Chapter : Sstems of Equations Section.: Sstems in Two Variables... 0 Section. Eercises... 9 Section.: Sstems in Three Variables... Section. Eercises... Section.: Linear Inequalities... Section.: Eercises. Section.: Solving Sstems using Matrices... Section. Eercises... Section.: Matrices and Matri Operations... 9 Section. Eercises... Section.: Sstems in Two Variables Back when studing linear equations, we found the intersection of two lines. Doing so allowed us to solve interesting problems b finding a pair of values that satisfied two different equations. While we didn't call it this at the time, we were solving a sstem of equations. To start out, we'll review an eample of the tpe of problem we've solved before. Eample A small business produces soap and lotion gift baskets. Labor, utilities, and other fied epenses cost $,000 a month. Each basket costs $8 to produce, and sells for $0. How man baskets does the compan need to sell each month to break even? In business terms, "break even" means for revenue (mone brought in) to equal costs. While this problem can be approached in several was, we'll approach here b first creating two linear functions, one for the costs, and another for revenue. Let's define n to be the number of gift baskets the compan sells in a month. There are $,000 of fied costs each month, and costs increase b $8 for each basket, so we can write the linear function for costs, C, as C( n) 000 8n Each sale brings in $0, so the revenue, R, after selling n baskets will be: R( n) 0n This chapter is part of Precalculus. b OpenSta College, 0 Rice Universit. This material is licensed under a Creative Commons CC-BY-SA license.
0 Chapter To find the break-even point, we are looking for the number of baskets where the revenue will equal the costs. In other words, if we were to graph the two linear functions, we are looking for the point that lies on both lines; the solution is the point that satisfies both equations. In this case we could probabl solve the problem from the graph itself, but we can also solve it algebraicall b setting the equations equal: R( n) C( n) 0n000 8n Subtract 8n from both sides n 000 Divide 000 n 00 Evaluate either function at this input R(00) C(00) 0,000 The break-even point is at 00 baskets. The compan must sell 00 baskets a month, at which point their revenue of $0,000 will cover their total costs of $0,000. The eample above illustrates one tpe of sstem of equations, one where both equations are given in functional form. When the equations are written this wa, it is eas to solve the sstem using substitution, b setting the two outputs equal, and solving for the input. However, man sstem of equations problems aren't written this wa. Eample A compan produces a basic and premium version of its product. The basic version requires 0 minutes of assembl and minutes of painting. The premium version requires 0 minutes of assembl and 0 minutes of painting. If the compan has staffing for,900 minutes of assembl and,00 minutes of painting each week. If the compan wants to full utilize all staffed hours, how man of each item should the produce? Notice first that this problem has two variables, or two unknowns - the number of basic products to make, and the number of premium products to make. There are also two constraints - the hours of assembl and the hours of painting available. This is going to give us two equations in two unknowns, what we call a b sstem of equations. We'll start b defining our variables: b: the number of basic products produced p: the number of premium products produced
Section. Sstems in Two Variables 0 Now we can create our equations based on the constraints. Each basic product requires 0 minutes of assembl, so producing b items will require 0b minutes. Each premium product requires 0 minutes of assembl, so producing p items will require 0p minutes. Together we have,900 minutes available, giving us the equation: 0b0 p 900 Using the same approach for painting gives the equation b0 p 00 Together, these form our sstem of equations. The are sometimes written as a pair with a curl bracket on the left to indicate that the should be considered as connected equations. 0b0 p900 b0 p00 As before, our goal is to find a pair of values, (b, p), that satisfies both equations. We'll return to this problem and solve it shortl. While it ma not be clear, the equation 0b0 p 900 we constructed above is a linear equation, like the linear equations from the first eample, it's just written differentl. We could, if desired, solve this equation for p to get it written in slopeintercept form: 0 p900 0b, so p0 b We tpicall don't do this, since it often makes the sstem harder to solve then when using other techniques. To dive into this further, let's first clarif what it means to find a solution to a sstem of linear equations. Sstem of Linear Equations A sstem of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the sstem are considered simultaneousl. A solution to a sstem is a set of numerical values for each variable in the sstem that will satisf all equations in the sstem at the same time. Not ever sstem will have eactl one solution, but we'll look more closel at that later. To check to see if an ordered pair is a solution to a sstem of equations, ou would:. Substitute the ordered pair into each equation in the sstem.. Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.
