MATH 2 part 2

Math Integration: What is the World’s Fastest Insect? Answer the following problems on geometric sequence to find out. Crossout the boxes that contain an answer. The remaining boxes will spell out thename of the world’s fastest insect, which can travel at a speed of around 60kilometers per hour. Really amazing!1. What is the common ratio of the geometric sequence 81, 243, 729, . . . ?2. What is the missing term of the sequence 5, 15, 45, , 405, . . .?3. What are the next two terms of the sequence 7, 49, 343, . . . ?4. What is the common ratio of the sequence 3 , 6 , 12 , . . . ? 4 12 365. Give the next two terms of the sequence 3 , 6 , 12 , . . .. 5 25 1256. Find the missing term of the sequence 5 , 10 , , 40 , . . .? 6 18 162BDURT3 9 20 3 2401;16807 54 2 A G T RO 24 ; 481625 3125 1 2 135 27 2 3 N F A L Y12401; 16807 24 ; 48 125 3 625 3125 20 4 56Answer: ____ ____ ____ ____ ____ ____ ____ ____ ____Source: Math Journal Vol. X, No. 4, 2002 – 2003 6

Lesson 2 Giving the Formula for the nth Term of a Geometric Sequence If the common ratio is greater than 1, the sequence grows at an increasingrate. Thus, the common ratio is known as the growth factor. The figure belowillustrates a geometric sequence formed by multiplying by 3. 27 9 31 7

If the common ratio is less than 1, the sequence shrinks at a decreasingrate. Thus, this common ratio is known as the decay factor. The figure illustratesa geometric sequence formed by multiplying by 1 . 216 8 4 2Examples:1. Consider the geometric sequence: 2, 6, 18, 54, 162, . . .The first term is 2 2 · 30 = 2The second term is 2 · 3 2 · 31 = 6The third term is 2 · 3 · 3 2 · 32 = 18The fourth term is 2 · 3 · 3 · 3 2 · 33 = 54The fifth term is 2 · 3 · 3 · 3 · 3 2 · 34 = 162If you want to continue, then you have the sixth term 2 · 3 · 3 · 3 · 3 · 3 =2 · 35 = 486 8

In the given example: 2, 6, 18, 54, 162, . . . , the common ratio is 3. If youpresent the first term as t1, using the pattern above, you have,1st term = t12nd term = t23rd term = t34th term = t45th term = t56th term = t6 Therefore, organizing the pattern, you have the number of terms andexpressions for the nth term as follows:1st term = t1 2 · 30 = t12nd term = t2 2 · 31 = t1 r13rd term = t3 2 · 32 = t1r24th term = t4 2 · 33 = t1r35th term = t5 2 · 34 = t1 r46th term = t6 2 · 35 = t1 r5nth term = tn 2 · 3n-1 = t1 rn-1 You will notice that the exponent of the common ratio is 1 less than thenumber of terms you are looking for. So, if you are looking the 3rd term, the exponent of the common ratio is 2;for the 4th term, the exponent of the common ratio is 3; for the 6th term, thecommon ratio is 5, etc… Therefore, for the nth term, the exponent of the common ratio is n – 1,where n is the number of terms.To find the nth term of a geometric sequence, the formula istn = t1 rn-1where n = the number of termsr = common ratiot1 = first termtn = last term 9

So, if you want to find the 10th term of the sequence 2, 6, 18, 54, 162, . . . ,then you have to use the formula for convenience tn = t1 rn-1.First identify the given before you substitute to the formula: t1 = 2 first term r =3 common ratio n = 10 number of terms tn = ? the term you are looking forSolution: tn = t1 rn-1 Substitute the given value t10 = 2 (3)10 -1 Simplify the exponent t10 = 2 (3)9 Simplify(3)9 . It is equal to 19, 683 t10 = 2 (19,683) t10 = 39,366Therefore the 10th term of the sequence 2, 6, 18, 54, 162, . . . is 39,366.2. Write the first terms of a geometric sequence in which t1 = 5 and r = 2.Since, the first term is given, write each term using the formula tn = t1 rn-1.t1 t2 t3 t4 t55 5(22 -1) 5(23 -1) 5(24 -1) 5(25 -1)5 5(21) 5(22) 5(23) 5(24)5 10 20 40 80The first five terms of the sequence are 5, 10, 20, 40, and 80.3. Find the seventh term t7, of a geometric sequence in which t3 = 96 and r = 4.Method 1:The general form of the third term of a sequence is t1 r2 .Find t1 Then find t7t3 = t1 r2 tn = t1 rn-1 10

