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6-1 Solving Systems by Graphing Warm Up Evaluate each expression for x = 1 and y =–3. 1. x – 4y 2. –2x + y –5 13 Write each expression in slopeintercept form. 3. y – x = 1 y=x+1 4. 2x + 3y = 6 y = x+2 5. 0 = 5y + 5x y = –x Holt Algebra 1 6-1 Solving Systems by Graphing Objectives Identify solutions of linear equations in two variables. Solve systems of linear equations in two variables by graphing. Holt Algebra 1 6-1 Solving Systems by Graphing A system of linear equations is a set of two or more linear equations containing two or more variables. A solution of a system of linear equations with two variables is an ordered pair that satisfies each equation in the system. So, if an ordered pair is a solution, it will make both equations true. Holt Algebra 1 6-1 Solving Systems by Graphing Example 1A: Identifying Systems of Solutions Tell whether the ordered pair is a solution of the given system. (5, 2); 3x – y = 13 0 3x – y 13 3(5) – 2 13 Substitute 5 for x and 2 for y in each equation in the system. 2–2 0 15 – 2 13 0 0 13 13 The ordered pair (5, 2) makes both equations true. (5, 2) is the solution of the system. Holt Algebra 1 6-1 Solving Systems by Graphing Helpful Hint If an ordered pair does not satisfy the first equation in the system, there is no reason to check the other equations. Holt Algebra 1 6-1 Solving Systems by Graphing Example 1B: Identifying Systems of Solutions Tell whether the ordered pair is a solution of the given system. x + 3y = 4 (–2, 2); –x + y = 2 x + 3y = 4 –x + y = 2 –2 + 3(2) 4 –(–2) + 2 –2 + 6 4 4 4 4 2 2 Substitute –2 for x and 2 for y in each equation in the system. The ordered pair (–2, 2) makes one equation true but not the other. (–2, 2) is not a solution of the system. Holt Algebra 1 6-1 Solving Systems by Graphing All solutions of a linear equation are on its graph. To find a solution of a system of linear equations, you need a point that each line has in common. In other words, you need their point of intersection. y = 2x – 1 y = –x + 5 The point (2, 3) is where the two lines intersect and is a solution of both equations, so (2, 3) is the solution of the systems. Holt Algebra 1 6-1 Solving Systems by Graphing Helpful Hint Sometimes it is difficult to tell exactly where the lines cross when you solve by graphing. It is good to confirm your answer by substituting it into both equations. Holt Algebra 1 6-1 Solving Systems by Graphing Example 2A: Solving a System Equations by Graphing Solve the system by graphing. Check your answer. y=x Graph the system. y = –2x – 3 The solution appears to be at (–1, –1). y=x Check Substitute (–1, –1) into the system. y = –2x – 3 y=x • (–1, –1) y = –2x – 3 (–1) –1 (–1) –1 (–1) –2(–1) –3 –1 2–3 –1 – 1 (–1, –1) is the solution of the system. Holt Algebra 1 6-1 Solving Systems by Graphing Example 2B: Solving a System Equations by Graphing Solve the system by graphing. Check your answer. y=x–6 y+ Graph using a calculator and then use the intercept command. x = –1 Rewrite the second equation in slope-intercept form. y+ − x = –1 x y= Holt Algebra 1 − x y=x–6 6-1 Solving Systems by Graphing Example 2B Continued Solve the system by graphing. Check your answer. Check Substitute into the system. y=x–6 –6 y=x–6 + –1 –1 –1 –1 Holt Algebra 1 – 1 The solution is . 6-1 Solving Systems by Graphing Example 3: Problem-Solving Application Wren and Jenni are reading the same book. Wren is on page 14 and reads 2 pages every night. Jenni is on page 6 and reads 3 pages every night. After how many nights will they have read the same number of pages? How many pages will that be? Holt Algebra 1 6-1 Solving Systems by Graphing Example 3 Continued 1 Understand the Problem The answer will be the number of nights it takes for the number of pages read to be the same for both girls. List the important information: Wren on page 14 Jenni on page 6 Holt Algebra 1 Reads 2 pages a night Reads 3 pages a night 6-1 Solving Systems by Graphing Example 3 Continued 2 Make a Plan Write a system of equations, one equation to represent the number of pages read by each girl. Let x be the number of nights and y be the total pages read. Total pages number is read every night plus already read. Wren y = 2 x + 14 Jenni y = 3 x + 6 Holt Algebra 1 6-1 Solving Systems by Graphing Example 3 Continued 3 Solve Graph y = 2x + 14 and y = 3x + 6. The lines appear to intersect at (8, 30). So, the number of pages read will be the same at 8 nights with a total of 30 pages. (8, 30) Nights Holt Algebra 1 6-1 Solving Systems by Graphing Example 3 Continued 4 Look Back Check (8, 30) using both equations. Number of days for Wren to read 30 pages. 