0 Chapter Eample (video eample here) Determine whether the ordered pair, is a solution to the given sstem of equations. 8 9 Substitute the ordered pair, into both equations. () () 8 8 8 True () 9 () = True The ordered pair, satisfies both equations, so it is the solution to the sstem. There are three common methods for solving sstems of linear equations with two variables. The first is solving b graphing. In the first eample above we graphed both equations, and the solution to the sstem was the intersection of the lines. Eample Solve the following sstem of equations b graphing. 8 Solve the first equation for. 8 8 Solve the second equation for. Graph both equations on the same set of aes. The lines appear to intersect at the point,. We can check to make sure that this is the solution to the sstem b substituting the ordered pair into both equations. ( ) ( ) 8 8 8 True ( ) ( ) True The solution to the sstem is the ordered pair,.
Section. Sstems in Two Variables 0 Tr it Now Solve the following sstem of equations b graphing. While this method can work well enough when the solution values are both integers, it is not ver useful when the intersection is not at a clear point. Additionall, it requires solving both equations for, which adds etra steps. Because of these limitations, solving b graphing is rarel used, but can be useful for checking whether our algebraic answers are reasonable. Solving a Sstem b Substitution Another method for solving a sstem of equations is the substitution method, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Solving a sstem using substitution. Solve one of the two equations for one of the variables in terms of the other.. Substitute the epression for this variable into the second equation, then solve for the remaining variable.. Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.. Check the solution in both equations. The problem we did in Eample was technicall done b substitution, but it was made easier since both equations were alread solved for one variable,. An eample of a more tpical case is shown net. Eample (video eample here) Solve the following sstem of equations b substitution. First, we will solve the first equation for. Now we can substitute the epression for in the second equation.
08 Chapter Now, we substitute 8 into the first equation and solve for. 8 Our solution is 8,. We can check the solution b substituting 8, into both equations. Tr it Now. Solve the following sstem of equations b substitution. 8 (8) () True 8 True Substitution can alwas be used, but is an especiall good choice when one of the variables in one of the equations has a coefficient of or -, making it eas to solve for that variable without introducing fractions. This is fairl common in man applications. Video Eample : Application of Sstems of Equations Eample Julia has just retired, and has $00,000 in her retirement account that she needs to reallocate to produce income. She is looking at two investments: a ver safe guaranteed annuit that will provide % interest, and a somewhat riskier bond fund that averages % interest. She would like to invest as little as possible in the riskier bond fund, but needs to produce $0,000 a ear in interest to live on. How much should she invest in each account? Notice there are two unknowns in this problem: the amount she should invest in the annuit and the amount she should invest in the bond fund. We can start b defining variables for the unknowns:
Section. Sstems in Two Variables 09 a: The amount (in dollars) she invests in the annuit b: The amount (in dollars) she invests in the bond fund. Our first equation comes from noting that together she is going to invest $00,000: ab 00,000 Our second equation will come from the interest. She earns % on the annuit, so the interest earned in a ear would be 0.0a. Likewise, the interest earned on the bond fund in a ear would be 0.0b. Together, these need to total $0,000, giving the equation: 0.0a0.0 b 0,000 Together, these two equations form our sstem. The first equation is an ideal candidate for the first step of substitution - we can easil solve the equation for a or b: a00,000 b Then we can substitute this epression for a in the second equation and solve. 0.0(00,000 b) 0.0b 0,000 8,000 0.0b 0.0b 0,000 0.0b,000 b 0,000 Now substitute this back into the equation a00,000 b to find a a 00, 000 0, 000 a 0, 000 In order to reach her goal, Julia will have to invest $0,000 in the bond fund, and $0,000 in the annuit. Solving a Sstem b the Addition Method A third method of solving sstems of linear equations is the addition method, also called the elimination method. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all sstems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations b multiplication so that one variable will be eliminated b addition.
