Solving Systems of Linear Equations by Graphing

1. Decide whether a given ordered pair is a solution of a system.

2. Solve linear systems by graphing.3. Solve special systems by graphing.4. Identify special systems without graphing.

Solving Systems of Linear Equations by Graphing

•A system of linear equations, often called a linear system, consists of two or more linear equations with the same variables. Examples of systems include 2 3 4

3 5

x y

x y

3 4

4 2

x y

y x

1

.3

x y

y

Linear systems

In the system on the right, think of y = 3 as an equation in two variables by writing it as 0x + y = 3.

Decide whether a given ordered pair is a solution of a system.

•A solution of a system of a linear equations is an ordered pair that makes both equations true at the same time. A solution is said to satisfy the equation.

EXAMPLE 1•Decide whether the ordered pair (4,−1) is a solution of each system.

Solution:

Determining whether an Ordered Pair Is a Solution

5 6 14

2 5 3

x y

x y

3

3

x y

x y

5 4 6 1 14

2 4 5 1 3

4 1 3

4 1 3

20 6 14

8 5 3

14 14

3 3

4 1 3

4 1 3

3 3

3 3

Yes No

Solve linear systems by graphing.

The set of all ordered pairs that are solutions of a system is its solution set.

Any intersection point would be on both lines and would therefore be a solution of both equations. Thus, the coordinates of any point at which the lines intersect give a solution of the system. Because the two different straight lines can intersect at no more then one point, there can never be more than one solution set for such a system.

One way to find the solution set of a system of two linear equations is to graph both equations on the same axes. The graph of each line shows points whose coordinates satisfy the equation of that line.

Solve linear systems by graphing. (cont’d)

To solve a system by graphing, follow these steps.

Step 1: Graph each equation of the system on the same coordinate axes.

Step 2: Find the coordinates of the point of intersection of the graphs if possible. This is the solution of the system.

Step 3: Check the solution in both of the original equations. Then write the solution set.

A difficulty with the graphing method is that it may not be possible to determine from the graph the exact coordinates of the point that represents the solution, particularly if those coordinates are not integers. The graphing method does, however, show geometrically how solutions are found and is useful when approximate answer will do.

EXAMPLE 2

•Solve the system by graphing.

Solution: {(3,2)}

Solve a System by Graphing

5 3 9

2 7

x y

x y

EXAMPLE 3

•Solve each system by graphing

Solution:

3 4

6 2 12

x y

x y

Solving Special Systems by Graphing

, 2 5 8x y x y

2 5 8

4 10 16

x y

x y

When a system has an infinite number of solutions, either equation of the system could be used to write the solution set. It’s best to use the equation (in standard form) with coefficients that are integers having no common factor (except 1).

Three Cases for Solutions of SystemsThe graphs intersect at exactly one point, which gives the (single)

ordered pair solution of the system. The system is consistent and the equations are independent. See figure (a).

The graphs are the same line. There is an infinite number of solutions, and the solution set is written in set-builder notation as {(x,y)|_________}, where one of the equations is written after the | symbol. The system is consistent and equations are dependent. See figure (c).

The graphs are parallel lines, so there is no solution and the solution set is Ø. The system is inconsistent and the equations are independent. See figure (b).

EXAMPLE 4Describe each system without graphing. State the number of solutions.

Solution:

a) The equations represent parallel lines. The system has no solution.

2 3 5

3 2 7

x y

y x

2 5

3 32 7

3 3

y x

y x

1 2

3 31 2

3 3

y x

y x

Identifying the Three Cases by Using Slopes

6 3

2 11

x y

x y

3 2

2 6 4

x y

x y

6 3

2 11

y x

y x

a) c)b)

b) The equations represent the same line. The system has an infinite number of solutions.

c) The equations represent lines that are neither parallel nor the same line. The system has exactly one solution.

EXAMPLE 5

•Solve the system by graphing.

Solution: {(3,2)}

Finding the Solution of a System by Graphing

5 3 9

2 7

x y

x y

Solving Systems of Linear Equations by Substitution

Solve linear systems by substitution.

Solve special systems by substitution.

Solve linear systems with fractions.

Solve linear systems by substitution.

Graphing to solve a system of equations has a serious drawback. It is difficult to find an accurate solution, such as

, from a graph. One algebraic method for solving a system of equations is the substitution method.

This method is particularly useful for solving systems in which

one equation is already solved, or can be solved quickly, for one of the variables.

