MATH 2 part 2

Mathematics II PART 2

Module 4 Radical ExpressionsWhat this module is about This module is about addition and subtraction of radical expressions.Adding and subtracting radical expressions is very much similar to adding andsubtracting similar terms of polynomials. Similar terms or like terms are thosewith the same literal parts or literal factors. Radicals are similar if they have thesame index and the same radicands. For example: 3 , 5 3 , -2 3 are similarradicals. All of the given radicals have the same index, 2 and radicand, 3. Also,expressions likeWhat you are expected to learn This module is designed for you to: 1. add and subtract radical expressions 2. solve simple problems involving radicals operations.How much do you knowSimplify the following radical expressions: 1. 5 5 + 6 5 2. 5 3 + 12 3. 1 3 + 1 3 23 4. 2 2a + 4 18a 5. 7 b − 5 b 1

6. 7 5 - 5 20 7. 16 2 - 3 32 8. 10x + 3 7 y + 5 10x − 9 7 y 9. 5 20 - 8 5 + 3 45 10. 72x + 2 8x - 5 2xWhat will you do Lesson 1 Addition of Radical Expressions Radicals may be combined into simple radical expression by addingsimilar terms. If radicals are not similar, they must be converted by simplification.Examples:Find the sum of the following radical expressions: 1. 4 2 + 5 2 + 6 2 Solution: Since they all have same index, add their coefficients and annex their common radical factor = (4 + 5 + 6) 2 = 15 2 2. 6 2 + 3 5 + 2 2 + 5 5 Solution: = (6 2 + 2 2 ) + (3 5 + 6 5 ) Group similar radicals, 2

=8 2 +9 5 Add their coefficients and annex their common radical factor.You cannot add the coefficient of different radicand.3. 2 50 + 8Solution:= 2 25 • 2 + 4 • 2 Factor the radicand such that one of the= 2(5 2 ) + 2 2 factors is a perfect square.= 10 2 + 2 2= 12 24. 6 27 + 2 75 Factor each radicandSolution: Simplify = 6 9 • 3 + 2 25 • 3 = 6(3 3 ) + 2(5 3 ) Add the coefficients and annex = 18 3 + 10 3 their common radical factor = 28 35. 3 32 + 3 108 Factor each radicand such that oneSolution: factor is a perfect cube. = 3 8 • 4 + 3 27 • 4 = 2 3 4 + 33 4 = (2 + 3) 3 4 3

= 53 4 Split the numerator and denominator6. 4 3 + 3 3 Simplify and add the coefficients 25 25 Split a7 such that the exponent is exactly Solution: divisible by the index, 2. Simplify =4 3 + 3 3 25 25 =4 3 + 3 3 55 =7 3 57. 9a7 + a3 Solution: = 9a6 • a + a2 • a = 3a2 a + a a = 3a2 + a a Keep in mind that our basic approach to these problems has been to firstput each term into simplest radical form before adding and /or subtracting.Try this outPerform the indicated operations.A. 1. 4 2 + 5 2 2. 5 3 + 2 3 + 10 3 3. 16 7 + 8 7 + 7 4. 15 11 + 3 11 + 5 11 + 6 11 4

5. 1 13 + 2 13 23B. 6. 2 27 + 3 167. 180 + 125 + 10 458. 3 36a + 100a9. 75 + 2 27 + 1210. 5 b + 4bC. 11. 80 + 3 45 + 2012. 12x + 27x + 48x13. 3 + 2 3 4414. 5x2 y + 5x2 y2 415. 3 5 + 5 5 16 16D. What’s the Message?Simplify the expression and find the correct answer in the box, then fill inthe small box with letter A. Once you have filled in all the small boxes with theappropriate letter, darken all the blank spaces to separate the words in the secretmessage. What’s the message? Have fun!___ __ __1. E : √7 + √3 + 4√7 9. U: 2√12 + √48__ __ __ ____ ___2. N : √18 + 2√32 + √50 10. H: 2√28y3 + √63y3__ _ __ ___ ____3. T: 3√81 + 2√9 + 5√16 11. O: 3√5/16 + 5√5/16__ ___ _____ _____4. S :6√10 + √10 12. L: 4√320a3 + 2√180a3___ __ __ ____ ____ 5

