Section 5.1 Systems of Linear Equations in Two Variables

Before Starting this Section, Review

  1. 1 Graphs of equations (Section 2.2 , page 186)

  2. 2 Graphs of linear equations (Section 2.3 , page 204)

Objectives

  1. 1 Verify a solution to a system of equations.

  2. 2 Solve a system of equations by the graphical method.

  3. 3 Solve a system of equations by the substitution method.

  4. 4 Solve a system of equations by the elimination method.

  5. 5 Solve applied problems by solving systems of equations.

Algebra–Iraq Connection

The most important contributions of medieval Islamic mathematicians lie in the area of algebra. One of the greatest Islamic scholars was Muhammad ibn Musa al-Khwarizmi (780–850). Al-Khwarizmi was one of the first scholars in the House of Wisdom, an academy of scientists, established by caliph al-Mamun in the city of Baghdad in what is now Iraq. Jews, Christians, and Muslims worked together in scholarly pursuits in the academy during this period. The eminent scholar al-Khwarizmi wrote several books on astronomy and mathematics. Western Europeans first learned about algebra from his books. The word algebra comes from the Arabic al-jabr, part of the title of his book Kitab al-jabr wal-muqabala.

This book was translated into Latin and was a widely used text. The Arabic word al-jabr means “restoration,” as in restoring broken parts. (At one time, it was not unusual to see the sign “Algebrista y Sangrador,” meaning “bone setter and blood letter,” at the entrance of a Spanish barber’s shop. The sign informed customers of the barber’s side business.) Al-Khwarizmi used the term in the mathematical sense of removing a negative quantity on one side of an equation and restoring it as a positive quantity on the other side.

System of Equations

  1. 1 Verify a solution to a system of equations.

A set of equations with common variables is called a system of equations. If each equation in a system of equations is linear, then the set of equations is called a system of linear equations or a linear system of equations. However, if at least one equation in a system of equations is nonlinear, then the set of equations is called a system of nonlinear equations. In this section, we solve systems of two linear equations in two variables, such as

{2x−y=5x+2y=5

Notice that we designate a system of equations by using a left brace. A system of equations is sometimes referred to as a set of simultaneous equations. A solution of a system of equations in two variables x and y is an ordered pair of numbers (a, b) such that when x is replaced with a and y is replaced with b, the resulting equations are true. The solution set of a system of equations is the set of all solutions of the system.

Example 1 Verifying a Solution

Check whether each ordered pair is a solution of the system of linear equations:

{2x−y=5(1)x+2y=5(2)

  1. (3, 1)

  2. (4, 3)

Solution

  1. Check (3, 1)

    Equation (1)Equation (2)2x−y=52(3)−1?=55=5✔x+2y=53+2(1)?=55=5✔

    Because (3, 1) satisfies both equations, it is a solution of the system.

  2. Check (4, 3)

    Equation (1)Equation (2)2x−y=52(4)−3?=55=5✔x+2y=54+2(3)?=510=5✕

    Because (4, 3) does not satisfy both equations, it is not a solution of the system.

Practice Problem 1

  1. Check whether each ordered pair is a solution of the system:

    {x+y=4(1)3x−y=0(2)

    1. (2, 2)

    2. (1, 3)

In this section, you will learn three methods of solving a system of two linear equations in two variables: the graphical method, the substitution method, and the elimination method.

Graphical Method

  1. 2 Solve a system of equations by the graphical method.

Recall that the graph of a linear equation

ax+by=c(a and b not both zero)

is a line. Consider the following system of linear equations:

{a1x+b1y=c1a2x+b2y=c2

A solution of this system is a point (an ordered pair) that satisfies both equations. Therefore, to find or estimate the solution set of this system, we graph both equations on the same coordinate axes and find the coordinates of any points of intersection. This procedure is called the graphical method.

Example 2 Solving a System by the Graphical Method

Use the graphical method to solve the system of equations.

{2x−y=4(1)2x+3y=12(2)

Solution

  1. Step 1 Graph both equations on the same coordinate axes.

    1. We first graph equation (1) by finding its intercepts.

      1. To find the y-intercept, set x=0 and solve for y. We have 2(0)−y=4, or y=−4; so the y-intercept is −4.

      2. To find the x-intercept, set y=0 and solve for x. We have 2x−0=4, or x=2; so the x-intercept is 2.

      The points (0, −4) and (2, 0) are on the graph of equation (1), which is sketched in red in Figure 5.1 .

