Math Refresher, Unit 3: Systems of Equations, Inequalities

RPAD Welcome Week August 18-24, 2012 Math Refresher, Unit 3: Systems of Equations, Inequalities Junesoo Lee PhD Candidate of Public Administration Instructor for Statistics (PUB316 and PAD505) ing1224@hotmail.com 1

Agenda Overview Chapter 25: Functions Chapter 26: Solving Systems of Equations Chapter 28: Solving Inequalities First Degree Book chapters from Forgotten Algebra (Barron s) On order at Mary Jane Books (see ad in Welcome Week booklet) 2

Overview Why math in MPA? o Policy and Management are basically numerical data driven o Must in RPAD 501 (budgets), 503 (economics), 504 (data), and 505 (statistics) You already know math well, but only forgot it o The only thing we should blame OBLIVION on is TIME Let s refresh what you must have known a million years before 3

Functions Terminology Takes a number and adds 3 X f (or g, or h as name of function) Domain Rule Input Independent variable X + 3 Y Range Output Dependent variable 𝑦 = 𝑓 𝑥 = 𝑥 + 3 𝑤𝑖𝑡ℎ 𝐷 = {5, 7, 8, 9} "𝑦 𝑒𝑞𝑢𝑎𝑙𝑠 𝑓 𝑜𝑓 𝑥 𝑒𝑞𝑢𝑎𝑙𝑠 𝑥 𝑝𝑙𝑢𝑠 3, 𝑤𝑖𝑡ℎ 𝑑𝑜𝑚𝑎𝑖𝑛 𝐷 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 5, 7, 8, 𝑎𝑛𝑑 9. " A function is a rule that assigns to each element in the domain one and only one element in the range. (Unless specified, the domain of a function is the set of all real numbers.) 5

Functions Examples Given this function, find each of the following: f x = x + 3 (i. e., y = x + 3) o f(7) "f of 7" o f(9) "f of 9" = 7 + 3 = 10 = 9 + 3 = 12 o f 1 (11) "f inverse of 11" 11 = x + 3 x = 11 3 = 8 "if the y value is 11, what is x? " 6

Functions Exercises Given this function, find each of the following: g x = 1 + x 2 (i. e., y = 1 + x 2 ) o g(3) = 1 + 3 2 = 1 + 9 = 10 o g( 2) = 1 + 2 2 = 1 + 4 = 5 o g 1 (10) 10 = 1 + x 2 x 2 = 9 "if the y value is 10, what is x? " x 2 = 9 x = ±3 7

Solving Systems of Equations An equation that has the form 𝑎𝑥 + 𝑏𝑦 = 𝑐, with 𝑎, 𝑏, and 𝑐 being real numbers, 𝑎 and 𝑏 not both zero, is a linear equation in two variables. The solutions to a system of equations are the pairs of values of 𝑥 and 𝑦 that satisfy all the equations in the system. Example Solve (i.e., find solutions): 𝑥 + 𝑦 = 8 o 𝑥 = 1 𝑎𝑛𝑑 𝑦 = 7. 𝑂𝑟 𝑠𝑖𝑚𝑝𝑙𝑦 1, 7 o 𝑥 = 2.5 𝑎𝑛𝑑 𝑦 = 5.5. 𝑂𝑟 2.5, 5.5 o 𝑥 = 3 𝑎𝑛𝑑 𝑦 = 11. 𝑂𝑟 ( 3, 11) o We can find infinitely many solutions to this equation. 9

Solving Systems of Equations A system of equations means that there is more than one equation related to one another. Example Solve: 2𝑥 + 𝑦 = 24 𝑥 𝑦 =6 o The solution to this system is 𝑥 = 10 𝑎𝑛𝑑 𝑦 = 4, or simply (10, 4). o Prove it by plugging these values in the system. 2(10) + 4 = 24 10 4 = 6 How can we find this solution? There are numerous ways to do it. But we will cover only two methods Elimination by addition (or by substitution.) 10

