systems of linear equations &...

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Systems of Linear Equations

& Inequalities

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Table of ContentsTe

ache

r not

e

Solve Systems by Graphing

Solve Systems by Substitution

Solve Systems by Elimination

Choosing your Strategy

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Strategy One:Graphing

Return to Table of Contents

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Some vocabulary...

The "solution" to a system is an ordered pair that will work in each equation. One way to find the solution is to graph the equations on the same coordinate plane and find the point of intersection.

A "system" is two or more linear equations.

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Suppose you are walking to school. Your friend is 5 blocks ahead of you. You can walk two blocks per minute, your friend can walk one block per minute. How many minutes will it take for you to catch up with your friend?

Consider this...

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Time (min.)

Friend's distance from

your start (blocks)

Your distance from your start

(blocks)

0 5 0

1 6 2

2 7 4

3 8 6

4 9 8

5 10 10

First, make a table to represent the problem.

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Next, plot the points on a graph.

Time (min.)

Blo

cks

05

20

15

10

1510

5

0

Time (min.)

Friend's distance from your

start (blocks)

Your distance from your

start(blocks)

0 5 0

1 6 2

2 7 4

3 8 6

4 9 8

5 10 10

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The point where they intersect is the solution to the system.

Time (min.)

Blo

cks

05

20

15

10

1510

5

0

(5,10) is the solution. In the context of the problem this means after 5 minutes, you will meet your friend

at block 10.

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Solve the system of equations graphically.

y = 2x -3y = x - 1

Solu

tion

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2x + y = 3x - 2y = 4

Solu

tion

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3x + y = 11x - 2y = 6

Solu

tion

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Solve using graphing

y = 4x+6movey = -3x-1moveWrite the equation forthe green line

Write the equation forthe blue line

What is this pointof intersection?(move the hand!)

(-1, 2)

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( , )-1 2y = 4x+6y = -3x-1

Now take the ordered pair we just found and substitute it into the equation to prove that it is a solution for both lines.

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y = 2x + 3

Solve by Graphingy = -4x - 3

(-1,1)

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y= x - 4y= -3x + 4

Solve by Graphing

(2,-2)

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What's the problem here? y= 2x - 4y= 2x + 4

Parallel lines do not intersect!

Therefore there is no solution.

No ordered pair that will work in BOTH equations

( )

click toreveal click to

reveal

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2y = -4x + 102 2 y = -2x + 5

2x + y = 5 -2x -2x y = -2x + 5

Solve by GraphingFirst - transform the equations into y = mx + b

form (slope-intercept form)

Now graph the two transformed lines.

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2y = 10 -4x becomesy = -2x + 5

2x + y = 5 becomesy = -2x + 5

What's the problem?

The equations

transform to the same

line.

So we have infinitely

many solutions.

click toreveal

click toreveal

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1 Solve the system by graphing.y = -x + 4y = 2x +1

A (3,1)

B (1,3)

C (-1,3)

D no solution

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2 Solve the system by graphing.y = 0.5x - 1y = -0.5x -1

A (0,-1)

B (0,0)

C infinitely many

D no solution

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3 Solve the system by graphing.2x + y = 3x - 2y = 4

A (2,4)

B (0.4, 2.2)

C (2, -1)

D no solution

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4 Solve the system by graphing.y = 3x + 3y = 3x - 3

A (0,0)

B (3,3)

C infinitely many

D no solution

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5 Solve the system by graphing.y = 3x + 44y = 12x + 16

A (3,4)

B (-3,-4)

C infinitely many

D no solution

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Strategy Two:Substitution


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y = x + 6.1y = -2x - 1.4

NO

TE

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Graphing can be inefficient or approximate.

Another way to solve a system is to use substitution.

Substitution allows you to create a one variable equation.

