Along the way, you learned many manipulative skills using the Properties of Real Numbers.

A LOOK at Algebra ============================= In a nutshell, you have learned how to: 1. solve linear equations and inequalities. solve quadratic equations and inequalities 3. solve systems of linear equations and inequalities. solve equations containing absolute value 5. solve equations containing radicals. graph equations/inequalities - 1 variable. graph equations - variables. graph absolute values 7. graph parabolas & exponentials 8. find equations of lines Along the way, you learned many manipulative skills using the Properties of Real Numbers. We also looked at graphs in one and two variables and learned we could graph them by inspection, if we took the time to see some patterns develop. And, because we enjoy problem solving so much, we studied formatted word problems so we might relate them to other problems we may encounter later. Please review the Algebra, you can do it sheets along with this packet. Copyright 003 Hanlonmath Page 1 of 15

Linear Equations The general strategy for solving linear equations and inequalities was to use the Gift Wrapping analogy. That is, to use the Order of Operations in reverse, using the opposite operation to isolate the variable. We learned to solve equations in the ax + b = c form. Equations that did not look like that, we used the Properties of Real Numbers, to change it into that form. Example 1 Solve: 3x = 10 3x + = 10 + 3x + 0 = 1 3x = 1 x = APE AI/Arithmetic fact Prop of Zero DPE Example Solve: 5(x 3) + 8 = 13 This problem looked different from the ax + b = c problems because of the parentheses. Using the Distributive Property, we get rid of the parentheses, combine terms, and make the problem look like one we solved before. 10x 15 + 8 = 13 10x 7 = 13 10x 7 + 7 = 13 + 7 10x = 0 x = Copyright 003 Hanlonmath Page of 15

Example 3 Solve: x/ x/3 = 10 This problem looks different because it has fractions. Using the Multiplication Property of Equality, I can multiply both sides of the equation by the common denominator, that eliminates the fractions and makes the problem look like one I have solved before.. { x/ x/3 } = (10) 3x x = 0 x = 0 Equations containing Absolute Value When solving equations containing absolute value, you must remember the expression on the inside of the absolute value signs could be positive or negative. When solving equations, we are looking for ALL the values of the variable that will make the open sentence true, we must solve the equation for both conditions, when the expression is positive and when the expression is negative. That means we should get two answers! Example 1 Solve: x + 3 = 11 If x + 3 is a positive number, then x + 3 = 11. Solving for x, we have x = 8, x = But, x + 3 could have been a negative number. If that were the case, then (x + 3) = 11 or x + 3 = 11 Solving for x, x = 1 x = 7 Notice, if x =, then x + 3 is 11 and the absolute value of 11 is 11. If x = 7, then x + 3 is negative 11, and the absolute value of a negative 11 is also 11. The solution set is {, 7} Copyright 003 Hanlonmath Page 3 of 15

Example Solve: 5x 3 > 7 Translated that means that if 5x 3 is positive, then 5x 3 > 7. If 5x 3 is negative, then (5x 3) > 7 Solving; 5x 3 > 7 OR (5x 3) > 7 5x > 30 5x 3 < 7 x > 5x < x < /5 The solution is {x/ x < /5 U x > } /5 0 Example 3 Solve: x 5 < 9 That means that x 5 < 9 AND (x 5) < 9 x 5 > 9 x < 1 x > x < 7 and x > The solution is{x/ < x < 7} 0 7 It might be helpful to keep in mind when the absolute value is less then, you have a conjunctive statement, when it s greater than, you have a disjunctive statement. Copyright 003 Hanlonmath Page of 15

Quadratic Equations Zero Product Property We looked at three ways of solving quadratic equations, the first using the Zero Product Property. That is, if AB = 0, then either A or B has to be zero. To solve quadratics using the Zero Product Property, 1. we put everything on one side of the equation, zero on the other side,. factor completely 3. set each factor equal to zero. solve the resulting equation Example 1 Solve: x + x 0 = 0 (x + 5)(x ) = 0 factoring In order for that product to be equal to zero, one of the factors has to be zero. Setting each factor equal to zero, we have x + 5 = 0 or x = 0 x = - 5 or x = Completing the Square We learned how to complete the square by noticing the relationship (pattern) between the linear and constant terms in perfect squares when the coefficient of the quadratic term was ONE. To solve quadratics by completing the square, 1. put the constant on the other side of the equation. made sure the leading coefficient is ONE 3. take half the linear term and squared it. added that number to both sides of the equation 5. factor into a perfect square. solved the resulting equation Copyright 003 Hanlonmath Page 5 of 15

Example 1 Solve: x + x 1 = 0 x + x = +1 x + x + _ = 1 + _ 3 x + x + 9 = 1 + 9 Take ½ the linear term () and square Adding 9 to both sides (x + 3) = 10 Factoring x + 3 = " ± 10 x = 3 " ± 10 Quadratic Formula The Quadratic Formula was derived from completing the square using ax + bx + c = 0. To use the quadratic formula, put everything on one side and zero on the other side to identify the values of a, b and c. Example 1 Solve: x + 5x = x + 5x = 0 Using the formula, a = 1, b = 5, and c =. Plugging those values into the Quadratic Formula, we have " x = b ± b ac " x = (5) ± a 5 1( ) 1 5 ± 5 +1 " x = " x = 5 ± 1 Copyright 003 Hanlonmath Page of 15