1 5,

3 6

Solve linear systems by substitution. (cont’d)

To solve a system by substitution, follow these steps:

Step 1: Solve one equation for either variable. If one of the variables has coefficient 1 or −1, choose it, since it usually makes the substitution method easier.

Step 2: Substitute for that variable in the other equation. The result should be an equation with just one variable.

Step 3: Solve the equation from Step 2.

Step 4: Substitute the result from Step 3 into the equation from Step 1 to find the value of the other variable.

Step 5: Check the solution in both of the original equations. Then write the solution set.

EXAMPLE 1 Using the Substitution Method

2 7 12

2

x y

x y

Solution:

The solution set found by the substitution method will be the same as the solution found by graphing. The solution set is the same; only the method is different.

•Solve the system by the substitution method.

2 72 12yy 4 7 12y y

3 1

3 3

2y

4y

2x y

8x

42x

8, 4

EXAMPLE 2 Using the Substitution Method

2 7 12

3 2

x y

x y

Solution: 2 73 2 12yy 6 4 7 12y y

Solve the system by the substitution method.

6 3 126 6y 3 1

3 3

8y

6y

63 2x 3 12x 15x

15, 6

Be careful when you write the ordered-pair solution of a system. Even though we found y first, the x-coordinate is always written first in the ordered pair.

EXAMPLE 3

Solution:

Using the Substitution Method

411 1x y

•Use substitution to solve the system.

4 1x y

2 14 5 11 yy 28 1 22 5 1y y

13 1

1

3

13 3y

1y

3, 1

4 11x

1 4

2 5 11

x y

x y

4 1x 3x

EXAMPLE 4

Solution:

Solving an Inconsistent Systemby Substitution

86 841 2 xx

•Use substitution to solve the system.

16 16 8 8x x 8 8

8 4

16 2 8

y x

x y

Since the statement is false, the solution set is Ø.

It is a common error to give “false” as the solution of an inconsistent system. The correct response is Ø.

EXAMPLE 5Solving a System with Dependent Equations by Substitution

Since the statement is true every solution of one equations is also a solution to the other, so the system has an infinite number of solutions and the solution set is {(x,y)|x + 3y = −7}.

•Solve the system by the substitution method.3 7

4 12 28

x y

x y

Solution:

3 33 7y yx y 7 3x y

4 127 3 28yy 28 12 12 28y y

28 28

It is a common error to give “true” as the solution of a system of dependent equations. Remember to give the solution set in set-builder notation using the equation in the system that is in standard form with integer coefficients that have no common factor (except 1).

EXAMPLE 6 Using the Substitution Method with Fractions as Coefficients

•Solve the system by the substitution method.1 1 1

2 3 31

2 22

x y

x y

Solution:

4 44 4y yx y

21

22

22x y

4 4x y

1 1 1

2 36

36 x y

3 2 2x y 2 243 4 yy

12 12 12 122 2y y

10 1

1

0

10 0y

1y

4 14x 4 4x

0x 0, 1

Solving Systems of Linear Equations by Elimination

• Solve linear systems by elimination.

• Multiply when using the elimination method.

• Use an alternative method to find the second value in a solution.

• Use the elimination method to solve special systems.

An algebraic method that depends on the addition property of equality can also be used to solve systems. Adding the same quantity to each side of an equation results in equal sums:

If A = B, then A + C = B + C.

Solve linear systems by elimination.

We can take this addition a step further. Adding equal quantities, rather than the same quantity, to each side of an equation also results in equal sums:

If A = B, then A + C = B + D.

Using the addition property to solve systems is called the elimination method. With this method, the idea is to eliminate one of the variables. To do this, one pair of variable terms in the two equations must have coefficients that are opposite.

EXAMPLE 1 Using the Elimination Method

Solution:

Solve the system.

3 7

2 3

x y

x y

2 33 7x yx y 2 2 3y 5 0

5 5

1x

2x

4 44 3y

1y

2, .1The solution set is

A system is not completely solved until values for both x and y are found. Do not stop after finding the value of only one variable. Remember to write the solution set as a set containing an ordered pair

Solving a Linear System by EliminationIn general, use the following steps to solve a linear system of

equations by the elimination method.Step 1: Write both equations in standard form, Ax + By = C.

Step 2: Transform the equations as needed so that the coefficients of one pair of variable terms are opposites. Multiply one or both equations by appropriate numbers so that the sum of the coefficients of either the x- or y-term is 0.

Step 3: Add the new equations to eliminate a variable. The sum should be an equation with just one variable.

Step 5: Substitute the result from Step 4 into either of the original equations, and solve for the other variable.