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Solve the system using substitution. Why was it difficult to solve this system by graphing?

y = x + 6.1y = -2x - 1.4

y = -2x - 1.4 -start with one equationx + 6.1 = -2x - 1.4 -substitute x + 6.1 for y in equation+2x -6.1 +2x - 6.1 3x = -7.5 -solve for x x = -2.5

Substitute -2.5 for x in either equation and solve for y. y = x + 6.1 y = (-2.5) + 6.1 y = 3.6

Since x = -2.5 and y = 3.6, the solution is (-2.5, 3.6)

CHECK: See if (-2.5, 3.6) satisfies the other equation. y = -2x - 1.43.6 = -2(-2.5) - 1.43.6 = 5 - 1.43.6 = 3.6?

?

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+ 3x = 21-3 y

y = -2x +14

Solve the system using substitution.

( )

Solu

tion

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= -y - 3x

x = -5y - 39

Solve the system using substitution.

( )

Solu

tion

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Examine each system of equations.Which variable would you choose to substitute?Why?

y = 4x - 9.6y = -2x + 9

y = -3x7x - y = 42

y = 4x + 1x = 4y + 1

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6 Examine the system of equations. Which variable would you substitute?

2x + y = 52y = 10 - 4x

A x B y

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2y - 8 = xy + 2x = 4

A x B y

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x - y = 202x + 3y = 0

A x B y

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Sometimes you need to rewrite one of the equations so that you can use the substitution method. For example:

The system: Is equivalent to:3x -y = 5 y = 3x -52x + 5y = -8 2x + 5y = -8

Using substitution you now have: 2x + 5(3x-5) = -8 -solve for x2x + 15x - 25 = -8 -distribute the 5 17x - 25 = -8 -combine x's 17x = 17 -at 25 to both sides x = 1 - divide by 17

Substitute x = 1 into one of the equations.2(1) + 5y = -8 2 + 5y = -8 5y = -10 y = -2

The ordered pair (1,-2) satisfies both equations in the original system. 3x -y = 5 2x + 5y = -83(1) - (-2) = 5 2(1) + 5(-2) = -8 3 + 2 = 5 2 - 10 = -8 -8 = -8

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Your class of 22 is going on a trip. There are four drivers and two types of vehicles, vans and cars. The vans seat six people, and the cars seat four people, including drivers. How many vans and cars does the class need for the trip?

Let v = the number of vansand c = the number of cars

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Set up the system:

Drivers: v + c = 4 People: 6v + 4c = 22

Solve the system by substitution. v + c = 4 -solve the first equation for v. v = -c + 4 -substitute -c + 4 for v in the 6(-c + 4) + 4c = 22 second equation -6c + 24 + 4c = 22 -solve for c -2c + 24 = 22 -2c = -2 c = 1

v + c = 4 v + 1 = 4 -substitute for c in the 1st equation v = 3 -solve for v

Since c = 1 and v = 3, they should use 1 car and 3 vans.

Check the solution in the equations: v + c = 4 6v + 4c = 22 3 + 1 = 4 6(3) + 4(1) = 22 4 = 4 18 + 4 = 22 22 = 22

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Now solve this system using substitution. What happens? x + y = 6 5x + 5y = 10

x + y = 6 -solve the first equation for x x = 6 - y5(6 - y) + 5y = 10 -substitute 6 - y for x in 2nd equation 30 - 5y + 5y = 10 -solve for y 30 = 10 -FALSE!

Since 30 = 10 is a false statement, the system has no solution.

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Now solve this system using substitution. What happens? x + 4y = -3 2x + 8y = -6

x + 4y = -3 - solve the first equation for x x = -3 - 4y2(-3 - 4y) + 8y = -6 - sub. -3 - 4y for x in 2nd equation -6 - 8y + 8y = -6 - solve for y -6 = -6 - TRUE! - there are infinitely many solutions

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How can you quickly decide the number of solutions a system has?