2(8) + 14 = 16 + 14 = 30 Number of days for Jenni to read 30 pages. 3(8) + 6 = 24 + 6 = 30 Holt Algebra 1 6-2 Solving Systems by Substitution Warm-Up Tell whether the ordered pair is a solution of the given system. 1. (–3, 1) 2. (2, –4) 3. Joy has 5 collectable stamps and will buy 2 more each month. Ronald has 25 collectable stamps and will sell 3 each month. After how many months will they have the same number of stamps? How many will that be? Holt Algebra 1 6-2 Solving Systems by Substitution Objective Solve linear equations in two variables by substitution. Holt Algebra 1 6-2 Solving Systems by Substitution Sometimes it is difficult to identify the exact solution to a system by graphing. In this case, you can use a method called substitution. The goal when using substitution is to reduce the system to one equation that has only one variable. Then you can solve this equation by the methods taught in Chapter 2. Holt Algebra 1 6-2 Solving Systems by Substitution Solving Systems of Equations by Substitution Step 1 Solve for one variable in at least one equation, if necessary. Step 2 Substitute the resulting expression into the other equation. Step 3 Solve that equation to get the value of the first variable. Step 4 Substitute that value into one of the original equations and solve. Step 5 Write the values from steps 3 and 4 as an ordered pair, (x, y), and check. Holt Algebra 1 6-2 Solving Systems by Substitution Example 1A: Solving a System of Linear Equations by Substitution Solve the system by substitution. y = 3x y=x–2 Step 1 y = 3x y=x–2 Both equations are solved for y. Step 2 Substitute 3x for y in the second equation. Solve for x. Subtract x from both sides and then divide by 2. y= x–2 3x = x – 2 Step 3 –x –x 2x = –2 2x = –2 2 2 x = –1 Holt Algebra 1 6-2 Solving Systems by Substitution Example 1A Continued Solve the system by substitution. Step 4 Step 5 Write one of the original equations. Substitute –1 for x. y = 3x y = 3(–1) y = –3 (–1, –3) Write the solution as an ordered pair. Check Substitute (–1, –3) into both equations in the system. y = 3x y=x–2 –3 3(–1) –3 –1 – 2 –3 Holt Algebra 1 –3 –3 –3 6-2 Solving Systems by Substitution Example 1B: Solving a System of Linear Equations by Substitution Solve the system by substitution. y=x+1 4x + y = 6 Step 1 y = x + 1 Step 2 4x 4x 5x Step 3 The first equation is solved for y. +y=6 Substitute x + 1 for y in the + (x + 1) = 6 second equation. Simplify. Solve for x. +1=6 –1 –1 Subtract 1 from both sides. 5x = 5 5x = 5 Divide both sides by 5. 5 5 x=1 Holt Algebra 1 6-2 Solving Systems by Substitution Example1B Continued Solve the system by substitution. Step 4 Step 5 y=x+1 y=1+1 y=2 (1, 2) Write one of the original equations. Substitute 1 for x. Write the solution as an ordered pair. Check Substitute (1, 2) into both equations in the system. y=x+1 4x + y = 6 2 1+1 4(1) + 2 6 2 2 6 6 Holt Algebra 1 6-2 Solving Systems by Substitution Sometimes you substitute an expression for a variable that has a coefficient. When solving for the second variable in this situation, you can use the Distributive Property. Holt Algebra 1 6-2 Solving Systems by Substitution Caution When you solve one equation for a variable, you must substitute the value or expression into the other original equation, not the one that had just been solved. Holt Algebra 1 6-2 Solving Systems by Substitution Example 2: Using the Distributive Property Solve y + 6x = 11 3x + 2y = –5 by substitution. Step 1 y + 6x = 11 – 6x – 6x y = –6x + 11 Solve the first equation for y by subtracting 6x from each side. Step 2 3x + 2y = –5 3x + 2(–6x + 11) = –5 Substitute –6x + 11 for y in the second equation. 