    2. We now graph equation (2) using intercepts.

      Here the x-intercept is 6 and the y-intercept is 4; so joining the points (0, 4) and (6, 0) produces the line sketched in blue in Figure 5.1 .

      Figure 5.1

      Graphical solution

  2. Step 2 Find the point(s) of intersection of the two graphs. We observe in Figure 5.1 that the point of intersection of the two graphs is (3, 2).

  3. Step 3 Check your solution(s). Replace x with 3 and y with 2 in equations (1) and (2).

    Equation (1)Equation (2)2x−y=42(3)−2?=44=4✔2x+3y=122(3)+3(2)?=1212=12✔

  4. Step 4 Write the solution set for the system.

    The solution set is {(3, 2)}.

Practice Problem 2

  1. Solve the system of equations graphically.

    {x+y=2(1)4x+y=−1(2)

If a system of equations has at least one solution (as in Example 2), the system is consistent. A system of equations with no solution is inconsistent, and its solution set is the empty set, ∅.

A system of two linear equations in two variables must have one of the following types of solution sets:

  1. One solution, also called unique solution (the lines intersect; see Figure 5.2(a)): the system is consistent, and the equations in the system are said to be independent.

    Figure 5.2

    Possible solution sets of a system of two linear equations

  2. No solution (the lines are parallel; see Figure 5.2(b)): the system is inconsistent.

  3. Infinitely many solutions (the lines coincide; see Figure 5.2(c)): the system is consistent, and the equations in the system are said to be dependent.

Substitution Method

  1. 3 Solve a system of equations by the substitution method.

Another way to solve a linear system of equations is with the substitution method.

Practice Problem 3

  1. Solve: {x−y=5(1)2x+y=7(2)

It is possible that in the process of solving the equation in Step 3, you obtain an equation of the form 0=k, where k is a nonzero constant. In such cases, the false statement 0=k indicates that the system is inconsistent.

Side Note

It is easy to make arithmetic errors in the many steps involved in the process of solving a system of equations. You should check your solution by substituting it into each equation in the original system.

Example 4 Attempting to Solve an Inconsistent System of Equations

Solve the system of equations.

{x+y=3(1)2x+2y=9(2)

Solution

  1. Step 1 Solve equation (1) for y in terms of x.

    y=3−xAdd −x to both sides.

  2. Step 2 Substitute this expression into equation (2).

    2x+2y=9Equation (2)2x+2(3−x)=9Replace y with 3−x (from Step 1).

  3. Step 3 Solve for x.

    2x+6−2x=9Distributive property0=3︸Subtract 6 from both sides and simplify.False

    Because equation 0=3 is false, the system is inconsistent. Figure 5.3 shows that the graphs of the two equations in this system are parallel lines. Because the lines do not intersect, the system has no solution. The solution set is ∅.

    Figure 5.3

    Inconsistent system

Practice Problem 4

  1. Solve the system of equations.

    {x−3y=1(1)−2x+6y=3(2)

In Step 3, it is also possible to end with an equation of the form k=k. In such cases, the equations are dependent and the system has infinitely many solutions.

Example 5 Solving a Dependent System

Solve the system of equations.

{4x+2y=12(1)−2x−y=−6(2)

Solution

  1. Step 1 Solve equation (2) for y in terms of x.

    −2x−y=−6Equation (2)−y=−6+2xAdd 2x to both sides.y=6−2xMultiply both sides by −1.

  2. Step 2 Substitute (6−2x) for y in equation (1).

    4x+2y=12Equation (1)4x+2(6−2x)=12Replace y with 6−2x (from Step 1).

  3. Step 3 Solve for x.

    4x+12−4x=12Distributive property0=0︸Subtract 12 from both sides and simplify.True

The equation 0=0 is true for every value of x. Thus, any value of x can be used in the equation y=6−2x for back-substitution.

The solutions of the system are of the form (x, 6−2x), and the solution set is

{(x, 6−2x)}.

In other words, the solution set consists of the infinite number of ordered pairs (x, y) lying on the line with equation 4x+2y=12, as shown in Figure 5.4. You can find particular solutions by replacing x with any real number. For example, if we let x=0 in (x, 6−2x), we find that (0, 6−2(0))=(0, 6) is a solution of the system. Similarly, letting x=1, we find that (1, 4) is a solution.