5. W: 3√25x +2 √4x +3 √9x 13. R: √44x5 + √99x5 ___ ___ ___ __ __6. I: √100 + √144 + √25 14. C: 2√27 + 3√16 __ __ _____7. Q: 8√12 + 2√3 15. F: √64/498. Y:√2 + √12 16. V: 3√200 + 2√147 + √338 ___ __ __ _ _ _ __√1/3 √14 + 3 7√ 10 8√3 12 + 6√2 12+6√2 5√7+√3 __ __ __ _ _ _ __ __ 277√10 7√10 5 √10 11√3 +√5 √7 +√11 √12 5 __ __ _ __ _ _ __ _ 10√2 44a√5a √2 +2 √3 4√10 + √ 37√10 18√2 +√5 2√58 __ ___ _ 53 __ _ 7y√7y 2√57 2√5 5x2√11x 4√3 __ __ _ _ _ __ __7√10 7y√7y 2√5 2/3√5 5√7 + √3 6√6 + 5√2 28√x __ __ _ __ _ _ __ ___15√3 +2√5 √2 + √3 10√2 5√7+√3 30√2+ 14√3 5√7+√3 5x2√11x _ _ __ 27 53 __ _100+√5 18√3 8√3 2/3√5 15√3+ __ +1/y√5 Lesson 2 6

Subtraction of Radicals Subtraction of radicals becomes possible by the rule of integral addition.The operation just like in addition is only possible for similar radical expressions.Consider the following examples: __ _Example 1. Simplify √3 - 2√ 5 + 5√3 __ _Solution: √3 - 2√5 + 5√3Group similar radical expression. __ _= (√3 + 5√3) – 2 √5 __= ( 1 + 5)√3 - 2√5 __= 6 √3 -2 √5Note that the last step can no longer be combined, they are not similarand are already in their simplified form. __ __ __Example 2. Simplify: √54 + 2 √40 - 3√96 __ __ __Solution: √54 + 2 √40 - 3√96Write each radical in simplest form ___ ____ ____= √9.6 + 2 √ 4.10 - 3√16.6 _ __ _= 3√6 + 2.2√10 - 3.4√6Group similar radicals. _ __= (3 - 12)√6 + 4√10 _ __= -9 √6 + 4√10 ____ ____ _____Example 3. Perform: √18x2 - √8x2y + √50x2y ____ ____ _____Solution: √18x2 - √8x2y + √50x2yWrite each radical in simplest form such that one factor is perfectsquare number. Leave the other factor as radicand. ______ _____ _____= √9.2.x2. y - √4.2x2y + √25x2y 7

__ ___ __ = 3x√2y - 2x√2y + 5x√2yGroup the radicals __ = (3x – 2x + 5x)√2y __ = 6x√2y _Example 4. Simplify: 2a√3 - 3a2 9Solution: _ 2a√3 - 3a2 9 _ = 2a√3 - 3a2. 1 9 _ = 2a√3 – a.1 3 3 __ = 2a√3 – a √3 3 = (2a – a )√3 3 _= (6a – a)√3 3 _= 5a √3 3Example 5. Simplify: 2 3 16 - 5 3 54Solution: 2 3 16 - 5 3 54Factor out the radicand. = 2 3 8.2 - 5 3 27.2Split the factored radicand. = 2 3 8 . 3 2 . 5 3 27 . 3 2 8

= 2.2 3 2 - 5.3 . 3 2Notice that 8 and 27 are perfect cubes hence their cube rootis 2 and 3 respectively. Simplifying further; = 4 3 2 - 15 3 2 = (4 – 15) 3 2 = -11 3 2Example 6. Simplify 2 3 x2y - 3 8x5 y4Solution: 2 3 x2y - 3 8x5 y4Rename the expression such that the radicand can be expressedas a perfect cube and the exponents are exactly divisible by theindex if possible then simplify. 33 = 2 x2y - (8x3y3)x2y 33 = 2 x2y - 2xy x2yExample 7. Perform: 3 Solution: = (2 - 2xy) x2y ___ __ √9a7 - √a3 ___ __ √9a7 - √a3 ____ ___ = √9a6.a - √a2.a __ = 3a3 √a - a√a = (3a3 – a) √aSubtraction of radical expressions with fraction. __Example 8.Perform 2 75 - 4 √ 8_ 16 √32Solution; 2 75 - 4 √ 8 16 √32 ___ = 2 25.3 - 4 √ 4.2 16 √16.2 9