0 Chapter Solving a Sstem b the Addition Method. Write both equations with - and -variables on the left side of the equal sign and constants on the right.. Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication b a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.. Solve the resulting equation for the remaining variable.. Substitute that value into one of the original equations and solve for the second variable.. Check the solution b substituting the values into the other equation. Eample Solve the given sstem of equations b addition. Both equations are alread set equal to a constant. Notice that the coefficient of in the second equation,, is the opposite of the coefficient of in the first equation,. We can add the two equations to eliminate without needing to multipl b a constant. Now that we have eliminated, we can solve the resulting equation for. Then, we substitute this value for into one of the original equations and solve for.
Section. Sstems in Two Variables The solution to this sstem is,. Check the solution in the first equation. True Often using the addition method will require multipling one or both equations b a constant so terms will eliminate. Eample 8 Solve the given sstem of equations b the addition method. Adding these equations as presented will not eliminate a variable. However, we see that the first equation has in it and the second equation has. So if we multipl the second equation b, the -terms will add to zero. Multipl both sides b. Use the distributive propert. Now, let s add them. For the last step, we substitute into one of the original equations and solve for. 0 9
Chapter Our solution is the ordered pair,. Check the solution in the original second equation. 8 True Tr it Now. Solve the sstem of equations b addition. 0 Eample 9 Solve the given sstem of equations in two variables b addition. 0 0 One equation has and the other has. The least common multiple is 0 so we will have to multipl both equations b a constant in order to eliminate one variable. Let s eliminate b multipling the first equation b and the second equation b. 0 80 0 0 00 0 Then, we add the two equations together. 0 80 00 0 0 Substitute into the original first equation.
Section. Sstems in Two Variables The solution is,. Check it in the other equation. 0 0 0 0 0 0 0 0 0 When one or both of the equations involve fractions, it can help to scale the equation b the least common denominator to eliminate the fractions first. Eample 0 Solve the given sstem of equations in two variables b addition. First clear each equation of fractions b multipling both sides of the equation b the least common denominator. 8 Now multipl the second equation b so that we can eliminate the -variable. Add the two equations to eliminate the -variable and solve the resulting equation. 8 Substitute into the first equation.
Chapter 8. The solution is,. Check it in the other equation. Tr it Now. Solve the sstem of equations b addition. 8 0 Using these approaches, we can revisit the equation from Eample Eample In Eample, we set up the sstem below. Solve it. 0b0 p900 b0 p00 Adding the equations would not eliminate a variable, but we notice that the coefficients on p are the same, so multipling one of the equations b - will change the sign of the coefficients. Multipling the second equation b - gives the sstem 0b0 p900 b 0 p 00 Adding these equations gives b 00 b 0 Substituting b = 0 into the first equation,
Section. Sstems in Two Variables 0(0) 0 p 900 00 0 p 900 0 p 00 p 0 The solution is b = 0, p = 0, meaning the compan should produce 0 basic products and 0 premium products to full utilize staffed hours. Checking our answer in the second equation: (0) 0(0) 00 800 00 00 00 00 Dependent and Inconsistent Sstems Up until now, we have onl considered cases where there is eactl one solution to the sstem. We can categorize sstems of linear equations b the number of solutions. A consistent sstem of equations has at least one solution. A consistent sstem is considered to be an independent sstem if it has a single solution, such as the eamples we just eplored. The two lines have different slopes and intersect at one point in the plane. A consistent sstem is considered to be a dependent sstem if the equations have the same slope and the same -intercepts. In other words, the lines coincide so the equations represent the same line. Ever point on the line represents a coordinate pair that satisfies the sstem. Thus, there are an infinite number of solutions. Another tpe of sstem of linear equations is an inconsistent sstem, which is one in which the equations represent two parallel lines. The lines have the same slope and different -intercepts. There are no points common to both lines; hence, there is no solution to the sstem. Tpes of Linear Sstems An independent sstem has eactl one solution pair,. The point where the two lines intersect is the onl solution. An inconsistent sstem has no solution. Notice that the two lines are parallel and will never intersect. A dependent sstem has infinitel man solutions. The lines are coincident. The are the same line, so ever coordinate pair on the line is a solution to both equations.