Figure 5.4

Dependent equations

Practice Problem 5

  1. Solve the system of equations.

    {−2x+y=−3(1)4x−2y=6(2)

Elimination Method

  1. 4 Solve a system of equations by the elimination method.

The elimination method of solving a system of equations is also called the addition method.

Practice Problem 6

  1. Solve the system.

    {3x+2y=3(1)9x−4y=4(2)

As in the substitution method, if you add the equations in Step 2 and the resulting equation becomes 0=k, where k≠0, then the system is inconsistent: It has no solutions. As an illustration, if you solve the system of equations in Example 4 by the elimination method, you will obtain the equation 0=3. You now conclude that the system is inconsistent. Similarly, in Step 2, if you obtain 0x+0y=0 (or 0=0), then the equations are dependent. See Example 5.

The next example illustrates how some nonlinear systems can be solved by making a substitution to create a linear system and then solving the new system by the elimination method. We then find the solution(s) to the original equations from the substitution formula.

Side Note

You should use the elimination method when the terms involving one of the variables can easily be eliminated by adding multiples of the equations. Otherwise, use substitution to solve the system. The graphing method is usually used to confirm or visually interpret the result obtained from the other two methods.

Example 7 Solving a Nonlinear System by a Linearizing Substitution

Solve the system.

{2x+5y=−5Equation (1)3x−2y=−17Equation (2)

Solution

Replace 1x with u and 1y with v. (Note that 2x=2(1x)=2u, and so on.) Equations (1) and (2) become

{2u+5v=−5Equation (3)3u−2v=−17Equation (4)

Now we solve the new system for u and v using the elimination method.

  1. Step 1 We choose to eliminate the variable u.

  2. Step 26u+15v=−15−6u+4v=3419v=19(5)(6)(7)Multiply equation (3) by 3.Multiply equation (4) by −2.Add equations (5) and (6).Step 3v=1Solve equation (7) for v.Step 42u+5v=−5Equation (3)2u+5(1)=−5Back-substitute v=1.u=−5Solve for u.

  3. Step 5 Now solve for x and y, the variables in the original system.

    u=1xandv=1yOriginal substitution−5=1x1=1yReplace u with −5 and v with 1.x=−15y=1Solve for x and y.

    The solution set of the original system is {(−15, 1)}.

  4. Step 6 Verify that the ordered pair (−15, 1) is the solution of the original system of equations (1) and (2).

Practice Problem 7

  1. Solve the system.

    {4x+3y=12x−6y=3

Applications

  1. 5 Solve applied problems by solving systems of equations.

We can deduce from Section 2.4 that as the price of a product increases, demand for it decreases and as the price increases, the supply of the product also increases. The equilibrium point is the ordered pair (x, p) such that the number of units, x, and the price per unit, p, satisfy both the demand and supply equations.

Side Note

The graph of a system of equations allows you to use geometric intuition to understand algebra. If the geometric and algebraic representations of a solution do not agree, you must go back and find the error(s).

Example 8 Finding the Equilibrium Point

Find the equilibrium point if the supply and demand functions for a new brand of digital video recorder (DVR) are given by the system

p=60+0.0012x(1)Supply equationp=80−0.0008x(2)Demand equation

where p is the price in dollars and x is the number of units.

Solution

We substitute the value of p from equation (1) into equation (2) and solve the resulting equation.

p=80−0.0008xDemand equation (2)60+0.0012x=80−0.0008xReplace p with 60+0.0012x in (1).0.002x=20Collect like terms and simplify.x=200.002=10,000Solve for x.

The equilibrium point occurs when the supply and demand for DVRs is 10,000 units. To find the price p, we back-substitute x=10,000 into either of the original equations (1) or (2).

p=60+0.0012xEquation (1)=60+0.0012(10,000)Replace x with 10,000.=72Simplify.

The equilibrium point is (10,000, 72). See Figure 5.5. This means that at a price of $72 per unit, the supply and demand for DVRs is 10,000 units.

Figure 5.5

Price of DVRs

Check. Verify that the ordered pair (10,000, 72) satisfies both equations (1) and (2).