__ Simplify further = 2 5√3 - 4 2√2 4 4√2 _ = 5√3 - 2 2 Find the common denominator. _ = 5√3 - 4 22 _ = 5√4 - 4 2Example 9. Perform 10 3 5 - 3 3 4 x6 x9Solution: 10 3 5 - 3 3 4 x6 x9Express the radicand fraction as separate radical expression. = 10 3 5 - 3 3 4 3 x6 - 3 x9 = 10 3- 5 - 3 3 4 x2 x3 Find the common denominator = 10x 3 5 - 3 3 4 x3 (if x ≠ 0) _____ 3 34Example 10. Simplify √8a3b3 + ab - 8a4b4 - 4a2b2 Solution: In this problem, no two radicands are identical and theradical expressions have indices, 2,3 and 4. The radical expression: can be convert ed to index 2. 4 4a2b2 ___ 4 4a2b2 = 4 (2ab)2 = 4 (2ab)2 = √2ab ___ Note that 4 (2ab)2 is the same as (2ab)2/4 or (2ab)1/2 = √2ab 10

You can now complete the solution as below. _____ √8a3b3 + 3 ab - 3 8a4b4 - 4 4a2b2Factoring powers of radicands 33 ___ ____ = (2ab)2 2ab + ab - (2ab)3 ab - √2abSimplifying the radicands ___ ___ - 2ab 3 ab - √2ab = 2ab√2ab + 3 abCombining 3 ab ___ ___ = 2ab √2ab - √2ab + 3 ab - 2ab ___ 3 = (2ab -1) √2ab + (1-2ab) ab Try this out: A. Perform the indicated operations. __A. 1. √3 - 2 √3 __ __ 2. 5√7x + 4√7x __ __ 3. 3 √6a + 7√6a __ _ 4. 4√x - 3√x + 2√x __ _ __ 5. √12 + 5√8 - 7√20 ___ ___B. 6. √36 - √100 __ __ __ 7. 6√18 - √32 +2√50 __ __ _ 8. -2√63 + 2√28 + 2√7 _ __ __ 9. 5√8 + 3 √72 -3√50 __ __ __ 10. -√27 + 2√48 - √75____ ____ ____ 11

C. 11. 3√72m2 - 5√32m2 - 3√18m2 33 12. - 54 + 2 16 33 13. 2 27x - 2 8x 44 14. 5 32 + 3 162 44 15. 3 x5y - 2x xyD. Mental MathWhy are Oysters greedy? _ __ __ 3√5 __ _ _-2√3 -18√5 5√7 4√3+5√7 2 3√35 - 6√5 5√7 _ __ _ __ __ _-5√3 -18√5 5√7 11+ 3√2 11+ 3√2 -5√5 10√10 -5√3 -18√5Perform and Simplify the following radicals. Write the letter in the box above itscorrect answer. Keep working and you will discover the answer the questions._ _ __ __ __ __H. √5 - 18√5 + 4√5 - 5√5 E. 4√7 + √7 L. 3√16 + 3 3√54__ _ __ __ __R. √35 - 6√5 I. 7√10 + √90 A. √20 + 5__ _ _ __ 24T. 1/3√27 - 3√3 F. 3√2 - 2√32__ __ __ __ _ __S. 2√27 - 3√48 + 1/5√75 Y. 4√27 + 5√7 - 2√48Let us summarizeOUTLINE 1. Put each radical into simplest form.FOR 2. Perform any indicated operations, if possible.WORKING 3. Make sure the1f2inal answer is also in simplest radicalWITHRADICALS form.

What have you learnedA. Simplify each radical and perform the indicated operations.__ __1. 2√3 + 4√3 13. 10 49 - √50__ 22. 5√5 - √5__ _ 14. 20 √3 - 3a23. 6√3 + 2√3 - 3√3 9_ __ ___ ____4. 2√7 - 5√7 -8√7 15. √2ab + 4√4a2b2_ __5. x√2 – 3x√2 + 5√2_ __ __6. 5 √3a + 5√3a -9√3a_ __7. 3√2 - 2√32__ _ __8. √12 + 5√8 -7√20_ _ __9. 6√2 -5√8 + 2√32__ __ _10. 7√72 + 3√20 - 4√5_11. √3 + 2 1 3 _12. 6 1 + 2 √2 23B. Solve. __1. Find the midpoint of the segment joining (4√3,5) and (-8√3, -11).2. Find the slope of the line containing the points with coordinates (√8, √50) 13