96 = t1 (4)2 t7 = 6(4)7-196 = t1 t7 = 6(4)616 t7 = 24,576 6 = t1Method 2: Begin with the third term and use r to find each successive term. t4 = t3 r4 t4 = 96 · 4 = 384 t5 = 384 · 4 = 1536 t6 = 1536 · 4 = 6144 t7 = 6144 · 4 = 24,576 Therefore, the seventh term is 24, 576. (using any method)4. Ryan read aloud a “foolproof” way to become a millionaire. He saved P1.00 on the first day. Then each day thereafter, he double the amount he saved the day before. Find the amount he should save on the 20th day of his plan. Solution: In his sequence, t1 = 1. Since the amount is twice that of the day before, r = 2. tn = t1 rn-1 t20 =1(20)20 -1 t20 = 1 (20)19 t20 = P524,288.00 On the 20th day, Fred should have P524, 288.00. 11

Try this outA. Give the first four items of each geometric sequence:1. a1 = 1 2 r=22. a1 = 3 r = -1 33. a1 = 3 r = -24. a1 = 4 r= 1 2Use tn = t1 rn-1 to find the nth term of each sequence.5. t1 = 3 n=4 r=26. t1 = 1 n=3 r=5 n=5 r= 17. t1 = 4 n=6 28. t1 = 16 n=3 r = -39. t1 = 1 r=- 3 81 2B. Find the first five terms of the given sequence and state whether the sequence is arithmetic, geometric, or neither.1. an = 2n + 52. an = n2 + 13. an = 2 · 3n4. an = 4 – 5n5. an = n + 1 n6. an = -2n · n 12

7. an = -2n8. an = 4 · 2n9. an = n + 1 n+210. an = n3 + 1Math Integration: The world’s tallest building is approximately 1483 feet high. It is found inKuala Lumpur, Malaysia. It is actually a twin tower, with each building having 88stories. (Source: Math Journal, Vol. X, No. 4) To find out its name, answer the following problems on geometricsequence.Encircle the letter that corresponds to the correct answer. The letters will spellout the name of the building.1. What is the third term of the geometric sequence, tn = (-2)n-1 ?P. 4 R. – 82. What is the second term of the geometric sequence, tn = (-1)n-1 ?E. – 1 O. 13. The fourth term of the geometric sequence, tn = 2(3)n-1 is ____.S. 6 T. 544. ______ is the first term of the geometric sequence, tn = 3(2)n-1R. 3 S. 65. In tn = 1 (4)n-1, t4 is equal to _____. 4I. 1 O. 16 46. In tn = 1 (-4)n-1, 8 is the _____ term. 2N. 3rd M. 4th 13

7. In tn = t1 rn-1, if t1 = 8 and r = 2 , then what is t6 ? 3A. 256 E. 16 243 2438. In tn = t1 rn-1, find the eighth term of the geometric sequence whose first term is 64 and whose ratio is - 1 ? 2R. 1 S. - 1 2 2Answer: ___ ___ ___ ___ ___ ___ ___ ___ Let’s summarize1. A geometric sequence or progression is a set of terms in which each term after the first is obtained by multiplying the preceding term by the same fixed number called the common ratio which is commonly represented by r.2. If the common ratio is greater than 1, the sequence grows at an increasing rate. Thus, the common ratio is known as the growth factor.3. If the common ratio is less than 1, the sequence shrinks at a decreasing rate. Thus, this common ratio is known as the decay factor. What have you learnedA. Tell whether or not each number sequence is a geometric sequence. Those that are geometric, give the common ratio. 1. 6, 24, 96, 384, . . . 2. 5, -10, 20, -40, . . . 3. 1 , 1, 2, 4, . . . 2 4. 3, 9, 27, . . . 14

5. 2, 4, 6, 8, 10, . . . 6. 1 , 3, 9, . . . 3 7. 7, 9, 11, 13, . . . 8. 15, 5, 5 , 5 , . . . 39 9. 4, 2, 1, 1 , . . . 2 10. 3, 7, 11, 15, . . .B. Given the following sequences, tell which are geometric and which are arithmetic sequences. State their common ratio or common difference. 1. 5, 20, 80, 320, . . . 2. 1, 4, 9, 16, . . . 3. 3, 8, 13, 18, . . . 4. 1, 4, 9, 16, . . . 5. 31, 38, 45, 52, . . . 6. 4, 12, 36, 108, . . . 7. 1, -3, 5, -7, . . . 8. 1, 6, 36, 216, . . . 9. 1 , 2 , 3 , 4 , . . . 2345 10. 2 , 3 , 4 , 5 , . . . 1 4 9 16C. Find the next four terms of the sequence. 1. 1 , 1 , 1, 2, . . . 42 2. - 1 , 1 , -1, 2, . . 42 3. 32, -16, 8, -4, . . . 4. - 1 , - 1 , - 1 , - 1 , . . 32 16 8 4 15

5. 2 , 2, 2 2 , 4, . . .6. 1 , 1 , 1 , 1, . . . 27 9 37. -1, 4, -16, 32, . . .8. 12, 24, 48, 96, . . .9. 10, 2, 2 , 2 , . . . 5 2510. 7, 14, 28, 56, . . . 16

