Section 2.0: Getting Started

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1 Solving Linear Equations: Graphically Tabular/Numerical Solution Algebraically Section 2.0: Getting Started Example #1 on page 128. Solve the equation 3x 9 = 3 graphically. Intersection X=4 Y=3 We are only interested in the value for x, so the solution is x = 4. On your graphing calculator do the following: Set Y1 = 3x 9 Set Y2 = 3 Graph the 2 lines and to find where the lines intersect select <2 nd > <TRACE> = <CALC> and then select option 5: Intersect. Then select <ENTER> <ENTER> Now move the cursor close to the intersection point using the Arrow Keys and then press <ENTER>. The solution will be at the bottom of the screen and it will look like the following: Finding the Intersection of two lines with a TI-84 Graphing Calculator: Example #2, page 129: Solve the following equation algebraically: x5 15 x5 10x 75 2x x 75 6x 69 4x 69 x To solve this same equation numerically: Perform the following steps on your TI-84 calculator. Input the left side of the equation into Y1 and the right side into Y2. Format your table by pressing <2 nd > <WINDOW> = <TBLSET> and then select TBLSTART and ΔTBL values. Press <2 nd > <GRAPH> = <TABLE> to view the values. The solution will be the x value where Y1=Y2. You may have to scroll up/down to see more values. A Linear Inequality in one variable can be written in one of the following forms: ax + b > 0 ax + b < 0 Whereas a and b are real numbers and a 0. ax + b 0 ax + b 0 1

2 Rules for solving linear inequalities: follow the same methods you apply to solving a linear equation in one variable: Isolate the variable, remember, whatever you do to one side of the equation you must do to the other side (addition, subtraction, multiplication and division). Note: you cannot divide by zero as that is undefined. If the inequality includes fractions, eventually you may need to clear the fractions by multiplying by the LCD. If you multiply or divide an inequality by a negative number then you must reverse (change the direction) of the inequality sign. Example 5 on page 133. Solve for x, given: x x 50 x 25 Example 5 on page 133. Solve for x, given: x x 50 x 25 2

3 Solving a Linear System of Two Variables: Graphically Numerically Algebraically Section 2.1: Graphical and Numerical Solutions Graphically: the solution represents the intersection of two lines, if there is no solution, then the lines are parallel and therefore don t intersect. Classification of Linear Systems: If there is a solution, whether it s exactly one point or an infinite number of points then we say the system is consistent. If the solution includes an Infinite number of points the lines coincide. If there is no solution, then we say the system is inconsistent parallel lines. Below is a summary: If only one solution consistent and independent. If an infinite number of solutions consistent and dependent. If no solution inconsistent and independent. 3

4 Section 2.2: Substitution and Elimination Substitution method: Solve for either x or y in the 1 st equation and then substitute that expression into the 2 nd equation. The second equation is then a one-variable equation, solve for that variable and then return to the 1 st equation substitute your result and now solve for the 2 nd variable. Your solution is an ordered pair (x, y). And then check your answer by substituting your solution into the 2 nd equation. Example: Solve the following system of equations via the substitution method. 2x + 3y = 5 x + 4y = 2 Use the 2 nd equation to solve for x x = 2 4y, now substitute this expression into the 1 st equation in place of x. 2(2 4y) + 3y = 5 y Now substitute this value in for y back into the 2 nd equation and solve for x 2x 3 5 2x 5 2x x Solution: 14 5,1 5 Check by substituting result into either equation or w/calculator. Elimination method: Multiply/divide either or both equations such that coefficients for x or y are of the same magnitude but opposite in sign. Then add the two equations which will eliminate one of the variables. Solve for the remaining variable and then go back to either of the original equations, substitute the value you just found and solve for the other variable. Write answer as an ordered pair, (x, y) Check your solution by substituting your result into the equation not used above. Example from above: 2x + 3y = 5 x + 4y = 2 Multiply both sides of the 2 nd equation by 2 and then add the 2 equations together to eliminate x. 2x + 3y = 5 2x + 3y = 5 2(x + 4y = 2) 2x 8y = 4 5y = 1 y Solve for x in equation #1, solution 14 5,1 and then check your answer using equation

5 Identifying Inconsistent and Dependent Systems Algebraically Example #8 on page 159: solve the following system of equations. 10x 6y 2 10x 6y 8 Add the 2 equations together to eliminate x 0 = 10. Both variables have been eliminated and the result is an invalid equation no solution. This is an example of an inconsistent system of equations. Example #9 on page 160: solve the following system of equations. 10x 6y 12 5x 3y 62 Multiply the 2 nd equation by 2 and then add the 2 equations together to eliminate x 0 = 0. In this case, all the x and y pairs that satisfy the 1 st equation, also satisfy the 2 nd equation infinitely many solutions consistent and dependent. Example: solve the following system of equations. 3x 5y = 9 2x + 4y = 8 Example: solve the following system of equations. 3x 5y = 9 2x + 4y = 8 2(3x 5y = 9) 6x 10y = 18 2( 2x + 4y = 8) 6x + 12y = 24 2y = 6 y = 3 Now solve for x using either original equation 3x 5y = 9 3x 5(3) = 9 3x = 6 x = 2 Therefore the solution is (2, 3). Which is where the 2 lines intersect. 5