__ __and ( √72, √32).3. Find the equation of the line containing the points in number 2. __ __4. Find the slope of the line containing the points with coordinates (√48, √27)__ ___and √75, √147 M5. __ \√50 Y m ∠ MXL =X|2√32 \LAnswer KeyIHow much do you know _ 1. 11√5 _ 2. 7√3 _ 3. 5/6 √3 __ _ 4. 6√10 - 6√7 _ 5. 14√2 _II 6. 2√5 _ 7. -3√5 _ 8. 6√2 _ 9. 2√5 _ 11. 2√2 14

Lesson 1Try this outA. _ 1. 9√2 _ 2. 17√3 _ 3. 25√7 __ 4. 19√11 __ 5. 7/6 √13 _B. 6. 12+6√3 _ 7. 39√5 _ 8. 28√a _ 9. 13√3 _ 10. 7√b _ 11. 15√5 __ 12. 9√3x _ 13. 3√3 2 _ 14. 4xy√5 3 _ 15. 2√5 C. What’s the Message S UCCE __ _ __ __ 7√ 10 8√3 12 + 6√2 12+6√2 5√7+√3 SS I __ __ 277√10 7√10 15

S ONL Y7√10 __ _ __ __ F 2√5 10√2 44a√5a √2 +2 √3 OR T H O8 __ ___ 53 __ _7 2√5 5x2√11x 7y√7y 2√5 S E WH O __ __ _ __ _7√10 5√7 + √3 28√x 7y√7y 2√5 NE V E R _ __ __ __ ___ 5√7+√3 5x2√11x 10√2 5√7+√3 30√2+ 14√3 QUI T _ __ 53 18√3 8√3 27The Message: SUCCESS IS ONLY FOR THOSE WHO NEVER QUITLesson 2.Try this outA. _1. -√3 __2. 9√7x __3. 10√6a _4. 3√x __ _5. 10√3 + 2√3 + 14√5B. 6. -4 _ 7. 24√2 16

8. 0 _ 9. 13√2 10. 0C. _ 11. -23√2 _ 12. 7 3√2 13. 0 _ 14. 19 4√2 __ 15. x 4√xy D. Mental HealthE. Why are Oysters greedy?THE Y ARE _ __ __ 3√5 __ _ _-2√3 -18√5 5√7 4√3+5√7 2 3√35 - 6√5 5√7SHE L L F I SH _ __ _ __ __ _-5√3 -18√5 5√7 11+ 3√2 11+ 3√2 -5√5 10√10 -5√3 -18√5What have you learned _ _ 14. 20 -a√3 or 60 -a√3A. _ 1. 6√3 3 3 _ 2. 4√5 ___ _ 15. 2√2ab 3. 5√3 17

_ B. _4. -11√7 1. (-2√3, -5) 2 _5. -2x + 5√2 2. -1/4 __ _6. √3a 3. y = -1 x+ 11√2 _7. -5√2 42 _ _ __ or 4y = -x + 22√28. 2√3 + 10√2 - 14√5 4. slope=4 _ _9. 4√2 5. 13√2 __10. 42√2 + 2√5 _11. 5√3 3_12. 11√2 3 _13. 30√2 18

Module 4 System of Linear Equations and Inequalities What this module is about This module is about solving systems of linear inequalities. As you go onwith the lesson, you will develop skills in solving systems of linear inequalities bygraphing. Learning how to graph linear inequalities will give you the idea that aregion contains many solutions to linear inequalities. What are you expected to learn This module is designed for you to:1. define a system of linear inequalities2. draw the graph of linear inequality in two variables.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 know Examine the inequalities on the left and the expressions describing thegraphs of the inequalities on the right. Match each inequality with one or more ofthe description.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 line

6. x ≥ -8 F. left of the line7. y < x + 88. y ≥ -2x + 1What you will do Lesson 1 Draw 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 not both equal to 0. The graph of a linearinequality is a 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 < 2-2y < -3x + 2 Division by a negative number y > 3 x -1 reverses the order of the inequality. 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. a) (2,1), a point above the dashed lineSubstitute, to the inequality: 3x – 2y < 2 3(-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 contained 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. Check. Choose points from both sides of the line and substitute to the inequality. 3