Answer KeyHow much do you knowA. 1. Geometric sequence, r = 2 2. not 3. Geometric sequence, r = 1 4. not 5. Geometric sequence, r = 3 2 6. Geometric sequence, r = 1 3 7. Geometric sequence, r = 2 8. not 9. not 10. Geometric sequence, r = 3B. 1. geometric, r = - 1 2 2. geometric, r = 2 3. geometric, r = 1 2 4. geometric, r = 5 5. geometric, r = 3 6. geometric, r = -3 7. arithmetic, d = -4 8. arithmetic, d = 11 9. arithmetic, d = 1 2 10. arithmetic, d = 2 3C. 1. 686, 4802 2. -54, -162 3. 40, 80 4. 64, 256 5. 0.875, 0.4375 6. 1 , 1 9 27 7. 0.0008, 0.00008 8. 1 , 1 16 32 9. 9, 9 3 10. 1331 a4, 14641 a5 17

Try this outLesson 1 A. 1. geometric, r = 5 2. not 3. geometric, r = 2 4. not 5. geometric, r = 3 6. not 7. geometric, r = 2 8. geometric, r = 1 2 9. geometric, r = 2 3 10. geometric, r = 3 2 B. 1. 10 , 10 39 2. 54, 162 3. 67.5, 101.25 4. 27, 9 5. 384, 1536 6. 189, 567 7. -40, 80 8. 4, 8 9. 1, 3 10. 2, -4 C. Math Integration 1. 3 2. 135 3. 2401, 16807 18

4. 2 Arithmetic 3 neither geometric 5. 24 , 48 arithmetic 625 3125 neither 6. 20 54 neither geometric Answer: Dragonfly geometricLesson 2 A. 1. 1 , 1, 2, 4 2 2. 3, -1, - 1 , 1 39 3. 3, -6, 12, -24 4. 4, 2, 1, 1 2 5. 32 6. 25 7. 1 4 8. 3888 9. 1 36 B. 1. 7, 9, 11, 13, 15 2. 2, 5, 10, 17, 26 3. 6, 18, 54, 162, 486 4. -1, -6, -11, -16, -21 5. 2, 3 , 4 , 5 , 6 2345 6. -2, 8, -24, 64, -160 7. -2, 4, -8, 16, -32 8. 8, 16, 32, 64, 128 19

9. 2 , 3 , 4 , 5 , 6 neither 34567 neither 10. 2, 9, 28, 65, 126C. 1. P 4 2. E -1 3. T 54 4. R 3 5. O 16 6. N 3rd 7. A 256 243 8. S - 1 2 Answer: PETRONASWhat have you learnedA. 1. geometric, r=4 2. geometric, r = -2 3. geometric, r=2 4. geometric, r=3 5. arithmetic, d2 6. geometric, r=3 7. arithmetic, d=2 r= 1 8. geometric, 3 9. geometric, r= 1 10. arithmetic, 2 d=4B. 1. geometric, r=42. neither arithmetic or geometric3. arithmetic, d=5 20

4. neither arithmetic or geometric5. arithmetic, d=76. geometric, r=37. neither arithmetic or geometric8. geometric, r=69. neither arithmetic or geometric10. neither arithmetic or geometricC. 1. 4, 8, 16, 32 2. -4, 8, -16, 32 3. 2, -1, 1 , - 1 24 4. - 1 , -1, 1 , - 1 2 24 5. 4 2 , 8, 8 1 , 16 2 6. 3, 9, 27, 81 7. -128, 512, -2048, 8192 8. 192, 384, 768, 1536 9. 2 , 2 , 2 , 2 125 625 3125 15625 10. 112, 224, 448, 896 21

Module 4 Systems of Linear Equations and Inequalities What this module is about This module is about solving systems of linear inequalities. In module 1,you learned how to graph systems of linear equations and find its solutions. Inthis module, you will discover now the graphs of linear equations is related tographs of systems of linear inequalities and how their solutions differ. What you are expected to learn This module is designed for you to: 1. draw the graph of linear inequality in two variables. 2. define a system of linear inequalities 3. represent the solution set of a system of linear inequalities by graphing. 4. translate certain situations in real life to linear inequalities.How much do you knowA. Match each inequality at the left with one or more descriptions at the right.1. x ≤ 3 A. includes the line2. y > -4 B. excludes the line3. y ≤ 2x + 3 C. above the line4. y > 3x + 5 D. below the line5. y ≤ -7 E. right of the line6. x ≥ -8 F. left of the line7. y < x + 8B. Translate into linear inequality: 8. The sum of 3 consecutive integers is greater than 30. 9. A number n is greater than twice another number y. 10. Imelda has two more shoes than Cristina. The total number of shoes is greater than 84.