Practice Problem 8

  1. Find the equilibrium point.

    {p=20+0.002x(1)Supply equationp=77−0.008x(2)Demand equation

In general, to solve an applied problem involving two unknowns, we need to set up two equations using two variables that represent the two unknowns. The general rule is that we need two equations to find two unknowns.

Example 9 Analyzing Investments

Last year Mrs. Rogers invested $40,000. Part was invested at 6% interest rate per year, and the rest was invested in a risky venture at 10% per year. Combined income for the year totaled $3,120. How much did she invest at each rate?

Solution

Two amounts are to be determined, so we will represent them using two variables. Let

x=amount invested at 6%y=amount invested at 10%

Because 6% of x is 0.06x and 10% of y is 0.10y, we have

0.06x=income from the 6% investment0.10y=income from the 10% investment

We now write our two equations in two variables:

x+y=40,000(1)Total amount invested0.06x+0.10y=3,120(2)Total income from both investments

Let’s use the elimination method to solve these equations.

−6x−6y=−240,0006x+10y=312,0004y=72,000y=72,0004=18,000Multiply equation (1) by −6.Multiply equation (2) by 100.Add.Solve for y.

Back-substitute y=18,000 in equation (1) to get

x+18,000=40,000x=40,000−18,000Solve for x.=22,000.

Mrs. Rogers invested $22,000 at 6% and $18,000 at 10%.

Check. 6% of $22,000 is 0.06×22,000=1320, and10% of $18,000 is 0.10×18,000 =1800.

Finally, $22,000+$18,000=$40,000, and $1320+$1800=$3120, as given.

Practice Problem 9

  1. A speculator invested part of $150,000 in a high-risk venture and received a return of 12% at the end of the year. The rest was invested at 8% annual return. The combined annual income from the two ventures was $15,400. How much was invested at each rate?

Section 5.1 Exercises

Concepts and Vocabulary

  1. The ordered pair (a, b) is a(n)                           of a system of equations in x and y provided that when x is replaced with a and y is replaced with b, the resulting equations are true.

  2. The two nongraphical methods for solving a system of equations are the                           and                           methods.

  3. If in the process of solving a system of equations you get an equation of the form 0=k, where k is not zero, then the system is                           .

  4. If in the process of solving a system of equations you get an equation of the form 0=0, then the system has                           .

  5. True or False. A system consisting of two identical equations has no solution.

  6. True or False. If in the process of solving a system of two equations in x and y you get the equation 5=5, then the system has exactly one solution.

  7. True or False. If x=4, y=−7 is the solution of a system of two equations in x and y, then the lines determined by the two equations intersect at the point (4, −7).

  8. True or False. When solving {3x−2y=149x+8y=23 by the addition method, we can eliminate x by multiplying the first equation by −3 and adding the equations.

Building Skills

In Exercises 9–14, determine which ordered pairs are solutions of each system of equations.

  1. {2x+3y=33x−4y=13(1, −3), (3, −1), (6, 3), (5, 12)

  2. {x+2y=63x+6y=18(2, 2), (−2, 4), (0, 3), (1, 2)

  3. {5x−2y=7−10x+4y=11(54, 1), (0, 114), (1, −1), (3, 4)

  4. {x−2y=−53x−y=5(1, 3), (−5, 0), (3, 4), (3, −4)

  5. {x+y=112x+13y=2(0, 1), (1, 0), (23, 32), (10, −9)

  6. {2x+3y=26x+18y=9(3, 2), (2, 3), (4, 3), (3, 4)

In Exercises 15–24, estimate the solution(s) (if any) of each system by the graphical method. Check your solution(s). For any dependent equations, write your answer with x being arbitrary.

  1. {x+y=3x−y=1

  2. {x+y=10x−y=2

  3. {x+2y=62x+y=6

  4. {2x−y=4x−y=3

  5. {3x−y=−9y=3x+6

  6. {5x+2y=10y=−52x−5

  7. {x+y=7y=2x 

  8. {y−x=2y+x=9

  9. {3x+y=12y=−3x+12 

  10. {2x+3y=66y=−4x+12

In Exercises 25–38, determine whether each system is consistent or inconsistent. If the system is consistent, determine whether the equations are dependent or independent. Do not solve the system.