Solving Systems of Equations Example Solve: 3𝑥 + 2𝑦 = 12 𝑦 = 2𝑥 1 1. Write them in standard form. 2. Multiply (the second equation by -2 so that the y-coefficients are the negatives of one another.) 3𝑥 + 2𝑦 = 12 2𝑥 + 𝑦 = 1 3𝑥 + 2𝑦 = 12 2 2𝑥 + 𝑦 = 2( 1) 3. 4. Add. Solve. 3𝑥 + 2𝑦 = 12 4𝑥 2𝑦 = 2 7𝑥 𝑥 = 14 =2 Five steps of elimination by addition: 1. Write the equations in standard form like 𝑎𝑥 + 𝑏𝑦 = 𝑐. 2. Multiply (if necessary) the equations by constants so that the coefficients of the 𝑥 or the 𝑦 variable are the negatives of one another. 3. Add the equations from step 1. 4. Solve the equations from step 2. 5. Substitute the answer from step 3 back into one of the original equations, and solve for the second variable. 5. Substitute back into an original equation. 𝐼𝑓 𝑥 = 2 𝑎𝑛𝑑 𝑦 = 2𝑥 1, 𝑡ℎ𝑒𝑛 𝑦 = 2 2 1, 𝑦 = 3. Solution to system is 𝒙 = 𝟐 𝒂𝒏𝒅 𝒚 = 𝟑. 11

Solving Systems of Equations Example (continued) 3𝑥 + 2𝑦 = 12 System: 𝑦 = 2𝑥 1 3𝑥 + 2𝑦 = 12 𝑦 = 2𝑥 1 (0, 6) From an algebraic point of (2, 3) view, 𝑥 = 2 𝑎𝑛𝑑 𝑦 = 3 is the 1 solution to this system. (2, 0) From a geometric point of (0, -1) (4, 0) view, (2, 3) is the point of intersection for two lines whose equations are given above. How to draw each line easily? Find x-intercept and y-intercept by plugging zero in x or y. And connect those intercepts. 12

Solving Systems of Equations Exercise Solve: 3𝑥 𝑦 = 7 5𝑦 + 5 = 5𝑥 1. Write them in standard form. 3𝑥 𝑦 = 7 5𝑥 + 5𝑦 = 5 2. Multiply (the second equation by 5 so that the y-coefficients are the negatives of one another.) 1 3𝑥 𝑦 = 7 5𝑥 5𝑦 5 + = 5 5 5 3. 4. 3𝑥 𝑦 = 7 𝑥 + 𝑦 = 1 Solve. 4𝑥 = 8 𝑥 = 2 Add. Five steps of elimination by addition: 1. Write the equations in standard form like 𝑎𝑥 + 𝑏𝑦 = 𝑐. 2. Multiply (if necessary) the equations by constants so that the coefficients of the 𝑥 or the 𝑦 variable are the negatives of one another. 3. Add the equations from step 1. 4. Solve the equations from step 2. 5. Substitute the answer from step 3 back into one of the original equations, and solve for the second variable. 5. Substitute back into an original equation. 𝐼𝑓 𝑥 = 2 𝑎𝑛𝑑 3𝑥 𝑦 = 7, 𝑡ℎ𝑒𝑛 3 2 𝑦 = 7, 𝑦 = 1. Solution to system is 𝒙 = 𝟐 𝒂𝒏𝒅 𝒚 = 𝟏. 13

Solving Systems of Equations Exercise (continued) 3𝑥 𝑦 = 7 System: 5𝑦 + 5 = 5𝑥 5𝑦 + 5 = 5𝑥 3𝑥 𝑦 = 7 (0, 7) Given the algebraic solution to system (𝑥 = 2 𝑎𝑛𝑑 𝑦 = 1), show and verify that the solution point (-2, 1) also (-2, 1) 7 ( 3, 0) (-1, 0) (0, -1) makes sense from a geometric point of view. How to draw each line easily? Find x-intercept and y-intercept by plugging14 zero in x or y. And connect those intercepts.