What you will do Lesson 1Draw the Graphs of Linear Inequalities The general form of linear inequality in two variables, x and y is ax + by >c or ax + by < c, where a and b are equal to 0. The graph of a linear inequality isa region.Example 1: Graph 3x – 2y < 2.Use the following steps to draw the graph.1. Transform the linear inequality in the y form:3x – 2y < 2 Division by a negative number-2y < -3x + 2 reverses the order of the inequality. y > 3 x -1 22. Locate the y-intercept, -1, then from it, move 3 units up then, 2 unitsright. Mark where you stop. Draw a dashed line passing through the markand the y-intercept.3. Check. Choose a test point from both sides of the dashed line.Test points (-2,1) and (4, -2)a) (-2,1), a point above the dashed lineSubstitute, to the inequality: 3x – 2y < 23(-2) – 2(1) ? 2 -6 – 2 ? 2 -8 ? 2 -8 < 2 True 2

b) (4, -2), a point below the line Substitute, to the inequality 3x – 2y < 2 3(4) – 2(-2) ? 2 12 + 4 ? 2 16 ? 2 16 < 2 FalseThe graph: y > 3/2x - 1 Notice that since y > 3 x – 1, the region above the line is shaded. Also a 2dashed line is used to indicate that the line is not a part of the graph. The graphwould include the line if the inequality contains the symbol ≥. When the line ispart of the graph, a solid line is used.Example 2. Graph x + y > 31. Solve for y.x+y >3 addition property y > -x + 32. Draw the graph of the inequality. Locate the y-intercept, 3, and then from it move 1 unit up and then one unit left. Mark where you stop. Draw a dashed line, passing through the mark and the y-intercept. 3

3. Check. Test points from both sides of the line and substitute to the inequality.Test points (3,3) and (-2,3): a) (3,3) a point on one side x+y>3 3+3?3 6?3 6 > 3 True b) (-2,3) False x+y>3 -2 + 3 ? 3 1?3 1> 34. Shade the region where the point that satisfies the inequality is located5. All the points in the shaded region are the solutions to the given inequality.The graph: x+y>3Example 3: Graph: 2x – 3y ≤ 0 dividing by negative number reverses the inequality. 1. Solve for y: -3y ≤ - 2x y ≥ 2x 3 4

2. Locate the y-intercept, this time, it is zero, or it is located at the origin. From the origin, move 2 units up, then 3 units right. Mark again, where you stop. This time, draw a solid line passing through your mark and the origin. Why do you think so?3. Choose again points from both sides of the line. Substitute to the inequality.a. (-1, 4) True or false? 2x – 3y ≤ 0 2(-1) – 3(4) ? 0 -2 – 12 ? 0 -14 ? 0 -14 ≤ 0b. (1,-2) True or false? 2x – 3y ≤ 0 2(1) – 3(-2) ? 0 2+6?0 8?0 8≤04. Shade the side of the line, where the point chosen satisfies the inequality. If you did right, your graph will appear like this.2x – 3y < 0 (-1,4) satisfies the inequality.You did? Congratulations. 5

Example 4: Draw the graph of y ≤ -2.Follow these steps: 1. Graph the equation: y = -2 2. Draw a solid horizontal line passing through y = -2. Pick a point above and below the line. 3. Substitute the points to the inequality: a) (2,2), a point above the line. y ≤ -2 2 ≤ -2 False b) (2,-2), a point along the line y ≤ -2 2 ≤ -2 True 4. Shade the region below the line y < -2Example 5: Draw the graph of x ≥ 2. Follow these steps:1. Graph the equation x = 2.2. Draw a vertical solid line passing through x = 2.3. Pick a point on both sides of the line, x = 2, means any value of y can be paired to x = 2 and it is called the x-intercept.a) (-2,2), a point on the left side of the line. x≥2 -2 ≥ 2 Falseb) (3,5) True x≥2 3≥2 6

4. Shade the region to the left of the line x = 2.The Graph: x>2 Now that you have graphed linear inequalities in two variables, you cannow shade the region by inspection. That is, the shade is on the left side of theline if the inequality symbol is ≤ or <, to the right if the inequality symbol is ≥ or >. Study how graphs of linear inequalities are described in the table belowInequalities Description y < mx + b The graph is the region or the half-plane below line y = mx + b. y ≤ mx + b y > mx + b The graph is the region below y = mx + b and includes line y =y ≥ mx + b mx + b.x>ax>a The graph is the region above line y = mx + b.x<ax<a The graph is the region above line y = mx + b and includes liney>a y = mx + b. The graph is the region to the right of line x = a. The graph is the region to the right of line x = a and includes line x = a. The graph is the region to the left of line x = a. The graph is the region to the left of line x = a and includes line x = a. The graph is the region above line x = a.y>a The graph is the region above line x = a and includes line x = a.y<a The graph is the region below line x = a.y<a The graph is the region below line x = a and includes line x = a. 7