3x + 2(–6x + 11) = –5 Distribute 2 to the expression in parenthesis. Holt Algebra 1 6-2 Solving Systems by Substitution Example 2 Continued Solve y + 6x = 11 3x + 2y = –5 by substitution. Simplify. Solve for x. Step 3 3x + 2(–6x) + 2(11) = –5 3x – 12x + 22 = –5 –9x + 22 = –5 – 22 –22 Subtract 22 from –9x = –27 both sides. –9x = –27 Divide both sides by –9. –9 –9 x=3 Holt Algebra 1 6-2 Solving Systems by Substitution Example 2 Continued Solve Step 4 y + 6x = 11 3x + 2y = –5 y + 6x = 11 y + 6(3) = 11 y + 18 = 11 –18 –18 by substitution. Write one of the original equations. Substitute 3 for x. Simplify. Subtract 18 from each side. y = –7 Step 5 Holt Algebra 1 (3, –7) Write the solution as an ordered pair. 6-2 Solving Systems by Substitution Example 2: Consumer Economics Application Jenna is deciding between two cell-phone plans. The first plan has a $50 sign-up fee and costs $20 per month. The second plan has a $30 sign-up fee and costs $25 per month. After how many months will the total costs be the same? What will the costs be? If Jenna has to sign a one-year contract, which plan will be cheaper? Explain. Write an equation for each option. Let t represent the total amount paid and m represent the number of months. Holt Algebra 1 6-2 Solving Systems by Substitution Example 2 Continued Total paid is signup fee payment for each plus amount month. Option 1 t = $50 + $20 m Option 2 t = $30 + $25 m Step 1 t = 50 + 20m t = 30 + 25m Both equations are solved for t. Step 2 50 + 20m = 30 + 25m Substitute 50 + 20m for t in the second equation. Holt Algebra 1 6-2 Solving Systems by Substitution Example 2 Continued Step 3 50 + 20m = 30 + 25m –20m – 20m 50 = 30 + 5m –30 –30 20 = 5m 20 = 5m 5 5 m=4 Solve for m. Subtract 20m from both sides. Subtract 30 from both sides. Step 4 t = 30 + 25m Write one of the original equations. Substitute 4 for m. Simplify. t = 30 + 25(4) t = 30 + 100 t = 130 Holt Algebra 1 Divide both sides by 5. 6-2 Solving Systems by Substitution Example 2 Continued Step 5 (4, 130) Write the solution as an ordered pair. In 4 months, the total cost for each option would be the same $130. If Jenna has to sign a one-year contract, which plan will be cheaper? Explain. Option 1: t = 50 + 20(12) = 290 Option 2: t = 30 + 25(12) = 330 Jenna should choose the first plan because it costs $290 for the year and the second plan costs $330. Holt Algebra 1 6-3 Solving Systems by Elimination Warm-Up Holt Algebra 1 6-3 Solving Systems by Elimination Objectives Solve systems of linear equations in two variables by elimination. Compare and choose an appropriate method for solving systems of linear equations. Holt Algebra 1 6-3 Solving Systems by Elimination Another method for solving systems of equations is elimination. Like substitution, the goal of elimination is to get one equation that has only one variable. To do this by elimination, you add the two equations in the system together. Holt Algebra 1 6-3 Solving Systems by Elimination Solving Systems of Equations by Elimination Step 1 Write the system so that like terms are aligned. Step 2 Eliminate one of the variables and solve for the other variable. Step 3 Substitute the value of the variable into one of the original equations and solve for the other variable. Step 4 Write the answers from Steps 2 and 3 as an ordered pair, (x, y), and check. Holt Algebra 1 6-3 Solving Systems by Elimination Example 1: Elimination Using Addition Solve 3x – 4y = 10 by elimination. x + 4y = –2 Step 1 Step 2 3x – 4y = 10 x + 4y = –2 4x + 0 = 8 4x = 8 4x = 8 4 4 x=2 Holt Algebra 1 Write the system so that like terms are aligned. Add the equations to eliminate the y-terms. Simplify and solve for x. Divide both sides by 4. 6-3 Solving Systems by Elimination Example 1 Continued Step 3 x + 4y = –2 2 + 4y = –2 –2 –2 4y = –4 4y –4 4 4 y = –1 Step 4 (2, –1) Holt Algebra 1 Write one of the original equations. Substitute 2 for x. Subtract 2 from both sides. Divide both sides by 4. Write the solution as an ordered pair. 6-3 Solving Systems by Elimination When two equations each contain the same term, you can subtract one equation from the other to solve the system. To subtract an equation add the opposite of each term. Holt Algebra 1 6-3 Solving Systems by Elimination Example 2: Elimination Using Subtraction Solve 2x + y = –5 by elimination. 