Solving Systems of Equations Example Solve: 1. 𝑦 + 8 = 2𝑥 3𝑥 + 2𝑦 = 12 Find (or make) any variable having a coefficient of 1, and isolate it. 𝑦 = 2𝑥 8 3𝑥 + 2𝑦 = 12 2. Use the isolated variable with a coefficient of 1 to replace that in the other equation. Three steps of elimination by substitution: 1. Find any variable with a coefficient of 1, or make any variable so, and isolate it in one equation like 𝑦 = 𝑎𝑥 + 𝑏, 𝑜𝑟 𝑥 = 𝑎𝑦 + 𝑏. 2. Use the equation having the variable with a coefficient of 1 to replace that variable in the other equation. 3. Finish the problem as before by substituting back into an original equation. 3𝑥 + 2(2𝑥 8) = 12 3. Finish the problem. 3𝑥 + 4𝑥 16 = 12 7𝑥 = 28 𝑥 =4 𝐼𝑓 𝑥 = 4 𝑎𝑛𝑑 𝑦 + 8 = 2(4), 𝑡ℎ𝑒𝑛 𝑦 = 2 4 8, 𝑦 = 0. Solution to system is 𝒙 = 𝟒 𝒂𝒏𝒅 𝒚 = 𝟎. 15

Solving Systems of Equations Exercise Solve: 1. 𝑦 1 = 3𝑥 3𝑥 + 4𝑦 = 26 Find (or make) any variable having a coefficient of 1, and isolate it. 𝑦 = 3𝑥 + 1 3𝑥 + 4𝑦 = 26 2. Use the isolated variable with a coefficient of 1 to replace that in the other equation. Three steps of elimination by substitution: 1. Find any variable with a coefficient of 1, or make any variable so, and isolate it in one equation like 𝑦 = 𝑎𝑥 + 𝑏, 𝑜𝑟 𝑥 = 𝑎𝑦 + 𝑏. 2. Use the equation having the variable with a coefficient of 1 to replace that variable in the other equation. 3. Finish the problem as before by substituting back into an original equation. 3𝑥 + 4 3𝑥 + 1 = 26 3. Finish the problem. 𝐼𝑓 𝑥 = 2 𝑎𝑛𝑑 𝑦 1 = 3( 2), 3𝑥 + 12𝑥 + 4 = 26 𝑡ℎ𝑒𝑛 𝑦 = 6 + 1, 𝑦 = 5. 15𝑥 = 30 𝑥 = 2 Solution to system is 𝒙 = 𝟐 𝒂𝒏𝒅 𝒚 = 𝟓. 16

Solving Systems of Equations Example Solve by using elimination by addition: 1. 3𝑥 + 4𝑦 = 2 4𝑦 = 8 3𝑥 Write them in standard form. 3𝑥 + 4𝑦 = 2 3𝑥 + 4𝑦 = 8 2. Multiply (the second equation by -1 so that the y-coefficients are the negatives of one another.) 3𝑥 + 4𝑦 = 2 1 3𝑥 + 4𝑦 = 1(8) 3. Add and solve. 3𝑥 + 4𝑦 = 2 3𝑥 4𝑦 = 8 0 = 6 False statement. There is no solution to the system. The equations are said to be inconsistent when there is no solution to the system. Geometrically, two lines are parallel so there is no point of intersection for the two lines. 17

Solving Systems of Equations Example (continued) System: From an algebraic point of view, there is no solution to this system. 3x + 4y = 2 4y = 8 3x 4y = 8 3x 3x + 4y = 2 (0, 2) (0, 1 2 ) ( 2 3, 0) ( 8 3, 0) From a geometric point of view, two lines are parallel so there is no point of intersection for the two lines. 18

Solving Systems of Equations Example Solve by using elimination by addition: 3𝑥 𝑦 = 5 6𝑥 2𝑦 10 = 0 1. Write them in standard form. 2. Multiply (the first equation by -2 so that the y-coefficients are the negatives of one another.) 2 3𝑥 𝑦 = 2(5) 3𝑥 𝑦 = 5 6𝑥 2𝑦 = 10 6𝑥 2𝑦 = 10 3. Add and solve. 6𝑥 + 2𝑦 = 10 6𝑥 2𝑦 = 10 0=0 True statement. All values of x and y satisfy this equation. The equations are said to be dependent when there are infinitely many solutions to the system. Geometrically, two lines coincide so there are infinitely many points of intersection for the two lines. 19