Try this outA. Use the steps learned in graphing inequalities. Use a test point to shade. 1. x + y ≥ 2 2. x – y ≥ 1 3. 3x – y ≥ 6 4. 3x + y ≤ 3 5. 2x + 3y > 12B. Graph each linear inequality with the use of a test point. 1. y > 2x – 1 2. y ≤ 3x + 2 3. y ≥ 2 x – 1 3 4. y < - 3 x + 2 4 5. –x – 2y ≥ 4 Lesson 2Define a System of Linear Inequalitiesand Represent its Solution by Graphing Now, that you have been exposed to graphing linear inequalities in twovariables, this time you are now ready to do graphing systems of linearinequalities. First, study how systems of linear inequalities is defined. Statements ofthe form1. a1x + b1y > c1 3. a1x + b1y ≤ c1 a2x + b2y > c2 a2x + b2y ≤ c22. a1x + b1y < c1 a2x + b2y < c2 or a combination of inequalities, where a1, b1, a2, b2 and c2 are realnumbers are called systems of linear inequalities. 8

You will notice that a system of linear inequalities is composed of two ormore inequalities. Each inequality represents a region each pair of inequalitiesrepresents two regions whose intersection is the solution of the system. Study the examples in graphing a system of linear inequalities. These willinvolve finding the region where the solution of the system lies.Example 1: Graph the solution of the system x + y > -13x – 2y > 4Steps: 1. Change each inequality in the form similar to the equation y = mx + b.(a) x + y > -1 y > -x – 1(b) 3x – 2y > 4 Multiplying inequality by negative -2y > -3x + 4 number reverses the inequality 2y < 3x – 4 y < 3 x–2 22. Graph inequality (a) x + y > -1a. Graph y = -x – 1b. Shade above the line to represent y > -x - 1 x + y > -13. Graph inequality (b) 3x – 2y > 4 9

a. Graph y < 3 x – 2 2 b. Shade below the line to represent y < 3 x – 2 2 y < 3/2x - 2 Combining the two graphs in one coordinate axes, you will have this. (4, 1) (1, -4) Choose two test points, one in each single shaded region and one fromthe double shaded region. Try (1, -4) and (4, 10). Substitute these points to each inequality. Point (1, -4) which is contained in a single shaded region. 10

Substitute in (1): x + y > -1 (1) + (-4) > -1 False -3 > -1Substitute in (2): 3x – 2y > 4 3(1) –2(-4) > 4 True 3+8>4 11 > 4Point (4,1) which is contained in the double shaded region.Substitute in (1): x + y > -1 4 + 1 > -1 True 5 > -1Substitute in (2): 3x – 2y > 4 3(4) – 2(1) > 4 True 12 – 2 > 4 10 > 4 The solution for the linear inequalities are the points that satisfy both linearinequalities. Which of these points satisfy the linear inequalities? Where do you findthem? You are right! The point (4, 1) satisfies the linear inequalities and it isfound in the double shaded region. Thus, the solution is the set of all points found in the double shadedregion.Example 2: Graph the solution of the system:2x + y > 2x–y≥3Inequality (1): 2x + y > 21. Graph y = -2x + 2. Use broken lines in the graph of y > -2x + 2. This does not include the line.2. Shade the region above the line to represent the graph of y > -2x + 2 11

y > -2x + 2Inequality (2): x – y ≥ 3 Reverse the direction of the inequality -y ≥ -x + 3 when you multiply by a negative number. y≤x–31) Graph y = x – 3 on the same coordinate plane where the inequality (1) was done.2) Shade the region below the line to represent the graph y ≤ x – 3. y > -2x + 2 y<x-3 Choose two test points, one in the single shaded region and one on thedouble shaded region. Use (7, 2) and (-7,2).Point (7,2): See if this point satisfies both the inequalities.Substitute: (1) 2x+ y > 2 True 2(7) + 2 > 2 14 + 2 > 2 16 > 2 12

(2) x – y ≥ 3 True 7–2≥3 5≥3Point (-7, 2), located in a single shaded regionSubstitute: (1) 2x+ y > 2 2(-7) + 2 > 2 -14 + 2 > 2 -12 > 2 False(2) x – y ≥ 3 -7 –2 ≥ 3 -9 ≥ 3 False Which point satisfies both the inequalities? If you did right, the point thatsatisfies both the inequality is the region containing all the solutions.Example 3: Solve the system by graphing:2x – y ≤ -3 x ≤ -2Steps: 1. Graph y ≥ 2x + 3 a. Graph y = 2x + 3. Use a solid line. b. Shade above the line.2. Graph x ≤ -2 a. graph x = -2. Use solid line. b. Shade the left side of the line. y = 2x + 3 x = -2Notice, that the graph of x ≤ -2 is the region to the left of line x = -2. 13