2x – 5y = 13 Step 1 2x + y = –5 –(2x – 5y = 13) Step 2 2x + y = –5 –2x + 5y = –13 0 + 6y = –18 6y = –18 y = –3 Holt Algebra 1 Add the opposite of each term in the second equation. Eliminate the x term. Simplify and solve for y. 6-3 Solving Systems by Elimination Example 2 Continued Step 3 2x + y = –5 2x + (–3) = –5 2x – 3 = –5 +3 +3 Write one of the original equations. Substitute –3 for y. 2x Simplify and solve for x. = –2 x = –1 Step 4 (–1, –3) Holt Algebra 1 Add 3 to both sides. Write the solution as an ordered pair. 6-3 Solving Systems by Elimination Remember! Remember to check by substituting your answer into both original equations. Holt Algebra 1 6-3 Solving Systems by Elimination In some cases, you will first need to multiply one or both of the equations by a number so that one variable has opposite coefficients. This will be the new Step 1. Holt Algebra 1 6-3 Solving Systems by Elimination Example 3A: Elimination Using Multiplication First Solve the system by elimination. x + 2y = 11 –3x + y = –5 Step 1 Step 2 Holt Algebra 1 x + 2y = 11 –2(–3x + y = –5) x + 2y = 11 +(6x –2y = +10) 7x + 0 = 21 7x = 21 x=3 Multiply each term in the second equation by –2 to get opposite y-coefficients. Add the new equation to the first equation. Simplify and solve for x. 6-3 Solving Systems by Elimination Example 3A Continued Step 3 x + 2y = 11 3 + 2y = 11 –3 –3 2y = 8 y=4 Step 4 Holt Algebra 1 (3, 4) Write one of the original equations. Substitute 3 for x. Subtract 3 from each side. Simplify and solve for y. Write the solution as an ordered pair. 6-3 Solving Systems by Elimination Example 3B: Elimination Using Multiplication First Solve the system by elimination. –5x + 2y = 32 2x + 3y = 10 Step 1 2(–5x + 2y = 32) 5(2x + 3y = 10) –10x + 4y = 64 +(10x + 15y = 50) Step 2 Holt Algebra 1 19y = 114 y=6 Multiply the first equation by 2 and the second equation by 5 to get opposite x-coefficients Add the new equations. Simplify and solve for y. 6-3 Solving Systems by Elimination Example 3B Continued Step 3 2x + 3y = 10 2x + 3(6) = 10 2x + 18 = 10 –18 –18 Step 4 Holt Algebra 1 2x = –8 x = –4 (–4, 6) Write one of the original equations. Substitute 6 for y. Subtract 18 from both sides. Simplify and solve for x. Write the solution as an ordered pair. 6-3 Solving Systems by Elimination Check It Out! Example 3b Solve the system by elimination. 2x + 5y = 26 –3x – 4y = –25 Step 1 Step 2 Holt Algebra 1 3(2x + 5y = 26) +(2)(–3x – 4y = –25) Multiply the first equation by 3 and the second equation by 2 to get opposite x-coefficients 6x + 15y = 78 +(–6x – 8y = –50) Add the new equations. 0 + 7y = 28 Simplify and solve for y. y =4 6-3 Solving Systems by Elimination Check It Out! Example 3b Continued Step 3 2x + 5y = 26 2x + 5(4) = 26 2x + 20 = 26 –20 –20 2X = 6 x=3 Step 4 (3, 4) Holt Algebra 1 Write one of the original equations. Substitute 4 for y. Subtract 20 from both sides. Simplify and solve for x. Write the solution as an ordered pair. 6-3 Solving Systems by Elimination Example 4: Application Paige has $7.75 to buy 12 sheets of felt and card stock for her scrapbook. The felt costs $0.50 per sheet, and the card stock costs $0.75 per sheet. How many sheets of each can Paige buy? Write a system. Use f for the number of felt sheets and c for the number of card stock sheets. 0.50f + 0.75c = 7.75 f + c = 12 Holt Algebra 1 The cost of felt and card stock totals $7.75. The total number of sheets is 12. 6-3 Solving Systems by Elimination Example 4 Continued Step 1 0.50f + 0.75c = 7.75 Multiply the second equation by –0.50 to get + (–0.50)(f + c) = 12 opposite f-coefficients. 0.50f + 0.75c = 7.75 Add this equation to the + (–0.50f – 0.50c = –6) first equation to 0.25c = 1.75 Step 2 eliminate the f-term. Simplify and solve for c. c=7 Step 3 f + c = 12 f + 7 = 12 –7 –7 f = 5 Holt Algebra 1 Write one of the original equations. Substitute 7 for c. Subtract 7 from both sides. 6-3 Solving Systems by Elimination Example 4 Continued Step 4 (7, 5) Write the solution as an ordered pair. Paige can buy 7 sheets of card stock and 5 sheets of felt. Holt Algebra 1 6-3 Solving Systems by Elimination Holt Algebra 1 6-4 Solving Special Systems Warm-Up Solve each system by elimination. 1. 