1 Solution Different slopes

No Solution Same slope; different y-intercept (Parallel Lines)

Infinitely Many Same slope; same y-intercept (Same Line)

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9 3x - y = -2 y = 3x + 2

A 1 solution

B no solution

C infinitely many solutions

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10 3x + 3y = 8 y = x

A 1 solution

B no solution


1 3

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11 y = 4x 2x - 0.5y = 0

A 1 solution

B no solution


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12 3x + y = 5 6x + 2y = 1

A 1 solution

B no solution


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13 y = 2x - 7 y = 3x + 8

A 1 solution

B no solution


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Teac

her n

ote

14 Solve each system by substitution.y = x - 3y = -x + 5

A (4,9)

B (-4,-9)

C (4,1)

D (1,4)

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15 Solve each system by substitution.y = x - 6y = -4

A (-10,-4)

B (-4,2)

C (2,-4)

D (10,4)

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16 Solve each system by substitution.y + 2x = -14y = 2x + 18

A (1,20)

B (1,18)

C (8,-2)

D (-8,2)

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17 Solve each system by substitution.4x = -5y + 50x = 2y - 7

A (6,6.5)

B (5,6)

C (4,5)

D (6,5)

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18 Solve each system by substitution.y = -3x + 23-y + 4x = 19

A (6,5)

B (-7,5)

C (42,-103)

D (6,-5)

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Strategy Three:Elimination


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When both linear equations of a system are in Standard Form, Ax + By = C, you can solve the system using elimination.

You can add or subtract the equations to eliminate a variable.

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How do you decide which variable to eliminate?

First, look to see if one variable has the same or opposite coefficients. If so, eliminate that variable.

Second, look for which coefficients have a simple least common multiple. Eliminate that variable.

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If the variables have the same coefficient, you can subtract the two equations to eliminate the variable.

If the variables have opposite coefficients, you add the two equations to eliminate the variable.

Sometimes, you need to multiply one, or both, equations by a number in order to create a common coefficient.

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5x + y = 44-4x - y = -34

Solve by Elimination - Click on the terms to eliminate and they will disappear, then add

the two equations together.

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3x + y = 15-3x -3y = -21

Solve by Elimination - Click on the terms and they will disappear then add the two

equations together.

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5x + y = 17-2x + y = -4

Solve by Elimination - There are 2 ways to complete this problem. See both examples.

Mul

tiplic

atio

n by

-1

Sub

tract

ion

5x + y = 17-2x + y = -4

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Solve the system by elimination.

4x + 3y = 16 2x - 3y = 8 Pu

llP

ull

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19

A (5,1)

B (-5,-1)

C (1,5)

D no solution

Solve each system by elimination.x + y = 6x - y = 4

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20 Solve each system by elimination.2x + y = -52x - y = -3

A (-2,1)

B (-1,-2)

C (-2,-1)

D infinitely many

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21 Solve each system by elimination.2x + y = -63x + y = -10

A (4,2)

B (3,5)

C (2,4)

D (-4,2)

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22 Solve each system by elimination.4x - y = 5x - y = -7

A no solution

B (4,11)

C (-4,-11)

D (11,-4)

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23 Solve each system by elimination.3x + 6y = 48-5x + 6y = 32

A (2,-7)

B (7,2)

C (2,7)

D infinitely many

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Sometimes, it is not possible to eliminate a variable by adding or subtracting the equations.

When this is the case, you need to multiply one or both equations by a nonzero number in order to create a common coefficient. Then add or subtract the equations.

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Examine each system of equations.Which variable would you choose to eliminate?What do you need to multiply each equation by?

2x + 5y = -1 x + 2y = 0

3x + 8y = 815x - 6y = -39

3x + 6y = 62x - 3y = 4

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In order to eliminate the y, you need to multiply first.

3x + 4y = -10 5x - 2y = 18

Multiply the second equation by 2 so the coefficients are opposites. 2(5x - 2y = 18) 10x - 4y = 36

Now solve by adding the equations together. 3x + 4y = -10 10x - 4y = 36 13x = 26 x = 2

Solve for y, by substituting x = 2 into one of the equations. 3x + 4y = -10 3(2) + 4y = -10 6 + 4y = -10 4y = -16 y = -4

So (2,-4) is the solution.