Step 4: Solve the equation from Step 3 for the remaining variable.

Step 6: Check the solution in both of the original equations. Then write the solution set.

It does not matter which variable is eliminated first. Choose the one that is more convenient to work with.

EXAMPLE 2 Using the Elimination Method

Solution:

Solve the system: 2

2 10

x y

x y

2 22x yy y 2x y

2 02 1y xx y 3 2

3 3

1x

4, .2The solution set is

2 10x yy y 2 10x y

4x

4 44 2y 2y

EXAMPLE 3 Multiplying Both Equations When Using the Elimination Method

Solution:

Solve the system: 4 5 18

3 2 2

x y

x y

4 18 252 x y 8 10 36x y

15 10 110 36 08 xx yy 2

23 23

3 46x

.2, 2The solution set is

3 25 52x y

15 10 10x y

2x

66 62 2y 2 4

2 2

y

3 22 2y

2y When using the elimination method, remember to multiply both sides of an equation by the same nonzero number.

EXAMPLE 4 Finding the Second Value by Using an Alternative Method

Solution:

Solve the system: 3 8 4

6 9 2

y x

x y

4 32 28x y 6 23 9 3x y

8 6 16x y 18 6 27x y

The solution set is

11

4 33 38x y

26x

12 9 24x y 12 4 18x y

6 2 292 x y

13 42y

+

11

26x

26

2 1

26

6 1x

+

13 1

3

3

1 42y

42

16, .

11

26

42

13y

EXAMPLE 5 Using the Elimination Method for an Inconsistent System or Dependent Equations

Solution:

Solve each system by the elimination method:

3 7

6 2 5

x y

x y

.The solution set is

2 5 1

4 10 2

x y

x y

72 3 2x y 6 2 5x y

6 2 14x y 6 2 5x y +

0 19

4 10 2x y 4 10 2x y +

0 0

2 5 212 x y 4 10 2x y

, 2 5 1 .x y x y

The solution set is

Applications of Linear Systems

1. Solve problems about unknown numbers.2. Solve problems about quantities and their

costs.3. Solve problems about mixtures.4. Solve problems about distance, rate (or

speed), and time.

Applications of Linear SystemsRecall from Section 2.4 the six step method for solving applied

problems. These slightly modified steps allow for two variables and two equations.

Step 1: Read the problem carefully until you understand what is given and what is to be found.

Step 2: Assign variables to represent the unknown values, using diagrams or tables as needed. Write down what each variable represents.

Step 3: Write two equations using both variables.

Step 5: State the answer to the problem. Is the answer reasonable?

Step 4: Solve the system of two equations.

Step 6: Check the answer in the words of the original problem.

Two top-grossing Disney movies in 2002 were Lilo and Stitch and The Santa Clause 2. Together they grossed $284.2 million. The Santa Clause 2 grossed $7.4 million less than Lilo and Stitch. How much did each movie gross? (Source: Variety.)

EXAMPLE 1 Solving a Problem about Two Unknown Numbers

Solution:Let x = gross of Lilo and Stitch in millions,and y = gross of The Santa Clause 2 in millions.

284.2x y 7.4x y

7.4 27 84.2. 7.44 y y 2 27

2 2

6.8y

138.4y

8.7.4 13 4x 145.8x

Lilo and Stitch grossed 145.8 million dollars and The Santa Clause 2 grossed 138.4 million dollars.

In 1997 – 1998, the average movie ticket (to the nearest U.S. dollar) cost $10 in Geneva and $8 in Paris. (Source: Parade, September 13, 1998.) If a group of 36 people from these two cities paid $298 for tickets to see The Rookie, how many people from each city were there?

EXAMPLE 2 Solving a Problem about Quantities and Costs

Solution:

36x y 8 10 298x y

8 10 29836 yy 22 888 88 21 9 80 2 8 8y y

2 0

2 2

1y

5y 31x

There were 5 people from Geneva, and 31 people from Paris that went to see The Rookie.

  Number of Price per Ticket Total Value

  Tickets (in dollars) (in dollars)

Paris x 8 8x

Geneva y 10 10y

Total 36 XXXXXXXX 298

6 53x

How many liters of a 25% alcohol solution must be mixed with a 12% solution to get 13 L of a 15% solution?

EXAMPLE 3 Solving a Mixture Problem Involving Percent

Solution:

13x y .12 .25 1.95x y

.12 .25 1100 100.95x y 12 25 19513 yy

156 12 25 15 195 566 1y y 13 1

3

3

1 39y 3 31x

To make 13 L of a 15% solution, 3 L of 25% solution, and 10 L of 12% solution must be used.