  1. {y=−2x+3y=3x+5

  2. {3x+y=52x+y=4

  3. {2x+3y=53x+2y=7

  4. {2x−4y=53x+5y=−6

  5. {3x+5y=76x+10y=14

  6. {3x−y=29x−3y=6

  7. {x+2y=−52x−y=4

  8. {x+2y=−22x−3y=5

  9. {2x−3y=56x−9y=10

  10. {3x+y=215x+5y=15

  11. {−3x+4y=592x−6y=152

  12. {6x+5y=119x+152y=21

  13. {7x−2y=311x−32y=8

  14. {4x+7y=1010x+352y=25

In Exercises 39–48, solve each system of equations by the substitution method. Check your solutions. For any dependent equations, write your answer in the ordered pair form given in Example 5.

  1. {y=2x+15x+2y=9x

  2. {x=3y−12x−3y=7

  3. {3x−y=5x+y=7

  4. {2x+y=23x−y=−7

  5. {2x−y=5−4x+2y=7

  6. {3x+2y=5−9x−6y=15

  7. {23x+y=33x+2y=1

  8. {x−2y=34x+6y=3

  9. {x−2y=5−3x+6y=−15

  10. {x+y=32x+2y=6

In Exercises 49–58, solve each system of equations by the elimination method. Check your solutions. For any dependent equations, write your answer as in Example 5.

  1. {x−y=1x+y=5

  2. {2x−3y=53x+2y=14

  3. {x+y=02x+3y=3

  4. {x+y=33x+y=1

  5. {5x−y=53x+2y=−10

  6. {3x−2y=1−7x+3y=1

  7. {x−y=2−2x+2y=5

  8. {x+y=52x+2y=−10

  9. {4x+6y=122x+3y=6

  10. {4x+7y=−3−8x−14y=6

In Exercises 59–76, use any method to solve each system of equations. For any dependent equations, write your answer as in Example 5.

  1. {2x+y=92x−3y=5

  2. {x+2y=10x−2y=−6

  3. {2x+5y=2x+3y=2

  4. {4x−y=63x−4y=11

  5. {2x+3y=73x+y=7

  6. {x=3y+4x=5y+10

  7. {2x+3y=93x+2y=11

  8. {3x−4y=0y=2x+13

  9. {x4+y6=1x+2(x−y)=7

  10. {x3+y5=12x−y=4

  11. {3x=2(x+y)3x−5y=2

  12. {y+x+2=0y+2x+1=0

  13. {0.2x+0.7y=1.50.4x−0.3y=1.3

  14. {0.6x+y=−1x−0.5y=7

  15. 73. {x2+y3=13x4+y2=1

  16. {x3−y5=26y−10x=25

  17. {x3−y2=13y8−x4=−34

  18. {x5−y2=115y−6x=−30

In Exercises 77–84, let u=1x and v=1y. Solve for u and v; then solve for x and y.

  1. {2x+5y=−53x−2y=−17

  2. {2x+1y=34x−2y=0

  3. {3x+1y=46x−1y=2

  4. {6x+3y=04x+9y=−1

  5. {5x+10y=32x−12y=−2

  6. {3x+4y=16x+4y=3

  7. {2x+1y=4x+2y=6xy

  8. {1x−2y=32x−y=5xy

Applying the Concepts

In Exercises 85–88, the demand and supply functions of a product are given. In each case, p represents the price in dollars per unit and x represents the number of units in hundreds. Find the equilibrium point.

  1. {2p+x=140Demand equation12p−x=280Supply equation

  2. {7p+x=150Demand equation10p−x=20Supply equation

  3. {2p+x=25Demand equationx−p=13Supply equation

  4. {p+2x=96Demand equationp−x=39Supply equation

In Exercises 89–108, use a system of equations to solve each problem.

  1. Diameter of a pizza. The sum of the diameters of the largest and smallest pizzas sold at the Monster Pizza Shop is 29 inches. The difference in their diameters is 13 inches. Find the diameters of the largest and smallest pizzas.

  2. Calories in hamburgers. The sum of the number of calories in a hamburger from Boston Burger and a hamburger from Carmen’s Broiler is 1130. The difference in the number of calories in the hamburgers is 40. If the Boston Burger has the larger number of calories, how many calories are in each restaurant’s hamburger?

  3. Trash composition. Paper and plastic together account for 48% (by weight) of the total trash collected. If the weight of paper trash collected is five times the weight of plastic trash, what percent of the total trash collected is paper and what percent is plastic?