Test points: a) (3,3) a point on one side x+y>3 3+3?3 6?3 6 > 3, true b) (-2,3) x+y>3 -2 + 3 ? 3 1?3 1> 3, false 4. Shade the region where the point satisfies the inequality. 5. All the points in the shaded region are the solutions to the given inequality.The graph: x+y>3Example 3: Graph: 2x – 3y ≤ 01. Solve for y:-3y ≤ - 2x dividing by negative number reversesy ≥ 2x the inequality 32. 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 4

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) 2x – 3y ≤ 0 2(-1) – 3(4) ? 0 -2 – 12 ? 0 -14 ? 0 -14 ≤ 0, True or False? b. (1,-2) 2x – 3y ≤ 0 2(1) – 3(-2) ? 0 2+6?0 8?0 8 ≤ 0, true or false. 4. 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.Example 4: Draw the graph of y ≤ -2. Follow the steps: 1. Graph the equation: y = -2 5

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 the 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, false b) (3,5) x≥2 3 ≥ 2, true. 4. Shade the region to the left of the line x = 2. 6

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 the table for graphing InequalitiesInequalities Descriptiony < mx + b The region or half-plane is below the line y = mx + by ≤ mx + b The region is below y = mx + b and the boundary line, y = mx +y > mx + b b.y ≥ mx + b The region is above the line y = mx + b The region is above the line, y = mx + b and the boundary line, y = mx + bTry this outA. Use the steps learned in graphing inequalities. Use a test point to shade.1. x + y ≥ 22. x – y ≤ 13. 3x – y ≥ 64. 3x + y ≤ 35. 2x + 3y > 12B. Graph each linear inequality with the use of a test point.1. y > 2x – 12. y ≤ 3x + 23. y ≥ 2/3 x – 14. y < -3/4 x + 25. –x – 2y ≥ 4 7

Lesson 2 Define a System of Linear Inequalities Now, that you have been exposed to graphing linear inequalities in twovariables, this time you are now ready to do graphing systems of linearinequalities. The following are examples of systems of linear inequalities in twovariables.1. x – y ≥ 2 4. 4x – 2y ≤ 8 x+y≤6 y–2≤0 x+y≥02. x + y < 4 x<23. x + y ≤ 3 5. x + y < 4 3x + 2y ≤ 6 x<2 x≥0 y≥0 What do you notice about the number of inequalities in a system? Noticethat a system of linear inequalities is composed of 2 or more inequalities. Anypoint contained in the double or more shaded region must satisfy each inequalityin a system.Examples: Which of the indicated points is contained in the shaded region of thefollowing inequalities?1. x – y < 2 Points: (-1,2), (10,5), (2,-4) x+y <6 Substitute each point in each inequality. If the point both satisfies theinequality, then it is contained in the double shaded region.Using: (-1,2) x–y>2 (-1) – 2 < 2 -3 < 2, true 8

x+y >6 (1) + (-2) < 6 -1 < 6, trueUsing: (10,5) x–y>2 (10) – (5) < 2 5 < 2. false x+ y<6 (10) + (5) < 6 15 < 6, falseUsing: (2,-4) x–y<2 (2) – (-4) < 2 2+4<2 6 < 2, false x+y <6 (2) + (-4) < 6 -2 < 6, true Therefore, (-1, 2) is contained in the double shaded region, since it bothsatisfies the two inequalities.2. x – y > 2 Points: (-1,3), (0,0), (2, -4) x+y <6Substitute the given points in the inequalities:Using: (-1,3) x–y>2 (-1) – (3) > 2 -4 > 2, false x+y <6 (-1) + (3) < 6 2 < 6, trueUsing: (0,0) x–y>2 (0) – (0) > 2 0 > 2, false x+y <6 (0) + (0) < 6 0 < 6, true 9

Using: (2, -4) x–y>2 (2) – (-4) > 2 6 > 2, true x+y <6 (2) + (-4) < 6 -2 < 6, trueTherefore, (2, -4) is contained in the double shaded region.3. 4x – 2y < 8 Points: (0,0), (1,1), (4,-1) x – 2y < 0 x+y>0Using: (0,0) 4x – 2y < 8 4(0) – 2(0) < 8 0 < 8, true x – 2y < 0 0 – 2(0) < 0 0 < 0, false x+y>0 0+0>0 0 < 0, falseUsing: (1,1) 4x – 2y < 8 4(1) – 2(1) < 8 4-2 < 8 2 < 8, true x – 2y < 0 1 – 2(1) < 0 -1 < 0, true x+y>0 1+1>0 2 > 0, trueUsing: (4, -1) 4x – 2y < 8 4(4) – 2(-1) < 8 16 + 2 < 8 18 < 8, false 10