Liters of Percent (as Liters of

Solution a decimal) pure alcohol

x .12 .12x

y .25 .25y

13 .15 1.95

3y 10x

Recall working these mixture problems! We used one variable.

In one hour, Abby can row 2 mi against the current or 10 mi with the current. Find the speed of the current and Abby’s speed in still water.

EXAMPLE 4 Solving a Problem about Distance, Rate, and Time

Solution:Let x = Abby’s speed in still water in mph,and y = the water speed of the current in mph.

10x y

2x y

22 1 20y y

Abby’s speed in still water is 6 mph, and the speed of the current is 4 mph.

2 8

2 2

y

4y

2 4x 6x

SYSTEMS OF LINEAR INEQUALITIES

Solving Linear Systems of Inequalities by Graphing

Solving Systems of Linear Inequalities

1. We show the solution to a system of linear inequalities by graphing them.

a) This process is easier if we put the inequalities into Slope-Intercept Form, y = mx + b.

Solving Systems of Linear Inequalities

2. Graph the line using the y-intercept & slope.

a) If the inequality is < or >, make the lines dotted.

b) If the inequality is < or >, make the lines solid.

Solving Systems of Linear Inequalities

3. The solution also includes points not on the line, so you need to shade the region of the graph:

a) above the line for ‘y >’ or ‘y ’.b) below the line for ‘y <’ or ‘y ≤’.

Solving Systems of Linear Inequalities

Example: a: 3x + 4y > - 4

b: x + 2y < 2

Put in Slope-Intercept Form: ) 3 4 4

4 3 4

31

4

a x y

y x

y x

) 2 2

2 2

11

2

b x y

y x

y x

Solving Systems of Linear Inequalities

a:

dotted

shade above

b:

dotted

shade below

Graph each line, make dotted or solid and shade the correct area.

Example, continued:

3: 1

4 a y x 1

: 12

b y x

Solving Systems of Linear Inequalities

a: 3x + 4y > - 4

3: 1

4 a y x

Solving Systems of Linear Inequalities

a: 3x + 4y > - 4

b: x + 2y < 2

3: 1

4 a y x

1: 1

2 b y x

Solving Systems of Linear Inequalities

a: 3x + 4y > - 4

b: x + 2y < 2

The area between the green arrows is the region of overlap and thus the solution.

Solving Systems of Three Linear Equations in Three

Variables

The Elimination Method

Solutions of a system with 3 equations

The solution to a system of three linear equations in three variables is an ordered triple.

(x, y, z)

The solution must be a solution of all 3 equations.

Is (–3, 2, 4) a solution of this system?

3x + 2y + 4z = 112x – y + 3z = 45x – 3y + 5z = –1

3(–3) + 2(2) + 4(4) = 112(–3) – 2 + 3(4) = 45(–3) – 3(2) + 5(4) = –1

Yes, it is a solution to the system because it is a solution to all 3 equations.

Use elimination to solve the following system of equations.

x – 3y + 6z = 213x + 2y – 5z = –302x – 5y + 2z = –6

Step 1

Rewrite the system as two smaller systems, each containing two of the three equations.

x – 3y + 6z = 213x + 2y – 5z = –302x – 5y + 2z = –6

x – 3y + 6z = 21 x – 3y + 6z = 213x + 2y – 5z = –30 2x – 5y + 2z = –6

Step 2

Eliminate THE SAME variable in each of the two smaller systems.

Any variable will work, but sometimes one may be a bit easier to eliminate.

I choose x for this system.

(x – 3y + 6z = 21) 3x + 2y – 5z = –30

–3x + 9y – 18z = –63 3x + 2y – 5z = –30

11y – 23z = –93

(x – 3y + 6z = 21) 2x – 5y + 2z = –6

–2x + 6y – 12z = –42 2x – 5y + 2z = –6

y – 10z = –48

(–3) (–2)

Step 3

Write the resulting equations in two variables together as a system of equations.

Solve the system for the two remaining variables.

11y – 23z = –93 y – 10z = –48

11y – 23z = –93 –11y + 110z = 528

87z = 435 z = 5

y – 10(5) = –48 y – 50 = –48

y = 2

(–11)

Step 4

Substitute the value of the variables from the system of two equations in one of the ORIGINAL equations with three variables.

x – 3y + 6z = 213x + 2y – 5z = –302x – 5y + 2z = –6

I choose the first equation.

x – 3(2) + 6(5) = 21x – 6 + 30 = 21 x + 24 = 21

x = –3

Step 5

CHECK the solution in ALL 3 of the original equations.