2x + y = 25 3y = 2x – 13 2. –3x + 4y = –18 x = –2y – 4 (2, –3) 3. –2x + 3y = –15 3x + 2y = –23 (–3, –7) (11, 3) 4. Harlan has $44 to buy 7 pairs of socks. Athletic socks cost $5 per pair. Dress socks cost $8 per pair. How many pairs of each can Harlan buy? 4 pairs of athletic socks and 3 pairs of dress socks Holt Algebra 1 6-4 Solving Special Systems Objectives Solve special systems of linear equations in two variables. Classify systems of linear equations and determine the number of solutions. Holt Algebra 1 6-4 Solving Special Systems In Lesson 6-1, you saw that when two lines intersect at a point, there is exactly one solution to the system. Systems with at least one solution are called consistent. When the two lines in a system do not intersect they are parallel lines. There are no ordered pairs that satisfy both equations, so there is no solution. A system that has no solution is an inconsistent system. Holt Algebra 1 6-4 Solving Special Systems Example 1: Systems with No Solution Solve y=x–4 –x + y = 3 Method 1 Compare slopes and y-intercepts. y=x–4 –x + y = 3 y = 1x – 4 Write both equations in slopeintercept form. y = 1x + 3 The lines are parallel because they have the same slope and different y-intercepts. This system has no solution so it is an inconsistent system. Holt Algebra 1 6-4 Solving Special Systems If two linear equations in a system have the same graph, the graphs are coincident lines, or the same line. There are infinitely many solutions of the system because every point on the line represents a solution of both equations. Holt Algebra 1 6-4 Solving Special Systems Example 2A: Systems with Infinitely Many Solutions Solve y = 3x + 2 3x – y + 2= 0 Method 1 Compare slopes and y-intercepts. y = 3x + 2 3x – y + 2= 0 y = 3x + 2 Write both equations in slopeintercept form. The lines y = 3x + 2 have the same slope and the same y-intercept. If this system were graphed, the graphs would be the same line. There are infinitely many solutions. Holt Algebra 1 6-4 Solving Special Systems Example 2A Continued Solve y = 3x + 2 3x – y + 2= 0 Method 2 Solve the system algebraically. Use the elimination method. y = 3x + 2 3x − y + 2= 0 y − 3x = 2 −y + 3x = −2 0 = 0 Write equations to line up like terms. Add the equations. True. The equation is an identity. There are infinitely many solutions. Holt Algebra 1 6-4 Solving Special Systems Caution! 0 = 0 is a true statement. It does not mean the system has zero solutions or no solution. Holt Algebra 1 6-4 Solving Special Systems Holt Algebra 1 6-4 Solving Special Systems Example 3A: Classifying Systems of Linear Equations Classify the system. Give the number of solutions. Solve 3y = x + 3 3y = x + 3 x+y=1 x+y=1 y= x+1 Write both equations in slope-intercept form. y= x+1 The lines have the same slope and the same y-intercepts. They are the same. The system is consistent and dependent. It has infinitely many solutions. Holt Algebra 1 6-4 Solving Special Systems Example 3B: Classifying Systems of Linear equations Classify the system. Give the number of solutions. Solve x+y=5 4 + y = –x x+y=5 y = –1x + 5 4 + y = –x y = –1x – 4 Write both equations in slope-intercept form. The lines have the same slope and different yintercepts. They are parallel. The system is inconsistent. It has no solutions. Holt Algebra 1 6-5 Solving Linear Inequalities Warm Up Graph each inequality. 1. x > –5 3. Write –6x + 2y = –4 in slope-intercept form, and graph. y = 3x – 2 Holt Algebra 1 2. y ≤ 0 6-5 Solving Linear Inequalities Objective Graph and solve linear inequalities in two variables. Holt Algebra 1 6-5 Solving Linear Inequalities A linear inequality is similar to a linear equation, but the equal sign is replaced with an inequality symbol. A solution of a linear inequality is any ordered pair that makes the inequality true. Holt Algebra 1 6-5 Solving Linear Inequalities Example 1A: Identifying Solutions of Inequalities Tell whether the ordered pair is a solution of the inequality. (–2, 4); y < 2x + 1 y < 2x + 1 4 2(–2) + 1 4 –4 + 1 4 < –3 (–2, 4) is not a solution. Holt Algebra 1 Substitute (–2, 4) for (x, y). 