Sstems of Linear Equations SUGGESTED LEARNING STRATEGIES: Shared Reading, Close Reading, Interactive Word Wall Have ou ever noticed that when an item is popular and man people want to bu it, the price goes up? Have ou ever noticed that items that no one wants are marked down to a lower price? The change in an item s price and the quantit available to bu are the basis of the concept of suppl and demand in economics. Demand refers to the quantit that people are willing to bu at a particular price. Suppl refers to the quantit that the manufacturer is willing to produce at a particular price. The final price that the customer sees is a result of both suppl and demand. Suppose that during a si-month time period, the suppl and demand for gasoline has been tracked and approimated b these functions, where Q represents millions of barrels of gasoline and P represents price per gallon in dollars. Demand function: P = -0.7Q + 9.7 Suppl function: P = 1.5Q - 10.4 To find the best balance between market price and quantit of gasoline supplied, find a solution of a sstem of two linear equations. The demand and suppl functions for gasoline are graphed below. 12 10 p ACTIVITY 1.2 MATH TERMS A point, or set of points, is a solution of a sstem of equations in two variables when the coordinates of the points make both equations true. Price (dollars) 8 6 4 2 5 10 Gasoline (millions of barrels) 1. Find an approimation of the coordinates of the intersection of the suppl and demand functions. Eplain what the point represents. Q Unit 1 Linear Sstems and Matrices 13
Sstems of Linear Equations SUGGESTED LEARNING STRATEGIES: Create Representations, Vocabular Organizer 2. What problem(s) can arise when solving a sstem of equations b graphing? TECHNOLOGY You can use a graphing calculator and its TRACE function to solve sstems of equations in two variables. 3. Graph each sstem. Determine the number of solutions. a. { =+1 = + 4 6 4 2 5 5 2 b. { = 5 + 2 = 2 6 MATH TERMS 4 Sstems of linear equations are classified b the number of solutions. 2 Sstems with no solution are inconsistent. A sstem with eactl one solution is independent. 2 c. { 2==22 ++ 14 6 4 A sstem with infinite solutions is dependent. 2 5 5 2 4. Graphing two linear equations illustrates the relationships of the lines. Classif the sstems in Item 3 as consistent and independent, consistent and dependent, or inconsistent. 14 SpringBoard Mathematics with Meaning Algebra 2 Sstems with one or man solutions are consistent. 5 5
Sstems of Linear Equations SUGGESTED LEARNING STRATEGIES: Note Taking, Look for a Pattern, Think/Pair/Share Investors tr to control the level of risk in their portfolios b diversifing their investments. You can solve some investment problems b writing and solving sstems of equations. One algebraic method for solving a sstem of linear equations is called substitution. EXAMPLE 1 MATH TERMS During one ear, Sara invested $5000 into two separate funds, one earning 2% and another earning 5% annual interest. The interest Sara earned was $205. How much mone did she invest in each fund? In the substitution method, ou solve one equation for one variable in terms of another. Then substitute that epression into the other equation to form a new equation with onl one variable. Solve that equation. Substitute the solution into one of the two original equations to find the other variable. Step 1: Let = mone in the first fund and = mone in the second fund. Write one equation to represent the amount of mone invested. Write another equation to represent the interest earned. + = 5000 0.02 + 0.05 = 205 Step 2: The mone invested is $5000. The interest earned is $205. Use substitution to solve this sstem. + = 5000 Solve the first equation for. = 5000-0.02 + 0.05(5000 - ) = 205 Substitute for in the second equation. 0.02 + 250-0.05 = 205 Solve for. -0.03 = -45 = 1500 Step 3: Substitute the value of into one of the original equations to find. + = 5000 1500 + = 5000 = 3500 Substitute 1500 for. Check our answer b substituting the solution (1500, 3500) into the second original equation. 0.02 + 0.05 = 205 Solution: Sara invested $1500 in the first fund and $3500 in the second fund. TRY THESE A Write our answers on notebook paper. Show our work. Solve each sstem of equations, using substitution. { = 25 3 a. 4 + 5 = 9 b. { 2+=2 =- 1410 c. { 3 +==416 5. When using substitution, how do ou decide which variable to isolate and which equation to solve? Eplain. Unit 1 Linear Sstems and Matrices 15
Sstems of Linear Equations SUGGESTED LEARNING STRATEGIES: Note Taking Another algebraic method for solving sstems of linear equations is the elimination method. MATH TERMS The elimination method is also called the addition-elimination or the linear combination method for solving a sstem of linear equations. MATH TERMS In the elimination method, ou eliminate one variable. Multipl each equation b a number so that the terms for one variable combine to 0 when the equations are added. Then use substitution with that value of the variable to find the value of the other variable. The ordered pair is the solution of the sstem. EXAMPLE 2 A stack of 20 coins contains onl nickels and quarters and has a total value of $4. How man of each coin are in the stack? Step 1: Step 2: Step 3: Step 4: Let n = number of nickels and q = number of quarters. Write one equation to represent the number of coins in the stack. Write another equation to represent the total value. n + q = 20 The number of coins is 20. 5n + 25q = 400 The total value is 400 cents. To solve this sstem of equations, first eliminate the n variable. -5(n + q) = -5(20) Multipl the first equation b -5. 5n + 25q = 400-5n - 5q = -100 5n + 25q = 400 Add the two equations to eliminate n. 20q = 300 Solve for q. q = 15 Find the value of the eliminated variable n b using the original first equation. n + q = 20 n + 15 = 20 Substitute 15 for q. n = 5 Check our answers b substituting into the original second equation. 5n + 25q = 400 5(5) + 25(15)? 400 Substitute 5 for n and 15 for q. 25 + 375? 400 400 = 400 Check. Solution: There are 5 nickels and 15 quarters in the stack of coins. 16 SpringBoard Mathematics with Meaning Algebra 2
