplify. Solve a Compound Inequality with OR x < 3 or x > 6 Simplify. The solutions are all real numbers less than 3 or greater than 6. than 6.

Solving an Absolute Value Equation The equation |ax + b| = c where c ≥ 0, is equivalent to the statement: is equivalent to the statement: ax + b = c or ax + b = -c ax + b = c or ax + b = -c

Solve 4|t + 9| - 5 = 19 Solve an Absolute Value Equation Solve an Absolute Value Equation First, rewrite the equation in the form ax + b = c. Next, solve the absolute value equation. 4 t + 9 – 5 = 19 4 t + 9 = 24 t + 9 = 6 t + 9 = 6 Write original equation. Add 5 to each side. Divide each side by 4. t + 9 = 6 t + 9 = 6 or t + 9 = –6 t = –3 or t = –15 Write absolute value equation. Rewrite as two equations. Addition & subtraction to each side

Solving an Absolute Value Inequality The inequality |ax + b| > c where c > 0, will result in a compound OR inequality. ax + b > c or ax + b c or ax + b < -c The inequality |ax + b| 0, will result in a compound AND inequality. -c < ax + b < c -c < ax + b < c

Solve |x - 5| > 7. Graph your solution Solve an Absolute Value Inequality Solve an Absolute Value Inequality Write original inequality. Rewrite as compound inequality. inequality. Add 5 to each side. x > 12 OR x 12 OR x < -2 |x - 5| > 7 +5 x - 5 > 7 x - 5 < -7 OR Simplify. Graph of x > 12 or x 12 or x <

Solve |-4x - 5| +3 < 9. Graph your solution Solve an Absolute Value Inequality Solve an Absolute Value Inequality Write original inequality. Rewrite as compound inequality. inequality. Add 5 to both sides. x > AND x AND x < 0.25 |-4x - 5| +3 < x – 5 < 6 -4x - 5 > -6 AND Simplify. Graph of < x < Subtract 3 from each side. -3 |-4x - 5| < 6 Divide by -4 to each side. -4x < 11 -4x > -1 AND

Graphing Linear Inequalities  Graphing Boundary Lines: Use a dashed line for. Use a dashed line for. Use a solid line for ≤ or ≥. Use a solid line for ≤ or ≥.

Graph an Inequality Graph the equation STEP 1 Graph the inequality y > 4x - 3. STEP 2 Test (0,0) in the original inequality. STEP 3 Shade the half-plane that contains the point (0,0), because (0,0) is a solution to the inequality.

Chapter 7 Systems of Equations and Inequalities

Equation 1 Equation 1 x + 2y = 11 x + 2y = 11 Equation 2 Equation 2 y = 3x + 2 y = 3x + 2 “Solve Linear Systems by Substituting” x + 2y = 11 x + 2y = 11 x + 2(3x + 2) = 11 x + 2(3x + 2) = 11 Substitute x + 6x + 4 = 11 x + 6x + 4 = 11 7x + 4 = 11 7x + 4 = 11 x = 1 x = 1 Equation 1 Equation 1 y = 3x + 2 y = 3x + 2 Substitute value for x into the original equation y = 3(1) + 2 y = 3(1) + 2 y = 5 y = 5 The solution is the point (1,5). Substitute (1,5) into both equations to check. (1) + 2(5) = 11 (1) + 2(5) = = 11 (5) = 3(1) + 2 (5) = 3(1) = 5

Section 7.3 “Solve Linear Systems by Adding or Subtracting”  ELIMINATION- adding or subtracting equations to obtain a new equation in one variable. Solving Linear Systems Using Elimination (1) Add or Subtract the equations to eliminate one variable. (2) Solve the resulting equation for the other variable. (3) Substitute in either original equation to find the value of the eliminated variable.

Equation 1 Equation 1 -3x + 2y = -9 -3x + 2y = -9 Equation 2 Equation 2 4x + 5y = 35 4x + 5y = 35 “Solve Linear Systems by Elimination Multiplying First!!” Equation 1 Equation 1 4x + 5y = 35 4x + 5y = 35 Substitute value for x into either of the original equations 4(5) + 5y = 35 4(5) + 5y = y = y = 35 The solution is the point (5,3). Substitute (5,3) into both equations to check. 4(5) + 5(3) = 35 4(5) + 5(3) = = 35 -3(5) + 2(3) = -9 -3(5) + 2(3) = = -9 MultiplyFirst + 23x = x = 115 x = 5 x = 5 y = 3 y = 3 Eliminated x (2) 15x - 10y = 45 8x + 10y = 70 8x + 10y = 70 x (-5)

Section 7.5 “Solve Special Types of Linear Systems” LINEAR SYSTEM- consists of two or more linear equations in the same variables. Types of solutions: (1) a single point of intersection – intersecting lines (2) no solution – parallel lines (3) infinitely many solutions – when two equations represent the same line

Equation 1 Equation 1 3x + 2y = 2 3x + 2y = 2 Equation 2 Equation 2 3x + 2y = 10 3x + 2y = 10 “Solve Linear Systems with No Solution” _ 0 = 8 0 = 8 No Solution No Solution Eliminated Eliminated -3x + (-2y) = -2 -3x + (-2y) = -2 + This is a false statement, therefore the system has no solution. By looking at the graph, the lines are PARALLEL and therefore will never intersect. “Inconsistent “InconsistentSystem”

Equation 1 Equation 1 Equation 2 Equation 2 x – 2y = -4 “Solve Linear Systems with Infinitely Many Solutions” -4 = = -4 Infinitely Many Solutions Infinitely Many Solutions y = ½x + 2 y = ½x + 2 This is a true statement, therefore the system has infinitely many solutions. By looking at the graph, the lines are the SAME and therefore intersect at every point, INFINITELY! “Consistent “ConsistentDependentSystem” Equation 1 Equation 1 x – 2y = -4 x – 2y = -4 Use ‘Substitution’ because we know what y is equals. x – x – 4 = -4 x – x – 4 = -4 x – 2(½x + 2) = -4 x – 2(½x + 2) = -4

Section 7.6 “Solve Systems of Linear Inequalities” SYSTEM OF INEQUALITIES- consists of two or more linear inequalities in the same variables. Inequality 1 Inequality 1 2x + y < 8 2x + y < 8 Inequality 2 Inequality 2 x – y > 7 x – y > 7 A solution to a system of inequalities is an ordered pair (a point) that is a solution to both linear inequalities.

y < x – 4 Inequality 1 y ≥ -x + 3 Inequality 2 Graph a System of Inequalities 0 < 5 – 4 ? 0 < 1 0 ≥ -2 0 ≥ ? Graph both inequalities in the same coordinate plane. The graph of the system is the intersection of the two half-planes, which is shown as the darker shade of blue. (5,0)

y ≥ -1 Inequality 1: x > -2 Inequality 2: Graph a System of THREE Inequalities y ≥ -1 ? 0 ≥ -1 0 > -2 x > -2 ? Graph all three inequalities in the same coordinate plane. The graph of the system is the triangular region, which is shown as the darker shade of blue. x + 2y ≤ 4 Inequality 3: x + 2y ≤ 4 ? ≤ 4 Check(0,0)

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