6 Section 2.3: Linear Inequalities in Two Variables A linear inequality in 2 variables can be written in one of the following forms: ax + by > c ax + by < c ax + by c ax + by c Where a, b and c are real numbers, where both a and b are not both zero. An ordered pair (x, y) will be part of the solution set if it satisfies the inequality. Technique for solving inequalities graphically Re-write the equation in slope-intercept form, isolate y (output). Graph the corresponding linear equation by replacing the inequality with an equal sign. If the inequality is either or, then the line will be part of the solution set. In this case, draw a solid line to indicate that the line is part of the solution set. If the inequality is a > or <, then the line will not be part of the solution set. In this case, draw a dashed line indicating that is not part of the solution set. Select any point that does not lie on the line to determine if that region satisfies the inequality, this is typically called a test point. Note (0, 0) can often be used and is a very easy test point. If the test point satisfies the inequality, then shade the region that contains the test point indicating all those ordered pairs are part of the solution. If it does not satisfy the inequality, then shade the opposite region. Also, if it s > or then you would shade the area above the line and vice versa. Example: graph 2y + 6x > 4 First solve for 2y > 6x 4 divide by 2 (flip the inequality sign) y < 3x + 2 Plot the line y = 3x + 2; since it s y < plot a dashed line Then shade the area below the line, since it s y <, but you can also verify that with a test point, pick (0, 0) 2(0) + 6(0) > _4? 0 > 4, therefore it s true and therefore (0,0) passes the test and therefore shade the area below the line. 6

7 Example 2 on page 166, you have $60 to purchase popcorn and sodas at a movie theater. PC costs $7.50 per bag and sodas cost $6. Graph the inequality. Let x = # of bags of popcorn and y = # of sodas 7.5x + 6y 60 y 7.5/6x + 10 y 1.25x + 10 Graph y = 1.25x + 10, but this time the line is part of the solution, use intercepts to graph line (0, 10) and (10/1.25, 0) = (8, 0) Since it is y, we will shade the area below the line. And you can also pick a test point, let s pick (0, 0) 7.5(0) + 6(0) 60?, yes, because it simplifies to 0 60, which is true and therefore passes the test, therefore we shade below line. And I stop the graph at x = 0 and x = 8, why is that? Section 2.4: System of Linear Inequalities A system of linear inequalities in two variables is a set of two or more linear inequalities that are to be solved simultaneously. Example #1 on page 177. Graph the following system of linear inequalities. y + 2x > 8 y x 1 Solution: 1) Graph each inequality as we did in section 2.3 including the shading. 2) Determine the point of intersection of the two boundary lines, this is called the vertex point (or corner point). To find the vertex graphically, we use the Intersect feature of the calculator. To find the point algebraically, we replace the inequality symbols with equal signs and solve the system of equations as we did in Section 2.1 and ) If the shaded areas overlap, then that is the region that satisfies both inequalities and therefore is the solution of the system of linear inequalities in two variables. 7

8 Solution: First find the vertex: y + 2x = 8 y x = 1 Multiply the 2 nd equation by 2 to eliminate the x and then add the 2 equations together. 3y = 10 y Solve for x Vertex, x1 x 1 x Plot the lines: since it s y + 2x > 8, use a dashed line as the line is not part of the solution and since it s y > shade the area above the line and graph y > 2x + 8. For y x 1, graph y x + 1 as a solid line and shade above the line. For the 1 st line, graph x and y intercepts (0, 8) and (4, 0) and for the 2 nd line, graph (0, 1) and the vertex point. Where the two shaded areas overlap, that is the solution region, verify w/test point (4, 6) 4 + 2(4) > 8? This is true because it is 12 > 8 and for the 2 nd equation, 6 4 1?, yes, because 2 1. Example 3 on page 181, word problem: Andrea makes $10 an hour walking dogs and $15 an hour bathing dogs. She doesn t want to work more than 12 hours per week and she wants to make at least $150 per week. How many hours must Andrea work, doing what, to earn at least $150 per week? Let x = # hours walking dogs Let y = # hours bathing dogs 8

9 Then we know the following: x + y < 12 10x + 15y 150 and we need to solve this system of inequalities. 10(x + y = 12) 10x + 15y = 150 5y = 30 y = 6 Solve for x, we know x + y = 12 x + 6 = 12 x = 6 Therefore (6, 6) is the vertex. Graph the lines y = x + 12 and y x10. Shade the appropriate regions. The y-intercepts are at (0, 10) and (0, 12), what do those points represent. The following vertices define our solution set: (6, 6); (0, 10) and (0, 12). Graphing a system of linear inequalities with a TI-84 Graphing Calculator: 9