Sstems of Linear Equations SUGGESTED LEARNING STRATEGIES: Close Reading, Vocabular Organizer TRY THESE B Solve each sstem of equations, using elimination. Write our answers in the space. Show our work. 2 3 = 5 a. { 5 + 3 = 40 b. { 5 + 6 = 14 2 = 10 c. { 3 + 3 = 21 5 = 17 Sometimes a situation has more than two pieces of information. For these more comple problems, ou ma need to solve equations that contain three variables. In Bisbee, Arizona, an old mining town, ou can bu souvenir nuggets of gold, silver, and bronze. For $20, ou can bu an of these mitures of nuggets: 14 gold, 20 silver, 24 bronze; or, 20 gold, 15 silver, 19 bronze; or, 30 gold, 5 silver, 13 bronze. What is the monetar value of each souvenir nugget? The problem above represents a sstem of linear equations in three variables. The sstem can be represented with these equations. 14g + 20s + 24b = 20 20g + 15s + 19b = 20 30g + 5s + 13b = 20 Although it is possible to solve sstems of equations in three variables with the substitution method, it can be difficult. It can also be ver challenging to solve this kind of sstem b graphing. Just as the ordered pair (, ) is a solution of a sstem in two variables, the ordered triple (,, z) is a solution of a sstem in three variables. Ordered triples are graphed in three-dimensional coordinate space. The point (3, -2, 4) is graphed below. z An ordered pair can also be the solution of a single equation in two variables. Likewise, an ordered triple can also be the solution of a single equation in three variables. (3, 2, 4) 4 units up O 2 units left 3 units forward Unit 1 Linear Sstems and Matrices 17
Sstems of Linear Equations SUGGESTED LEARNING STRATEGIES: Note Taking The elimination method is usuall the easiest method for solving sstems of equations in three variables. EXAMPLE 3 You can use either elimination or substitution to solve a sstem of three equations in three variables. Use elimination if the terms easil add to 0. Use substitution if one equation has onl one variable on one side, such as = 2 + z. 2 + 7 + z = -53 Solve this sstem, using elimination. -2 + 3 + z = -13 6 + 3 + z = -45 Step 1: Add the first and second equations to eliminate. 2 + 7 + z = -53-2 + 3 + z = -13 10 + 2z = -66 Step 2: Use the second and third equations to eliminate again. Multipl the second equation b 3 so that the -terms add to zero. 3(-2 + 3 + z) = 3(-13) $ -6 + 9 + 3z = -39 $ 6 + 3 + z = -45 6 + 3 + z = -45 12 + 4z = -84 Use the two equations ou found to write a new sstem in two variables. Multipl the second equation b -2 so that the z-terms add to zero. New sstem Multipl the first equation b -2. Add the equations to eliminate z. 10 + 2z = -66 $ -2(10 + 2z) = -2(-66) $ -20-4z = 132 $ 12 + 4z = -84 12 + 4z = -84 $ 12 + 4z = - 84-8 = 48 Solve for. = -6 Step 4: Substitute the -value into one of the two new equations to find z. 10 + 2z = -66 10(-6) + 2z = -66-60 + 2z = -66 2z = -6 z = -3 Step 5: As a final step, check our ordered triple solution in another original equation to be sure that our solution is correct. Substitute the - and z-values into one of the original equations to find. -2 + 3 + z = -13-2 + 3(-6) + -3 = -13-2 - 21 = -13-2 8 = -4 Solution: The solution of the sstem is (-4, -6, -3). 18 SpringBoard Mathematics with Meaning Algebra 2 Step 3:
Sstems of Linear Equations SUGGESTED LEARNING STRATEGIES: Note Taking, Vocabular Organizer, Look for a Pattern TRY THESE C Solve each sstem of equations using elimination. Write our answers on notebook paper. Show our work. a. + + z = 6 2 - + 3z = 9 - + 2 + 2z = 9 b. - 3 + 2z = -8 2 + 3 + z = 17 5-2 + 3z = 5 The graph of an equation in three variables is a plane in coordinate space. z O You can represent a sstem of equations in three variables as three planes in the same coordinate space. The graph of the solutions of a sstem in three variables is the intersection of the three planes. Onl those points that form the intersection of all three planes represent the solution. 6. Classif each sstem as consistent and independent, consistent and dependent, or inconsistent. a. b. Sstems of linear equations in three variables can be classified in the same wa as sstems in two variables. See Math Terms on page 14. Unit 1 Linear Sstems and Matrices 19
Sstems of Linear Equations SUGGESTED LEARNING STRATEGIES: Look for a Pattern 6. () c. d. CHECK YOUR UNDERSTANDING Writeour ouranswers answersononnotebook notebook paper. Show our work. Write paper or grid 4. Solve the sstem, using elimination. paper. Show our work. 3-2 + 7z = 13 + 8-6z = -47 1. Solve the sstem b graphing. 7-9 - 9z = -3 2 + 9 = 5. MATHEMATICAL Which solution method for = -4-3 R E F L E C T I O N solving sstems of 2. Solve the sstem, using substitution. equations do ou find easiest to use? Which 4 + 19 = method do ou find most difficult to use? 3 - = -13 Eplain wh. 3. Solve the sstem, using elimination. { { 43 -+ 22 == 417 20 SpringBoard Mathematics with Meaning Algebra 2 {