  4. Cost of food and clothing. The average monthly combined cost of food and clothing for the Martínez family is $1000. If they spend four times as much for food as they do for clothes, what is the average monthly cost of each?

  5. Mardi Gras parade “throws.” Levon paid 40¢ for each string of beads and 30¢ for each doubloon (a special coin) he bought to throw from his Mardi Gras float. He paid a total of $265 for the two items. His doubloons and beads combined to a total of 770 items. How many doubloons and how many strings of beads did Levon buy?

  6. Halloween candy. Janet bought 135 pieces of candy to give away on Halloween. She bought two kinds of candy, paying 24¢ apiece for one kind and 18¢ apiece for the other. If she spent $26.70 for the candy, how many pieces of each kind did she buy?

In Exercises 95 and 96, use the information in the following table.

Food Description McDonald’s Fat (gm) Carb (gm) Protein (gm)
Breakfast Burrito 20 21 13
Egg McMuffin 12 27 17

Suppose you ate breakfast at McDonald’s during one workweek (Monday–Friday).

  1. McDonald’s breakfast. If you consumed 123 grams of carbohydrates and 77 grams of protein, how many of each breakfast item did you eat during the workweek?

  2. McDonald’s breakfast. If you consumed 68 grams of fat and 129 grams of carbohydrates, how many of each breakfast item did you eat during the workweek?

  3. Investment. Last year Mrs. García invested $50,000. She put part of the money in a real estate venture that paid 7.5% for the year and the rest in a small-business venture that returned 12% for the year. The combined income from the two investments for the year totaled $5190. How much did she invest at each rate?

  4. Investment. Mr. Sharma invested a total of $30,000 in two ventures for a year. The annual return from one of them was 8%, and the other paid 10.5% for the year. He received a total income of $2550 from both investments. How much did he invest at each rate?

  5. Tutoring other students. A student earns twice as much per hour for tutoring as she does working at McDougal’s. If her average wage is $11.25 per hour, how much does she earn per hour at each job?

  6. Plumber’s earnings. A plumber earns $15 per hour more than her apprentice. In each 40-hour week, their combined earnings are $2200. What is the hourly rate for each?

  7. Mixture problem. An herb that sells for $5.50 per pound is mixed with tea that sells for $3.20 per pound to produce a 100-pound mix that is worth $3.66 per pound. How many pounds of each ingredient does the mix contain?

  8. Mixture problem. A chemist had a solution of 60% acid. She added some distilled water, reducing the acid concentration to 40%. She then added 1 more liter of water to further reduce the acid concentration to 30%. How much of the 30% solution did she then have?

  9. Airplane speed. With the help of a tail wind, a plane travels 3000 kilometers in 5 hours. The return trip against the wind requires 6 hours. Assume that the direction and the wind speed are constant. Find both the speed of the plane in still air and the wind speed. [Hint: Remember that the wind helps the plane in one direction and hinders it in the other.]

  10. Motorboat speed. A motorboat travels up a stream a distance of 12 miles in 2 hours. If the current had been twice as strong, the trip would have taken 3 hours. How long should it take for the return trip down the stream?

  11. Break-even analysis. A local publishing company publishes City Magazine. The production and setup costs are $30,000, and the cost of producing each magazine is $2. Each magazine sells for $3.50. Assume that x magazines are published and sold.

    1. Write the total cost function y=C(x) and the revenue function y=R(x).

    2. Graph both functions from part (a) on the same coordinate plane.

    3. How many magazines must be sold to break even?

  12. Break-even analysis. An electronic manufacturer plans to make digital portable music players (MP4s). Fixed costs will be $750,000, and it will cost $50 to manufacture each MP4, which will be sold for $125 to the retailers. Assume that x MP4s are manufactured and sold.

    1. Write the total cost function y=C(x) and the revenue function y=R(x).

    2. Graph both functions from part (a) on the same coordinate plane.

    3. How many MP4s must be sold to break even?

  13. Making a job decision. Shanaysha has job offers from department stores A and B to sell major appliances. Store A offers her a fixed salary of $400 per week. Store B offers her $150 per week plus a commission of 4% of her weekly sales. How much should Shanaysha’s weekly sales be for the offer from Store B to be better than that from Store A?