x – 2y < 0 4 – 2(-1) < 0 4+2<0 6 < 0, false x+y>0 1+1>0 2 < 0, false Point (1,1) is contained in the tripled shaded region, since there are threeinequalities involved. If in testing of points, it appears false in the first testing, do not continueanymore. One false conclusion is enough to stop the substitution. It is just donefor you for the sake of complete illustration.Try this out Match Column A with Column B. Determine the point in Column B whichsatisfies the inequalities in Column A.Column A Column BInequalities Points1. y + 2 > 0 a. (-3, -2) y–x>32. y > 2x + 3 ` b. (1,5) y < 5x + 63. y – 2x < 1 c. (5,4) y + 2x > 1 x<24. x + y < 3 d. (3,6) x<y+45. x < y + 2 e. (0,0) x + 2y > 8 y <5How many did you get right? 11

Lesson 3Represent the Solution Set of a System of Linear Inequalities by Graphing. Since you have learned how to graph linear inequalities in two variables,you are now ready to graph a system of linear inequalities. These will involvefinding the region where the solution of the system lies.Example 1: Graph the solution of the systemx + y > -1x – 2y > 4Steps: Change each equation in the form y = mx + b(1) x + y > -1 y > -x – 1(2) 3x – 2y > 4 Multiplying inequality by negative -2y > -3x + 4 number reverses the inequality 2y < 3x – 4 y <3x–2 2Graph the inequality (1) x + y > -1a. First graph y = -x – 1b. Shade above the line to represent y > -x - 1 x + y > -1 12

Graph the inequality (2) 3x – 2y > 4 a. First graph y < 3 x – 2 2 c. 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 the points to eachinequality. Point (1, -4), contained in a single shaded region. 13

Substitute in (1): x + y > -1 (1) + (-4) > -1 -3 > -1, FalseSubstitute in (2): 3x – 2y > 4 3(1) –2(-4) > 4 3+8>4 11 > 4, truePoint (4,1), contained in the double shaded region.Substitute in (1): x + y > -1 4 + 1 > -1 5 > -1, trueSubstitute in (2): 3x – 2y > 4 3(4) – 2(1) > 4 12 – 2 > 4 10 > 4, true The true solution for the linear inequalities is the point that satisfies bothlinear inequalities. 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 ordered pairs found in the doubleshaded region.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 14

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 2(7) + 2 > 2 14 + 2 > 2 16 > 2, true 15

(2) x – y ≥ 3 7–2≥3 5 ≥ 3, true Point (-7, 2), located in a single shaded region Substitute: (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 = -2 Notice, that the graph of x ≤ -2 is a vertical line. 16

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) 3x + 2y < 6 (2) x≥0 (3) y≥0 (4)Steps: Using the skills gained,1. graph the lines x + y = 3, use solid line2. graph 3x + 2y < 6, using dashed line, 17

3. graph x = 0, using solid line, or the y-axis4. graph y = 0, using solid line, or the x-axis. 3x + 2y = 6 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 3x + 2y < 6 is satisfied by the set of points in the regionbelow the line 3x + 2y = 6 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:(1) 4x – 2y ≤ 8 a. Set x to 0 in the equation 4x – 2y = 8 to find the y-intercept. 4(0) –2y = 8 -2y = 8 y=4 b. Set y to 0 using the same equation to find the x intercept. 4x – 2(0) = 8 4x = 8 x=2 18

(2) y – 2 ≤ 0, c. Solving for y, the y-intercept in this inequality is y = 2 d. 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 = -½ y-intercept = 1 (2) y + 2x ≥ 1 x-intercept = ½ y-intercept = 1 (3) x ≤ 2 x-intercept = 2 y-intercept = 0 2. Graph the system. Connect the x and y intercept of each inequality and shade the region of each inequality. 19

The solution to the system of inequalities is the intersection of all theshades 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 ≥ 2/5x + 6 2x – y ≤ 1 6. x ≥ 02. y < x + 3 4. x + y > 6 x–y≥0 y < 5 – 2x 2x – y < 4 x+y≤4 Lesson 4Translate 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 20

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 in 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 pants?Translation: 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 is at least P10 000. let x represents his sales.a. x < 10 000 b. x > 10 000 c. x ≤ 10 000 d. x ≥ 10 0002. Mother told the son, save at most P5 from your weekly allowance for Christmas.a. x < 5 b. x > 5 c. x ≤ 5 d. x ≥ 5 21