6-5 Solving Linear Inequalities Example 1B: Identifying Solutions of Inequalities Tell whether the ordered pair is a solution of the inequality. (3, 1); y > x – 4 y>x−4 1 3–4 1> –1 Substitute (3, 1) for (x, y). (3, 1) is a solution. Holt Algebra 1 6-5 Solving Linear Inequalities A linear inequality describes a region of a coordinate plane called a half-plane. All points in the region are solutions of the linear inequality. The boundary line of the region is the graph of the related equation. Holt Algebra 1 6-5 Solving Linear Inequalities Holt Algebra 1 6-5 Solving Linear Inequalities Graphing Linear Inequalities Step 1 Solve the inequality for y (slopeintercept form). Step 2 Graph the boundary line. Use a solid line for ≤ or ≥. Use a dashed line for < or >. Shade the half-plane above the line for y > Step 3 or ≥. Shade the half-plane below the line for y < or y ≤. Check your answer. Holt Algebra 1 6-5 Solving Linear Inequalities Example 2B: Graphing Linear Inequalities in Two Variables Graph the solutions of the linear inequality. 5x + 2y > –8 Step 1 Solve the inequality for y. 5x + 2y > –8 –5x –5x 2y > –5x – 8 y> x–4 Step 2 Graph the boundary line y = dashed line for >. Holt Algebra 1 x – 4. Use a 6-5 Solving Linear Inequalities Example 2B Continued Graph the solutions of the linear inequality. 5x + 2y > –8 Step 3 The inequality is >, so shade above the line. Holt Algebra 1 6-5 Solving Linear Inequalities Example 2B Continued Graph the solutions of the linear inequality. 5x + 2y > –8 Substitute ( 0, 0) Check y > x–4 for (x, y) because it is not on the 0 (0) – 4 boundary line. 0 –4 The point (0, 0) 0 > –4 satisfies the inequality, so the graph is correctly shaded. Holt Algebra 1 6-5 Solving Linear Inequalities Helpful Hint The point (0, 0) is a good test point to use if it does not lie on the boundary line. Holt Algebra 1 6-5 Solving Linear Inequalities Example 2C: Graphing Linear Inequalities in two Variables Graph the solutions of the linear inequality. 4x – y + 2 ≤ 0 Step 1 Solve the inequality for y. 4x – y + 2 ≤ 0 –y –1 ≤ –4x – 2 –1 y ≥ 4x + 2 Step 2 Graph the boundary line y ≥= 4x + 2. Use a solid line for ≥. Holt Algebra 1 6-5 Solving Linear Inequalities Example 2C Continued Graph the solutions of the linear inequality. 4x – y + 2 ≤ 0 Step 3 The inequality is ≥, so shade above the line. Holt Algebra 1 6-5 Solving Linear Inequalities Example 2C Continued Check y ≥ 4x + 2 3 4(–3)+ 2 3 –12 + 2 3 ≥ –10 Substitute ( –3, 3) for (x, y) because it is not on the boundary line. The point (–3, 3) satisfies the inequality, so the graph is correctly shaded. Holt Algebra 1 6-5 Solving Linear Inequalities Check It Out! Example 2a Graph the solutions of the linear inequality. 4x – 3y > 12 Step 1 Solve the inequality for y. 4x – 3y > 12 –4x –4x –3y > –4x + 12 y< –4 Step 2 Graph the boundary line y = Use a dashed line for <. Holt Algebra 1 – 4. 6-5 Solving Linear Inequalities Check It Out! Example 2a Continued Graph the solutions of the linear inequality. 4x – 3y > 12 Step 3 The inequality is <, so shade below the line. Holt Algebra 1 6-5 Solving Linear Inequalities Check It Out! Example 2a Continued Graph the solutions of the linear inequality. 4x – 3y > 12 Check y< –6 –6 –6 < –4 (1) – 4 –4 Substitute ( 1, –6) for (x, y) because it is not on the boundary line. Holt Algebra 1 The point (1, –6) satisfies the inequality, so the graph is correctly shaded. 6-5 Solving Linear Inequalities Check It Out! Example 2c Graph the solutions of the linear inequality. Step 1 The inequality is already solved for y. Step 2 Graph the boundary line = . Use a solid line for ≥. Step 3 The inequality is ≥, so shade above the line. Holt Algebra 1 6-5 Solving Linear Inequalities Check It Out! Example 2c Continued Graph the solutions of the linear inequality. Substitute (0, 0) for (x, y) because it is not on the boundary line. Check y≥ 0 x+1 (0) + 1 0 0+1 0 ≥ 1 A false statement means that the half-plane containing (0, 0) should NOT be shaded. (0, 0) is not one of the solutions, so the graph is shaded correctly. Holt Algebra 1 6-5 Solving Linear Inequalities Example 3a: Application Ada has at most 285 beads to make jewelry. A necklace requires 40 beads, and a bracelet requires 15 beads. Let x represent the number of necklaces and y the number of bracelets. Write an inequality. Use ≤ for “at most.” Holt Algebra 1 6-5 Solving Linear Inequalities Example 3a Continued Necklace beads 40x plus bracelet beads is at most 285 beads. + 15y ≤ 285 Solve the inequality for y. 40x + 15y ≤ 285 –40x –40x 15y ≤ –40x + 285 Subtract 40x from both sides. Divide both sides by 15. Holt Algebra 1 6-5 Solving Linear Inequalities Example 3b b. Graph the solutions. Step 1 Since Ada cannot make a negative amount of jewelry, the system is graphed only in Quadrant I. Graph the boundary line for ≤. Holt Algebra 1 = . Use a solid line 6-5 Solving Linear Inequalities Example 3b Continued b. Graph the solutions. Step 2 Shade below the line. Ada can only make whole numbers of jewelry. All points on or below the line with whole number coordinates are the different combinations of bracelets and necklaces that Ada can make. Holt Algebra 1 6-5 Solving Linear Inequalities Example 3c c. Give two combinations of necklaces and bracelets that Ada could make. Two different combinations of jewelry that Ada could make with 285 beads could be 2 necklaces and 8 bracelets or 5 necklaces and 3 bracelets. (2, 8) (5, 3) Holt Algebra 1 6-5 Solving Linear Inequalities Example 4A: Writing an Inequality from a Graph Write an inequality to represent the graph. y-intercept: 1; slope: Write an equation in slopeintercept form. The graph is shaded above a dashed boundary line. Replace = with > to write the inequality Holt Algebra 1 6-5 Solving Linear Inequalities Example 4B: Writing an Inequality from a Graph Write an inequality to represent the graph. y-intercept: –5 slope: Write an equation in slopeintercept form. The graph is shaded below a solid boundary line. Replace = with ≤ to write the inequality Holt Algebra 1 6-6 Solving Systems of Linear Inequalities Warm-Up 1. You can spend at most $12.00 for drinks at a picnic. Iced tea costs $1.50 a gallon, and lemonade costs $2.00 per gallon. Write an inequality to describe the situation. Graph the solutions, describe reasonable solutions, and then give two possible combinations of drinks you could buy. 2. Write an inequality to represent the graph. Holt Algebra 1 6-6 Solving Systems of Linear Inequalities Objective Graph and solve systems of linear inequalities in two variables. Holt Algebra 1 6-6 Solving Systems of Linear Inequalities A system of linear inequalities is a set of two or more linear inequalities containing two or more variables. The solutions of a system of linear inequalities consists of all the ordered pairs that satisfy all the linear inequalities in the system. Holt Algebra 1 6-6 Solving Systems of Linear Inequalities Example 1A: Identifying Solutions of Systems of Linear Inequalities Tell whether the ordered pair is a solution of the given system. (–1, –3); y ≤ –3x + 1 y < 2x + 2 (–1, –3) y ≤ –3x + 1 –3 –3(–1) + 1 –3 3+1 –3 ≤ 4 (–1, –3) y < 2x + 2 –3 2(–1) + 2 –3 –2 + 2 –3 < 0 (–1, –3) is a solution to the system because it satisfies both inequalities. Holt Algebra 1 6-6 Solving Systems of Linear Inequalities Check It Out! Example 1b Tell whether the ordered pair is a solution of the given system. y > –x + 1 (0, 0); y>x–1 (0, 0) y > –x + 1 0 –1(0) + 1 0 0+1 0 > 1 (0, 0) y>x–1 0 0–1 0 ≥ –1 (0, 0) is not a solution to the system because it does not satisfy both inequalities. Holt Algebra 1 6-6 Solving Systems of Linear Inequalities System of Inequalities Graph the following inequalities on the same graph: y<2 x ≥ -1 y > x -2 Holt Algebra 1 The Graph of the system is the overlap, or intersection, of the 3 half-planes shown. 6-6 Solving Systems of Linear Inequalities A Triangular Solution Region A solution of a system of inequalities is an ordered pair that is a solution of each inequality in the Holt Algebra 1 (2,1) is a solution because: (2,1) y<2 1<2 x ≥ -1 2 ≥ -1 y > x -2 1 > 2 -2 6-6 Solving Systems of Linear Inequalities Solution Region Between Parallel Lines What is the system of inequalities represente d by this graph? Holt Algebra 1 The graph of one inequality is the half-plane below the line y = 3 The graph of the other inequality is the half-plane above the line y = 1 6-6 Solving Systems of Linear Inequalities Solution Region Between Parallel Lines The shaded region of the graph is the horizontal band that lies between the two horizontal lines, but not on the lines. Holt Algebra 1 The solution of this system of inequalities can be represented by this compound inequality: 1<y<3 6-6 Solving Systems of Linear Inequalities Example 