3. Choose a point in the double shaded region. Does this point satisfy both the inequality? If so, then the solution to the inequalities are the set of points in the double shaded region.Example 4: Graph -1 ≤ x ≤ 4 Solution: This is actually a system of inequalities. Splitting theinequalities, you have: x ≥ -1 x ≤4Steps: 1. Graph x = -1. Using a solid line you can see that the graph of x ≥ - 1 is the region on the right side of line x = –1. 2. Graph x = 4. Using a solid line you can see that the graph of x ≤ I is the region on the left side of line x = 4.The graph: x=4 x=-1The solution is the intersection of the shades of the two inequalities.Example 5: Graph the solution set of the system of inequalities: x + y ≤ 3 (1) 2x + 3y < 9 (2) x≥0 (3) y≥0 (4)Steps: Using the skills gained,1. graph the lines x + y = 3, use solid line2. graph 2x + 3y < 9, using dashed line, 14

3. graph x = 0, using solid line, or the y-axis4. graph y = 0, using solid line, or the x-axis. 2x + 3y = 9 x+y=3Solution: The inequality x + y ≤ 3 is satisfied by the set of points found in the regionbelow the line x + y = 3. The inequality 2x + 3y < 9 is satisfied by the set of points in the regionbelow the line 2x + 3y = 9. The set of inequalities x ≥ 0 and y ≥ 0 is satisfied by points in the firstquadrant or points on the portions of the axis bounding the first quadrant. The solution set is the shaded region plus the solid portions of theboundary lines.Example 6: Graph the following system inequalities, by determining the x and y intercepts. 4x – 2y ≤ 8 (1) y – 2 ≤ 0 (2)Steps: a. Solve for the x and y intercepts of the two inequalities(1) 4x – 2y ≤ 8 To find the y-intercept, substitute x to 0 in the inequality. 4(0) –2y = 8 The y-intercept is the value of y -2y = 8 at point (0, y). y = -4 To find the x intercept, substitute y to 0 in the same inequality. 4x – 2(0) = 8 The x-intercept is the value of x 4x = 8 at point (x, 0). x=2 15

(2) y – 2 ≤ 0, The y-intercept in this inequality is y = 2. It has no y-intercept since the graph of the equation y – 2 = 0 is a vertical line.b. Then graph the system: Connect the x and y intercepts of each equation and shade the regionwhere the points are found satisfying each inequality. 4x – 2y = 8 y=2Example 7: Graph the system by determining the x and y intercepts of eachinequality. y – 2x ≤ 1 y + 2x ≥ 1 x≥2Solution: 1. Determine the x and y intercept of each inequality. (1) y - 2x ≤ 1 x-intercept = - 1 2 y-intercept = 1 (2) y + 2x ≥ 1 x-intercept = 1 2 y-intercept = 1 (3) x ≥ 2 x-intercept = 2 y-intercept = 0 2. Connect the intercepts of each inequality and shade the solution. 16

The solution to the system of inequalities is the intersection of all theshaded region of each inequality.The graph: y + 2x = 1 y – 2x = 1 x=2Try this out Graph the system of inequalities and indicate the solution of the system bydouble or multiple shades. You can use any method.1. x ≤ 1 3. y > 5x – 1 5. x + y > 2 y<4 y ≥ -2x + 6 2x – y ≤ 1 6. x ≥ 02. y < x + 3 4. x + y > 5 x–y≥0 y<5– 1x 2x – y < 4 x+y≤4 2 Lesson 3Translate Certain Situations in Real Life to Linear Inequalities Inequality symbols may be used also to describe mathematical situations.Refer to the table for mathematical translation of symbols of inequalities. Symbol Translation in words. < Less than ≤ Less than or equal to, at most > Greater than, more than ≥ Greater than or equal, at least 17

Example 1. Write an inequality to represent the following situations.a. The survey shows that less than 20 students have parents with houseof their own.Let x be the number students with parents owning house of their own.Translation: x < 20b. A man traveled a distance m km by walking and 8 times as far as bybus. He covered more than 100 km.Let m = the distance traveled by walkingTranslation: m + 8m > 100c. Errold bought 3 pairs of pants. He gave P2000.00 and got a change ofless than P80.00. What are the possible prices of the pair of pants?Let x = the price of the pantsTranslation: 2000 – 3x < 80Example 2. Mother gave me at most P200 allowance in a week. Which of thefollowing amount could she give?a. P200 or less b. P200 or more c. exactly P 200. What is your guess? What symbol is appropriate for the situation if n is theamount for allowance intended for a week?a. n ≤ 200 b. n < 200 e. n = 200 d . n > 200The right response is a and the guess is a.Try this outA. Which of the given inequality symbols describes each statement?1. Larry is an industrious appliance salesman. His average sales in a week isat least P10 000. let x represents his sales.a. x < 10 000 b. x > 10 000 c. x ≤ 10 000 d. x ≥ 10 0002. A son’s savings x is at most P5.a. x < 5 b. x > 5 c. x ≤ 5 d. x ≥ 53. The sum of the ages of Jennifer (m) and Roy (n) is not more than 32.a. m + n < 32 b. m + n < 32 c. m + n ≤ 32 d. m + n ≥ 32 18