Write the solution as an ordered triple.

x – 3y + 6z = 213x + 2y – 5z = –302x – 5y + 2z = –6

–3 – 3(2) + 6(5) = 213(–3) + 2(2) – 5(5) = –302(–3) – 5(2) + 2(5) = –6

The solution is (–3, 2, 5).

It is very helpful to neatly organize yourwork on your paper in the following manner.

(x, y, z)

Solve the system.Solve the system.1. x+3y-z=-111. x+3y-z=-11

2x+y+z=12x+y+z=1z’s are easy to cancel!z’s are easy to cancel!

3x+4y=-103x+4y=-102. 2x+y+z=12. 2x+y+z=1

5x-2y+3z=215x-2y+3z=21Must cancel z’s again!Must cancel z’s again!

-6x-3y-3z=-3-6x-3y-3z=-35x-2y+3z=215x-2y+3z=21 -x-5y=18-x-5y=18

2(2)+(-4)+z=12(2)+(-4)+z=1 4-4+z=14-4+z=1

3. 3x+4y=-103. 3x+4y=-10 -x-5y=18-x-5y=18

Solve for x & y.Solve for x & y.3x+4y=-103x+4y=-10-3x-15y+54-3x-15y+54

-11y=44-11y=44 y=- 4y=- 4

3x+4(-4)=-103x+4(-4)=-10 x=2x=2

(2, - 4, 1)(2, - 4, 1)

x+3y-z=-11x+3y-z=-112x+y+z=12x+y+z=1

5x-2y+3z=215x-2y+3z=21

z=1z=1

2. 2x+2y+z=52. 2x+2y+z=54x+4y+2z=64x+4y+2z=6

Cancel z’s again.Cancel z’s again.-4x-4y-2z=-10-4x-4y-2z=-104x+4y+2z=64x+4y+2z=6 0=- 40=- 4

Doesn’t make sense!Doesn’t make sense! No solutionNo solution

Solve the system.Solve the system.

1. -x+2y+z=31. -x+2y+z=32x+2y+z=52x+2y+z=5

z’s are easy to cancel!z’s are easy to cancel!-x+2y+z=3-x+2y+z=3-2x-2y-z=-5-2x-2y-z=-5-3x=-2-3x=-2x=2/3x=2/3

-x+2y+z=3-x+2y+z=32x+2y+z=52x+2y+z=5

4x+4y+2z=64x+4y+2z=6

3. x+y=33. x+y=32x+2y=62x+2y=6

Cancel the x’s.Cancel the x’s.-2x-2y=-6-2x-2y=-62x+2y=62x+2y=6 0=00=0

This is true.This is true.¸ ¸ many solutionsmany solutions

Solve the system.Solve the system.1.1. -2x+4y+z=1-2x+4y+z=1

3x-3y-z=23x-3y-z=2z’s are easy to cancel!z’s are easy to cancel!

x+y=3x+y=32.2. 3x-3y-z=23x-3y-z=2

5x-y-z=85x-y-z=8Cancel z’s again.Cancel z’s again.

-3x+3y+z=-2-3x+3y+z=-25x-y-z=85x-y-z=82x+2y=62x+2y=6

-2x+4y+z=1-2x+4y+z=13x-3y-z=23x-3y-z=25x-y-z=85x-y-z=8

Solve:

x – 6y – 2z = –8–x + 5y + 3z = 23x – 2y – 4z = 18

Answer:

x – 6y – 2z = –8–x + 5y + 3z = 23x – 2y – 4z = 18

(4, 3, –3)

Solve:

–5x + 3y + z = –1510x + 2y + 8z = 1815x + 5y + 7z = 9

Answer:

–5x + 3y + z = –1510x + 2y + 8z = 1815x + 5y + 7z = 9

(1, –4, 2)

Application

Courtney has a total of 256 points on three Algebra tests. His score on the first test exceeds his score on the second by 6 points. His total score before taking the third test was 164 points. What were Courtney’s test scores on the three tests?

Explore

Problems like this one can be solved using a system of equations in three variables. Solving these systems is very similar to solving systems of equations in two variables. Try solving the problemLet f = Courtney’s score on the first testLet s = Courtney’s score on the second testLet t = Courtney’s score on the third test.

Plan

Write the system of equations from the information given.

f + s + t = 256 f – s = 6 f + s = 164

The total of the scores is 256.