Solving Systems of Equations Example (continued) System: 3x y = 5 6x 2y 10 = 0 From an algebraic point of view, there are infinitely many solutions to the system. 3x y = 5 6x 2y 10 = 0 ( 5 3, 0) From a geometric point of view, two lines coincide so all points of the two lines intersect. (0, -5) 20

Solving Systems of Equations Exercise Solve, and figure out if two equations are inconsistent (i.e., no solution) or dependent (i.e., infinite solutions). 5𝑥 𝑦 = 4 2𝑦 = 10𝑥 6 1. Write them in standard form. 2. Multiply (the first equation by -2 so that the y-coefficients are the negatives of one another.) 2 5𝑥 𝑦 = 2(4) 5𝑥 𝑦 = 4 10𝑥 2𝑦 = 6 10𝑥 2𝑦 = 6 3. Add and solve. 10𝑥 + 2𝑦 = 8 10𝑥 2𝑦 = 6 0 = 2 False statement. There is no solution to the system. 21

Solving Systems of Equations Exercise (continued) System: Given the algebraic solution to system (no solution), show and verify that this result also makes sense from a geometric point of view. 5x y = 4 2y = 10x 6 ( 3, 0) ( 4, 0) 5 5 (0, -3) (0, -4) 5x y = 4 2y = 10x 6 22

Solving Inequalities First Degree Examples for inequality signs 2 < 3 is read 2 is less than 3. 5 > 1 is read 5 is greater than 1. a 4 is read a is less than or equal to 4. b 7 is read b is greater than or equal to 7. Both expressions 2 < 3 and 3 > 2 have the same meaning. But 2 < 3 is a better way because it clearly visualizes the direction of difference like the number line. Number line -2 < 3 24

Solving Inequalities First Degree Examples To solve a first-degree inequality, find the values of 𝑥 that satisfy the inequality. The basic strategy is the same as that used to solve first-degree equations. Solve: 𝑥 + 5 < 7 then 𝑥 < 7 5 Rule 1: A term may be transposed from one side of the inequality to the other by changing its sign as it crosses the inequality sign. and 𝑥 < 2. Solve: 1 𝑥 2 Graphically represent the solutions then 1 + 2 𝑥 and 3 𝑥. The heavy line indicates that all numbers to the left of 2 (or to the right of 3) are part of the answer. The open circle indicates that 2 is not part of the answer. The closed circle indicates that 3 is a part of 25 the answer.

Solving Inequalities First Degree Examples 6 < 15, divided by 3 6 then 3 < 15 3 Rule 2: Reverse the direction of an inequality symbol whenever an inequality is multiplied or divided by the same negative number. then and 2 < 5. 1 4 < 12, multiplied by 4 then 4 1 4 10 < 15, divided by -5 < 4 12 and 1 < 48. 10 5 > 15 5 and 2 > 3. (Or 3 < 2). 𝑥 2 8, multiplied by -2 then 2 𝑥 2 2(8) and 𝑥 16. (Or 16 𝑥). 26

Solving Inequalities First Degree Example Solve: 4𝑥 3 > 6𝑥 + 1 o 4𝑥 3 > 6𝑥 + 1 4𝑥 6𝑥 > 1 + 3 2𝑥 > 4 Rule 1: A term may be transposed from one side of the inequality to the other by changing its sign as it crosses the inequality sign. Rule 2: Reverse the direction of an inequality symbol whenever an inequality is multiplied or divided by the same negative number. Graphically represent the solution 2𝑥 4 < 2 2 𝑥 < 2 27

Solving Inequalities First Degree Exercise Solve: 4 𝑥 3 8𝑥 4 o 4𝑥 12 8𝑥 4 4𝑥 8𝑥 4 + 12 4𝑥 8 Rule 1: A term may be transposed from one side of the inequality to the other by changing its sign as it crosses the inequality sign. Rule 2: Reverse the direction of an inequality symbol whenever an inequality is multiplied or divided by the same negative number. Graphically represent the solution 4𝑥 8 4 4 𝑥 2 28

We are ready to start MPA! 29