4. A number x added to three times the number is less than 12.a. x + 3x < 12 b. x + 3x > 12 c. x + 3x ≤ 12 c. x + 3 ≤ 125. Seven times a number is at least 30. Let x be the number. a. 7x < 30 b. 7x > 30 c. 7x ≤ 30 d. 7x ≥ 30B. How to graph a Linear Inequality:Graph the linear inequality, x + 2 < yLook here my students, and you shall seehow to graph an inequality.Here’s a simple inequality to try; x+2<yx + 2 is less than y,First, Make the “less than “ “equal to”; y=x+2So now y equals x plus 2.Then test a number for x; say, 10; x = 10Now substitute that single constant in y=x+2Add 10 to 2 and you’ll get y; y = 10 + 2See if this pair will satisfy (10.12)x:10, y:12; you’ll find its right,So graph this point to expedite. FigureNow find a second ordered pair (3,5)that fits in your equation therex:3, y:5 will do quite well.And its correct, as you can tell.Plot this point, and then you’ve got Figureto draw a line from dot to dot.Make it neat and make it straight;a ruler’s edge I’d advocate.The next step’s hard! You’ve got to choosewhich side of this line you must use.Change “equal to” back to “less than,” x+2<yjust as it was when you began.Test a point on one side. 19

Use 3 for x and 1 for y. (3,1) 3+2<1Is 1 greater than 3 plus 2? (1,4)No! This side will never do! 1+2 <4 FigureOn the other side, let’s try x+2=y1 for x and 4 for y. (1,3) 1+2=31 plus 2 (which equals 3)is less than 4, as you can see.Shade in the side that dot is on;We’ve got one more step to come upon.Do the points upon your linefor the equation I assigned.Use 1 and 3 for this test;They’re “equal to” , soMake your line dotted to show this is trueand that is all you have to do!Let ‘s summarize Graphs of linear inequalities are described in the table below.Inequalities Description y < mx + b The graph is the region or the half-plane below line y = mx + b. y ≤ mx + b y > mx + b The graph is the region below y = mx + b and includes line y =y ≥ mx + b mx + b.x>ax>a The graph is the region above line y = mx + b.x<a The graph is the region above line y = mx + b and includes line y = mx + b. The graph is the region to the right of line x = a. The graph is the region to the right of line x = a and includes line x = a. The graph is the region to the left of line x = a. 20

x<a The graph is the region to the left of line x = a and includes liney>a x = a.y>a The graph is the region above line x = a.y<ay<a The graph is the region above line x = a and includes line x = a. The graph is the region below line x = a. The graph is the region below line x = a and includes line x = a. Steps in graphing the solution set of a system of linear inequalities:1. Graph first the equation. Use any of the two methods presented.2. Test a point on one side of the line by substituting the values in the inequality. If the point satisfies the inequality, shade the region that contains the point.3. The common point is the region where the graphs of the inequalities intersect. What have you learnedA. Tell which region A, B, C or D is the graph of the system of inequalities.1. x < 0 2. x > 2 y<0 y<3 A ACCD 21

3. x > -1 4. x < -2 y>1 y > -2B. Determine which of the following ordered pair is a solution of the system: x+y≤6 x–y<1 5. (1,3) 6. (1,2) 7. (0,0) 8. (6,0) 9. (-2,0) 10. (7,1)C. Translate the following inequality statements: 1. Jean is 3 years older than Jericho. The sum of their ages is less than 15 years. 2. Twice a number added to 4 is greater than 50. 22

Answer Key x–y>1How much do you knowA. 1. A, F 2. B, C 3. A, D 4. B, C 5. A, D 6. A, E 7. B, DB. 8. x + (x + 1) + (x + 2) , 30 or 3x + 3 < 30 9. n > 2y 10. 2x + 2 > 84Try this outLesson 11. 2. x+y>23. 4. 3x + y < 33x – y > 6 23

5. B. 1. 2x + 3y > 12 y > 2x - 1B. 2. 3. y > 2/3x - 1 y < 3x + 24. 5. y < -3/4x + 2 -x – 2y > 4 24

Lesson 2 2.1. y=4 y=x+3 y=5– 1x 2 x=13. 4. y = 5x - 1 x+y=6 y = -2x + 6 2x – y = 45. 6. x + y = 4 x–y=0 2x – y = 1 x+y=2 25

Lesson 3 yA. 1. d • (10, 12) 2. c 3. c (1,4) • (3,5) 4. a • 5. dB. Figure (3,1) x • 26

What have you learnedA. Tell which region A, B, C or D is the graph of the system of inequalities.D 1. x < 0 C 2. x > 2 y<0 y<3 3.A 3. x ≥ -1 C 4. x < -2 y≥1 y > -2 27