Chapter Independent Sstem Inconsistent Sstem Dependent Sstem We can use substitution or addition to identif inconsistent sstems. Recall that an inconsistent sstem consists of parallel lines that have the same slope but different - intercepts. The will never intersect. When searching for a solution to an inconsistent sstem, we will come up with a false statement, such as 0. Eample (video eample here) Solve the following sstem of equations. 9 We can approach this problem in two was. Because one equation is alread solved for, the most obvious step is to use substitution. 9 9 0 9 Clearl, this statement is a contradiction (a false statement) because 9. Therefore, the sstem has no solution, and the sstem is inconsistent. The second approach would be to first manipulate the equations so that the are both in slope-intercept form. We manipulate the first equation as follows. 9 9 9 We then convert the second equation epressed to slope-intercept form.
Section. Sstems in Two Variables Comparing the equations, we see that the have the same slope but different - intercepts. Therefore, the lines are parallel and do not intersect. 9 Recall that a dependent sstem of equations in two variables is a sstem in which the two equations represent the same line. Dependent sstems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identit, such as 0 0. Eample Find a solution to the sstem of equations using the addition method. 9 With the addition method, we want to eliminate one of the variables b adding the equations. In this case, let s focus on eliminating. If we multipl both sides of the first equation b, then we will be able to eliminate the -variable. ( ) Multipl both sides of the equation b. 9 9 9 0 0 We can see that there will be an infinite number of solutions that satisf both equations. In some cases, realizing there are an infinite number of solutions is enough, and we can stop there. In other cases, we will want to describe the set of solutions. One wa is to simpl sa it's the set of points that satisf, but often we would solve that equation for and describe the solution as set of points,. Like with the inconsistent sstem, if we rewrote both equations in the slope-intercept form we might know what the solution would look like before adding. Let s look at what happens when we convert the sstem to slope-intercept form.
8 Chapter 9 9 9 9 Notice the results are the same. The general solution to the sstem is,. Tr it Now. Solve the sstems: a. b. Tr it Now Answers. The solution to the sstem is the ordered pair,..,., ( ). 0,- a. No solution. The sstem is inconsistent. b. The sstem is dependent so there are infinite solutions of the form (, ).
Section. Sstems in Two Variables 9 Section. Eercises For the following eercises, determine whether the given ordered pair is a solution to the sstem of equations.. and (,0). 0 and (, ). 0 and (,). 9 and (, ). 8 and (,) For the following eercises, solve each sstem b substitution... 8 0 0 8. 0 9 0 9..8 9. 0...8. 0. 0. 9 0 0 90. 8. 9. 8 For the following eercises, solve each sstem b addition.. 0.
Chapter 0 8... 9.. 0. 0. 8 8. 0 8 0. 9. 0. 0. 0.. 0 0. 0. 0. For the following eercises, solve each sstem b an method.. 9. 0.9 0. 8 8.. 9. 0.. 9.. 8.... 0.... 0 0. 0. 0. 0.
Section. Sstems in Two Variables For the following eercises, graph the sstem of equations and state whether the sstem is consistent, inconsistent, or dependent and whether the sstem has one solution, no solution, or infinite solutions.. 0... 8. 9. 0. 9 For the following eercises, solve for the desired quantit.. A stuffed animal business has a total cost of production C = + 0 and a revenue function R 0. Find the break-even point.. A fast-food restaurant has a cost of production C() = +0 and a revenue function R( ). When does the compan start to turn a profit?. A cell phone factor has a cost of production C( ) 0 0,000 and a revenue function R( ) 00. What is the break-even point?. A musician charges C( ) 0,000, where is the total number of attendees at the concert. The venue charges $80 per ticket. After how man people bu tickets does the venue break even, and what is the value of the total tickets sold at that point?. A guitar factor has a cost of production C( ) 0,000. The price of each guitar is $. Write the revenue function. Determine the number of guitars the compan needs to sell to break even? What is the cost and the revenue when the break even?. For the following eercises, use a sstem of linear equations with two variables and two equations to solve.. A moving compan charges a flat rate of $0, and an additional $ for each bo. If a tai service would charge $0 for each bo, how man boes would ou need for it to be cheaper to use the moving compan, and what would be the total cost? 8. If a scientist mied 0% saline solution with 0% saline solution to get gallons of 0% saline solution, how man gallons of 0% and 0% solutions were mied?