  14. Making a job decision. Sheena has job offers from companies A and B to sell insurance. Company A offers her $25,000 per year plus 2% of her yearly sales premiums. Company B offers her $30,000 per year plus 1% of her yearly sales premiums. How much must Sheena earn in yearly sales premiums for the offer from Company B to be the better offer?

Beyond the Basics

In Exercises 109–112, solve each system for x and y.

  1. {2 log3 x+3 log3 y=83 log3 x−log3 y=1

  2. {3 log2(x+y)−5 log2(x−y)=22 log2(x+y)+3 log2(x−y)=14

  3. {3ex−4ey=42ex+5ey=18

  4. {ex+2y−3ex−y=163ex+2y−12ex−y=46

  5. For what value of the constant c is the following system consistent?

    {x+2y=73x+5y=11cx+3y=4

    [Hint: Solve the first two equations for x and y and then substitute the solution into the third equation.]

  6. Show that the lines x−y=6, 4x−3y=20, and 6x+5y+8=0 are concurrent (all intersect at one point).

    [Hint: Solve the first two equations. Verify that the point of intersection lies on the third line.]

  7. Find the value of c for which the following system of equations is inconsistent.

    {2x+cy=115x−7y=5

  8. If (x−2) is a factor of both f(x)=x3−4x2+ax+b and g(x)=x3−ax2+bx+8, find the values of a and b.

  9. Let l1:a1x+b1y+c1=0 and l2:a2x+b2y+c2=0 be two nonparallel lines. Then for any constant k, show that the line (a1x+b1y+c1)+k(a2x+b2y+c2)=0 is an equation of the line passing through the point of intersection of l1 and l2.

In Exercises 118–120, write your answer in slope–intercept form. [Hint: Use Exercise 117.]

  1. Find an equation of the line that passes through the point of intersection of the lines with equations 2x+y=3 and x−3y=12 and that is parallel to the line with equation 3x+2y=8.

  2. Find an equation of the line that passes through the point of intersection of the lines with equations 5x+2y=7 and 6x−5y=38 and that passes through the point (1, 3).

  3. Find an equation of the line that passes through the point of intersection of the lines with equations 2x+5y+7=0 and 13x−10y+3=0 and that is perpendicular to the line with equation 7x+13y=8.

Critical Thinking / Discussion / Writing

  1. Consider the system of equations

    {a1x+b1y=c1a2x+b2y=c2

    where a1, b1, c1, a2, b2, and c2 are constants. Find a relationship (an equation) among these constants such that the system of equations has

    1. Only one solution. Find the solution in terms of the constants.

    2. No solution.

    3. Infinitely many solutions.

  2. Let 3x+4y−12=0 be the equation of a line l1 and let P(5, 8) be a point.

    1. Find an equation of a line l2 passing through the point P and perpendicular to l1.

    2. Find the point Q of the intersection of the two lines l1 and l2.

    3. Find the distance d(P, Q). This is the distance from the point P to the line l1.

  3. Use the procedure outlined in Exercise 122 to show that the (perpendicular) distance from a point P(x1, y1) to the line ax+by+c=0 is given by |ax1+by1+c|√a2+b2.

  4. Use the formula from Exercise 123 to find the distance from the given point P to the given line l.

    1. P(2, 3), l: x+y−7=0

    2. P(−2, 5), l: 2x−y+3=0

    3. P(3, 4), l: 5x−2y−7=0

    4. P(0, 0), l: ax+by+c=0

  5. Find the reflection of the point P=(2, 1) about the line l:2x+3y=20. [Let Q=(a, b) be the reflection point; then you get two equations in a and b. (i) The midpoint of ¯PQ lies on l; (ii) l is perpendicular to the line containing the segment ¯PQ.]

  6. Find the reflection of the point P=(2, 1) about the line l: 3x+y=12.

Getting Ready for the Next Section

In Exercises 127–132, find the requested variable.

  1. Find y if y+2z=7 and z=2.

  2. Find x if x−3y=5 and y=−3.

  3. Find x if 2x+3y+5z=21, y=1, and z=2.

  4. Find y if 3x−5y+z=2, x=−2, and z=3.

  5. Find x if 2x+3y−2z=5, y−z=4, and z=−3.

  6. Find x if 3x+2y+z=2, y+2z=1, and z=2.

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