2A: Solving a System of Linear Inequalities by Graphing Graph the system of linear inequalities. Give two ordered pairs that are solutions and two that are not solutions. y≤3 y > –x + 5 (–1, 4) Graph the system. y≤3 y > –x + 5 (8, 1) and (6, 3) are solutions. (–1, 4) and (2, 6) are not solutions. Holt Algebra 1 (2, 6) (6, 3) (8, 1) 6-6 Solving Systems of Linear Inequalities Example 2B: Solving a System of Linear Inequalities by Graphing Graph the system of linear inequalities. Give two ordered pairs that are solutions and two that are not solutions. –3x + 2y ≥ 2 y < 4x + 3 –3x + 2y ≥ 2 2y ≥ 3x + 2 Holt Algebra 1 Write the first inequality in slopeintercept form. 6-6 Solving Systems of Linear Inequalities Example 2B Continued Graph the system. y < 4x + 3 (2, 6) and (1, 3) are solutions. (0, 0) and (–4, 5) are not solutions. Holt Algebra 1 (–4, 5) (2, 6) (1, 3) (0, 0) 6-6 Solving Systems of Linear Inequalities In Lesson 6-4, you saw that in systems of linear equations, if the lines are parallel, there are no solutions. With systems of linear inequalities, that is not always true. Holt Algebra 1 6-6 Solving Systems of Linear Inequalities Example 3A: Graphing Systems with Parallel Boundary Lines Graph the system of linear inequalities. y ≤ –2x – 4 y > –2x + 5 This system has no solutions. Holt Algebra 1 6-6 Solving Systems of Linear Inequalities Example 3B: Graphing Systems with Parallel Boundary Lines Graph the system of linear inequalities. y > 3x – 2 y < 3x + 6 The solutions are all points between the parallel lines but not on the dashed lines. Holt Algebra 1 6-6 Solving Systems of Linear Inequalities Check It Out! Example 3a Graph the system of linear inequalities. y>x+1 y≤x–3 This system has no solutions. Holt Algebra 1 6-6 Solving Systems of Linear Inequalities Example 4: Application In one week, Ed can mow at most 9 times and rake at most 7 times. He charges $20 for mowing and $10 for raking. He needs to make more than $125 in one week. Show and describe all the possible combinations of mowing and raking that Ed can do to meet his goal. List two possible combinations. Earnings per Job ($) Holt Algebra 1 Mowing 20 Raking 10 6-6 Solving Systems of Linear Inequalities Example 4 Continued Step 1 Write a system of inequalities. Let x represent the number of mowing jobs and y represent the number of raking jobs. x≤9 y≤7 20x + 10y > 125 Holt Algebra 1 He can do at most 9 mowing jobs. He can do at most 7 raking jobs. He wants to earn more than $125. 6-6 Solving Systems of Linear Inequalities Example 4 Continued Step 2 Graph the system. The graph should be in only the first quadrant because the number of jobs cannot be negative. Solutions Holt Algebra 1 6-6 Solving Systems of Linear Inequalities Example 4 Continued Step 3 Describe all possible combinations. All possible combinations represented by ordered pairs of whole numbers in the solution region will meet Ed’s requirement of mowing, raking, and earning more than $125 in one week. Answers must be whole numbers because he cannot work a portion of a job. Step 4 List the two possible combinations. Two possible combinations are: 7 mowing and 4 raking jobs 8 mowing and 1 raking jobs Holt Algebra 1 6-6 Solving Systems of Linear Inequalities Helpful Hint An ordered pair solution of the system need not have whole numbers, but answers to many application problems may be restricted to whole numbers. Holt Algebra 1 6-6 Solving Systems of Linear Inequalities 1. Graph Lesson Quiz: Part I y<x+2 . 5x + 2y ≥ 10 Give two ordered pairs that are solutions and two that are not solutions. Possible answer: solutions: (4, 4), (8, 6); not solutions: (0, 0), (–2, 3) Holt Algebra 1 6-6 Solving Systems of Linear Inequalities Lesson Quiz: Part II 2. Dee has at most $150 to spend on restocking dolls and trains at her toy store. Dolls cost $7.50 and trains cost $5.00. Dee needs no more than 10 trains and she needs at least 8 dolls. Show and describe all possible combinations of dolls and trains that Dee can buy. List two possible combinations. Holt Algebra 1 6-6 Solving Systems of Linear Inequalities Lesson Quiz: Part II Continued Reasonable answers must be whole numbers. Possible answer: (12 dolls, 6 trains) and (16 dolls, 4 trains) Solutions Holt Algebra 1