Systems of Linear Equations We used two methods to solve equations simultaneously, the addition method and substitution. To solve a system of equations of two equations with two variables, we will reduce it to one equation with one variable a problem we already know how to solve. Addition Method To do that by using the Addition Method, we will make the coefficients of one of the variables the same, but opposite in sign. Then, by adding the equations together, one of the variables will fall out and the resulting equation will be linear in one variable an equation we can solve. Example 1 Solve: 3x + y = x y = 1 By adding these equations together by the APE, we have equation that we know how to solve. 9x = 18 x = If x =, then by plugging that value back into one of the equations, we have 3() + y = + y = y = 0 y = 0 These two lines intersect at the point (,0). Example Solve: 3x + y = 5 x + 3y = Copyright 003 Hanlonmath Page 7 of 15

In this example, none of the coefficients are the same. Using the Properties of Real Numbers, I make that happen by multiplying both sides of an equation by the same number. 3x + y = 5 (x3) 9x + y = 15 x + 3y = ( ) 8x y = 1 adding the equations x = 3, and y =, the solution (3, ) Substitution To solve equations simultaneously using substitution, you solve for one variable in one equation and plug that expression into the other equation. Generally, the only time you would use the substitution method is when one of the coefficients on one the variables in 1. Example 3 Solve: y = x + 1 x + 3y = 19 y was already solved for in the first equation, so plug (x + 1) into the second equation, we have: x + 3(x + 1) = 19 x + x + 3 = 19 8x + 3 = 19 8x = 1 x = If x =, then y = 5 so the point of intersection is (, 5) Equations containing Radicals To solve equations containing radicals, isolate the radical, then get rid of it. Solve the resulting equation. Be careful, when you are raising powers, you might introduce extraneous solutions, so you MUST check your answers. Copyright 003 Hanlonmath Page 8 of 15

Example 1 Solve Solve: x = x = 10 x = 100 x = 10 Example Solve: y y +1 = 0 y = y +1 ( y) = ( y +1) y + y = (y +1) y 8y = 0 y(y 8) = 0 y = 0 or y = 8 IMPORTANT check your answers, 8 does not work when you plug it back into the original equation, so y = 0 is the only answer. You could have an equation that contains more than one radical, if so, isolate one at a time and get rid of that radical. Example 3 x 5 = x 1 x 5 = x -= x 3 = x 9 = x x +1; squaring both sides Note well when you squared the right side, the radical did NOT disappear. The radical needs to be isolated again! Copyright 003 Hanlonmath Page 9 of 15

Graphing Slope Intercept y = mx + b 1. Plot b, the yint. From b, use m to find second point 3. Connect the points. Example 1 Graph y = x + 5 0 18 Start at (0, 5), then go up and over 1 1 1 1 10 y = x + 5 8 10 5 5 10 15 0 Example Graph y = ( /3) x + 1 Start at (0, 1), then go down and over 3 8 y = x 3 + 1 10 5 5 Copyright 003 Hanlonmath Page 10 of 15

Graphing General Form Ax + By = C (Cover-up Method) 1. Let x = 0, find y int. Let y = 0, find x int 3. Connect Points Example 1 Graph 3x + y = 3x + y = 5 5 Example Graph 5x 3y = 15 8 5x 3y = 15 5 5 8 10 Copyright 003 Hanlonmath Page 11 of 15

Finding Equations of Lines y y1 = m(x x1) Point - Slope To find an equation of a line, you need two pieces of information, the slope and a point. Substitute those values into the Point-Slope form of an equation of a line. If you are not given slope explicitly, you might have to find the slope using the slope formula if you are given two points. Or, if you ar given an equation, then remember parallel lines have the same slope and perpendicular lines have negative reciprocal slopes. y y1 = m(x x1) Example 1 Find an equation of a line passing through (, 3) and slope 5 y ( 3) = 5(x ) y + 3 = 5x 10 y = 5x 13 Example Find an equation of a line passing through (, 5) and (, 7) The slope is 1/ =. Pick either point - (, 5) and substitute into the Point-Slope y 5 = (x ) y 5 = x + 1 y = x + 17 Example 3 Find an equation of a line that passes through ( 1, ) and is parallel to y = 3x The slope is 3. Substitute into equation y = 3(x ( 1)) y = 3(x + 1) y = 3x + 3 y = 3x + 5 Copyright 003 Hanlonmath Page 1 of 15

Graph Absolute Value y = x Graph by plotting points or use the parent function to translate the vertex. The vertex is at the origin. 3 y = x 1 1 In the graph of the absolute value, y = x a + b, the b moves the V shape up or down. The a moves the graph horizontally. Example 1 Graph y = x 3 + Go over +3 and up from the origin f( x) = x 3 + 5 Copyright 003 Hanlonmath Page 13 of 15

Example Graph y = x + 3 Go over and down 3 y = x + 3 0 5 Graph Parabolas Graph by plotting points, using parent function or vertex and symmetry. This is just like how we graphed absolute value. The vertex is at the origin. 10 Parent Equation Vertex at (0, 0) y = x 8 5 5 Copyright 003 Hanlonmath Page 1 of 15

In the graph y = (x a) + b, the b moves the the graph up and down and the a moves the graph horizontally. Example 1 Graph y = (x 3) + 10 8 Go over 3 and up from the parent function y = ( x 3) + 5 When the equation is not in vertex format, use b/a to find the x- coordinate, use that to find the y-coordinate,then pick a convenient point(s) and symmetry to find other point(s) Example Graph y = x +x + 10 10 Use b/a to find the x -coordinate = / = 3 Substitute 3 into the equation, then y = 1. The vertex is at ( 3, 1). Substitute another value of x, let x = 0, then y= 10, a second point is (0, 10). Using symmetry, a third point would be (, 10) 8 y = x + x + 10 10 5 5 Copyright 003 Hanlonmath Page 15 of 15