3. 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 ≥ 324. 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: Problem: Graph the linear inequality, x + 2 < ySolution: x+y<yLook here, my children, and you shall see y=x+2how to graph an inequality. x = 10Here’s a simple inequality to try; y = 10 + 2 (10.12)x + 2 is less than y, FigureFirst, Make the “less than “ “equal to”; (3,5)So now y equals x plus 2.Then pick a point for x; say, 10; FigureNow plug that single constant inadd 10 plus 2 and you’ll get y; x+2<ySee if this pair will satisfyx:10, y:12; you’ll find its right, (3,1)So graph this point to expedite. 3+2<1Now find a second ordered pair (1,4)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 gotto 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,”just as it was when you began.Pick a point on one side!Use 3 for x and 1 for y.Is 1 greater than 3 plus 2?No! This side will never do!On the other side, let’s try1 for x and 4 for y. 22

1 plus 2 (which equals 3) 1+2=3<4 Figureis less than 4, as you can see.Shade in the side that dot is on; (1,3)We’ve got one more step to come upon. 1+2=3Do the points upon your linefor the equation I assigned?Use 1 and 3 for this test;They’re “equal to” but they’re “lessMake your line dotted to show this is trueand that is all you have to do! Let ‘s summarize To show the graph of the solution set of a system of linearinequalities, follow these steps. 1. Graph first the equation. Use any of the methods presented. 2. Test the 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 solution is the region where the shaded portion intersects. 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 > 2y<0 y<3 23

3. x > -1 4. x < -2 y>1 y > -2B. Determine whether the ordered pair is a solution of the system: x+y≤6 x–y<1 5. (1,3) 5. (1,2) 6. (0,0) 7. (6,0) 8. (-2,0) 9. (7,1) 24

Answer KeyHow much do you know 2.1. A, F2. B, C3. A, D, E4. B, C,I5. S, D6. A, E7. B, D, E8. A, C, ETry this outLesson 11.x+y>23. 4. x–y<1 3x – y > 6 3x + y < 3 25

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 26

Lesson 2: 2.1. b 2. e 3. d 4. a 5.cLesson 3 y=x+31. y = 5 – 2x y=4 x=13. 4. y = 5x - 1 x+y=6 y = 2/5x + 6 2x – y = 45. 6. x + y = 4 x–y=0 2x – y = 1 x+y=2 27

Lesson 4A. 1. d 2. c 3. c 4. b 5. d y • (10, 12)B. Figure (1,4) • (3,5) • (3,1)x•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 > 2y<0 y<3 3. 28

3. x ≥ -1 4. x < -2 y≥1 y > -2B. Determine whether the ordered pair is a solution of the system: x+y≤6 x–y<1 5. (1,3) yes 6. (1,2) yes 7. (,0,0) yes 8. (6,0) no 9. (-2,0) yes 10. (7,1) no 29

Module 4 Radical Expressions What this module is about This module is about addition and subtraction of radical expressions.Adding and subtracting radical expressions is very much similar to adding andsubtracting similar terms of polynomials. Similar terms or like terms are thosewith the same literal parts or literal factors. Radicals are similar if they have thesame index and the same radicands. For example: 3x and 5 3x are similarradicals whose index is 2 and whose radicand is 3x. What you are expected to learn This module is designed for you to: 1. add and subtract radical expressions 2. solve simple problems involving radical operations. How much do you knowSimplify and perform the necessary operations:1. 5 5 + 6 52. 5 3x + 12x3. 1 3 + 1 3 234. 2 2a + 4 18a5. 7 b − 5 b6. 7 5 - 5 20

7. 16 2 - 3 328. 10x + 3 7 y + 5 10x − 9 7 y9. 5 20 - 8 5 + 3 4510. 72x + 2 8x - 5 2xWhat will you do Lesson 1 Addition of Radical Expressions Radicals may be combined into simple radical expression by addingsimilar terms. The sum of two radicals cannot be simplified if the radicals havedifferent indices or different radicands.3+ 33 cannot be simplified (indices are different5+ 2 cannot be simplified (radicands are different It is important that the radicals are written in simplest form before addingand subtracting.3 + 27 can further be simplified. 27 = 3 3= 3 +3 3 add =4 3Examples:Find the sum of the following radical expressions:1. 4 + 5 2 + 6 2 2