The difference between the 1st and 2nd is 6 points.

The total before taking the third test is the sum of the first and second tests..

Solve

Now solve. First use elimination on the last two equations to solve for f. f – s = 6

f + s = 164 2f = 170 f = 85 The first test score is 85.

Solve

Then substitute 85 for f in one of the original equations to solve for s. f + s = 164

85 + s = 164 s = 79 The second test score is 79.

Solve

Next substitute 85 for f and 79 for s in f + s + t = 256. f + s + t = 256

85 + 79 + t = 256 164 + t = 256

t = 92 The third test score is 92.

Courtney’s test scores were 85, 79, and 92.

ExamineNow check your results against the original problem. Is the total number of points on the three tests 256

points?85 + 79 + 92 = 256 ✔

Is one test score 6 more than another test score?79 + 6 = 85 ✔

Do two of the tests total 164 points? 85 + 79 =164 ✔

Our answers are correct.

Solutions?

You know that a system of two linear equations doesn’t necessarily have a solution that is a unique ordered pair. Similarly, a system of three linear equations in three variables doesn’t always have a solution that is a unique ordered triple.

Graphs

The graph of each equation in a system of three linear equations in three variables is a plane. Depending on the constraints involved, one of the following possibilities occurs.

Graphs

1. The three planes intersect at one point. So the system has a unique solution.

2. The three planes intersect in a line. There are an infinite number of solutions to the system.

Graphs

3. Each of the diagrams below shows three planes that have no points in common. These systems of equations have no solutions.

Solve this system of equations

123

13

92

z

zy

zyx

Answer:

Substitute 4 for z and 1 for y in the first equation, x + 2y + z = 9 to find x. x + 2y + z = 9 x + 2(1) + 4 = 9 x + 6 = 9 x = 3 Solution is (3, 1, 4)Check:1st 3 + 2(1) +4 = 9 ✔2nd 3(1) -4 = 1 ✔3rd 3(4) = 12 ✔

123

13

92

z

zy

zyx

Solve the third equation, 3z = 123z = 12

z = 4Substitute 4 for z in the second equation 3y – z = -1 to find y.3y – (4) = -1 3y = 3 y = 1

Solve this system of equations

1423

1123

32

zyx

zyx

zyx

Answer:

Set the next two equations together and multiply the first times 2.

2(x + 3y – 2z = 11)2x + 6y – 4z = 223x - 2y + 4z = 15x + 4y = 23

Next take the two equations that only have x and y in them and put them together. Multiply the first times -1 to change the signs.

1423

1123

32

zyx

zyx

zyx

Set the first two equations together and multiply the first times 2.

2(2x – y + z = 3)4x – 2y +2z = 6

x + 3y -2z = 11 5x + y = 17

Answer:

Now you have y = 2. Substitute y into one of the equations that only has an x and y in it.

5x + y = 17 5x + 2 = 17

5x = 15 x = 3

Now you have x and y. Substitute values back into one of the equations that you started with.2x – y + z = 32(3) - 2 + z = 36 – 2 + z = 34 + z = 3z = -1

1423

1123

32

zyx

zyx

zyx

Next take the two equations that only have x and y in them and put them together. Multiply the first times -1 to change the signs.-1(5x + y = 17)-5x - y = -175x + 4y = 23 3y = 6

y = 2

Check your work!!!

Solution is (3, 2, -1)Check:1st 2x – y + z =2(3) – 2 – 1 = 3 ✔2nd x + 3y – 2z = 113 + 3(2) -2(-1) = 11 ✔3rd 3x – 2y + 4z3(3) – 2(2) + 4(-1) = 1 ✔

1423

1123

32

zyx

zyx

zyx

Competition Problems

Find the solution to the linear system:

Answer:

x = 3y = -4

If A x B =8, B x C = 28, and A x C = 14,

find A x B x C

Answer:

56

A polygon is formed by the intersections of

y ≥ 3x – 3y ≤ 3

x ≥ –2 and3x + 2y ≥ –6

What is the area of the polygon?

Answer:

15

Given the system

graph the figure formed by the solution of the system?

2

25.0

5.0

xy

xy

xy

Answer:

scalene triangle

Line m has the equation 3y = 6 + y. Line n is perpendicular to line m and contains

the point (4, –2). When graphed, what is the point of

intersection for lines m and n?

Answer:

(4,3)

Find the area bounded byx = –1, x = 5, y = 0 and y = –2x + 13.

Answer:

54