B. Determine whether the ordered pair is a solution of the system:5. (1,3) 6. (1,2) 7. (0,0) and 8. (6, 0)C. 1. Jericho is x, Jean is x + 3, The sum of their ages is 2x + 3, The inequality: 2x +3 < 15. 2. 2x + 4 > 50 28

Module 4 Rational Algebraic Expressions What this module is about This module is about equations and solving problems involving rational algebraicequations. As you go over the exercises, you will develop skills in solving equationsand problems in real life which involves rational algebraic equations., You will also useproperties in simplifying rational algebraic expressions What you are expected to learnThis module is designed for you to: 1. solve rational algebraic equations. 2. solve problems involving rational algebraic equations. 3. use and solve formulas for a specified variable.How much do you knowFind or solve what is asked.1. Solve for k: 9 = 3 k42. In solving a − 4 = a + 8 , we will multiply both sides by the LCD which is ___. 4 163. The value of a is _____.4. An airplane flies 800 km/hr for t hours. It travels a distance of ____ km.a. 400t b. 800t c. 400d d. 800d5. If the airplane flies for 3 hours at the rate of 950 km/hr, find the distance it traveled.

Use this problem to solve # 6,7 and 8 Working alone, it will take Peter 8 hours to wall paper his room. Working withVhernie, Peter takes only 5 hours to wall paper the room.Let x = time it will take Vhernie working by himself6. What is the work rate of Vhernie?7. What is the work done by Peter?8. How long will it take Vhernie to finish the job working by himself?a. 12 1 b. 13 1 c. 12 1 d. 13 1 hours 3 3 5 5Use this problem to solve # 9, 10, and 11. The rate of the plane is P km/hr in still air. The rate of the wind is W km/hr. Find therate of the plane under the conditions described. Express your answers in terms of Pand W.9. Flying with the wind.10. Flying against the wind11. Flying against the wind if the rate of the wind increases by 15 km/hr.12. Flying with the wind if the rate of the wind decreases by 20 km/hr.13. Solve for Q in P = Q + R DD 2

What you will do Lesson 1 Solving Rational EquationsRational equations are equations containing rational expressions.Example 1. Solve 11 − 2 = 1 2x 3x 611 − 2 = 1 The LCD is 6x2x 3x 66x  11 − 2  = 1 (6x) Multiply both sides of the equation by the LCD.  2x 3x  6 Simplify3(11) – 2(2) = x33 – 4 = x 29 = xCheck: 11 − 2 = 1 2x 3x 611 − 2 = 12(29) 3(29) 611 − 2 = 158 87 611(87) − 2(58) = 158(87) 6 3

957 − 116 = 158(87) 6841 = 15046 61 =166Example 2. Solve x + x = 10 + x 34x + x = 10 + x The LCD is 123412  x + x  = (10 + x)12 Multiply both sides of the equation by the LCD 3 412x + 12x = 120 + 12x 344x + 3x = 120 + 12x Simplify7x = 120 + 12x7x – 12x = 120– 5x = 120x = – 24Check: x + x = 10 + x 34− 24 + − 24 = 10 + (−24) 34–8+–6 – 14 –14 – 14 4

Example 3. Solve y − y −1 = 1 23 y − y −1 = 1 The LCD IS 6 236  y − y −1 = 1(6) Multiply both sides of the equation by the LCD 2 3 6 y − 6( y −1) = 623 Be careful to put parentheses around y – 1 ; otherwise an incorrect solution maybe found.6y − 6y −6 = 6233y – 2y + 2 = 6 y=4Check: y − y −1 = 1 23 4 − 4−1 = 1 23 4−3 = 1 23 2–1 = 1Example 4. Solve x = 2 + 2 x−2 x−2 x = 2 +2 The LCD IS (x – 2) x−2 x−2 5

(x – 2)  x  =  2 + 2 (x – 2) Multiply both sides by the LCD x−2 x−2  x = 2 + 2x – 4 –x =–2 x= 2Check: x = 2 +2 x−2 x−2 2 = 2 +2 2−2 2−2 2=2 + 2 00 The proposed solution 2 cannot be a solution because 2 makes bothdenominators equal to 0. The equation is meaningless and has no solution.Example 5. Solve 2b + 1 = 2 b2 − 4 b − 2 b+2 Since b2 – 4 = (b – 2)(b + 2), use (b – 2)(b + 2) as the LCD. Multiply both sidesby (b – 2)(b + 2). (b – 2)(b + 2)  b 2b 4 + 1 = 2   2− b−2 b+2(b – 2)(b + 2)  2b 4  + (b – 2)(b + 2)  1  =  2  (b – 2)(b + 2)  b2 −  b−2 b+2 2b + (b + 2)(1) = 2(b – 2) 2b + b + 2 = 2b – 4 b = –6Checking is left for you to do. 6