Check: 3x + 4y = -10 5x - 2y = 183(2) + 4(-4) = -10 5(2) - 2(-4) = 18 6 + -16 = -10 10 + 8 = 18 -10 = -10 18 = 18

+

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Now solve the same system by eliminating x. What do you multiply the two equations by?

3x + 4y = -10 5x - 2y = 18

Multiply the first equation by 5 and the second equation by 3 so the coefficients will be the same 5(3x + 4y = -10) 3(5x - 2y = 18) 15x + 20y = -50 15x - 6y = 54

Now solve by subtracting the equations. 15x + 20y = -50 15x - 6y = 54 26y = -104 y = -4

Solve for x, by substituting y = -4 into one of the equations. 3x + 4y = -10 3x + 4(-4) = -10 3x + -16 = -10 3x = 6 x = 2

So (2,-4) is the solution. Check: 3x + 4y = -10 5x - 2y = 183(2) + 4(-4) = -10 5(2) - 2(-4) = 18 6 + -16 = -10 10 + 8 = 18 -10 = -10 18 = 18

-

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24 Which variable can you eliminate with the least amount of work?

A x

B y 9x + 6y = 15-4x + y = 3

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A x

B y 3x - 7y = -2-6x + 15y = 9

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A x

B y x - 3y = -72x + 6y = 34

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27 What will you multiply the first equation by in order to solve this system using elimination?

2x + 5y = 203x - 10y = 37

Now solve it.... (11, ) 25-

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3x + 2y = -19x - 12y = 19

Now solve it.... (-5,-2)


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x + 3y = 43x + 4y = 2

Now solve it.... (-2,2)


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Choose Your Strategy


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Altogether 292 tickets were sold for a basketball game. An adult ticket costs $3. A student ticket costs $1.

Ticket sales were $470.

Let a = adults s = students

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Set up the system: money collected: 3a + s = 470 number of tickets sold: a + s = 292

First eliminate one variable. 3a + s = 470 - in both equations s has the same a + s = 292 coefficient so you subtract the 2 2a+ 0 = 178 equations in order to eliminate it. a = 89 -solve for a

Then, find the value of the eliminated variable. a + s = 29289 + s = 292 -substitute 89 for a in 1st equation s = 203 -solve for s

There were 89 adult tickets and 203 student tickets sold.

(89, 203)

Check: a + s = 292 3a + s = 47089 + 203 = 292 3(89) + 203 = 470 292 = 292 267 + 203 = 470 470 = 470

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30 A piece of glass with an initial temperature of 99 F is cooled at a rate of 3.5 F/min. At the same time, a piece of copper with an initial temperature of 0 F is heated at a rate of 2.5 F/min. Let m = the number of minutes and t = the temperature in F. Which system models the given information?

A B Ct = 99 + 3.5mt = 0 + 2.5m

t = 99 - 3.5mt = 0 + 2.5m

t = 99 + 3.5mt = 0 - 2.5m

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31 Which method would you use to solve the system?

A graphing

B substitution

C elimination

t = 99 - 3.5mt = 0 + 2.5m

Now solve it...m = 16.5 t = 41.25

This means that in 16.5 minutes, the temperatures will both be 41.25℃.

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32 What method would you choose to solve the system?

A graphing

B substitution

C elimination

4s - 3t = 8t = -2s -1

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D (-2, )

33 Now solve the system!

A ( , -2) 4s - 3t = 8t = -2s -1

1 2

B ( , 2)

1 2

C (2 , -2)

1 2

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A graphing

B substitution

C elimination

y = 3x - 1y = 4x

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35 Now solve it!

A (1, 4)

B (-4, -1)

C (-1, 4)

y = 3x - 1y = 4x

D (-1, -4)

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A graphing

B substitutionC elimination

3m - 4n = 13m - 2n = -1

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37 Now solve it!

A (-2, -1)

B (-1, -1)

C (-1, 1)

3m - 4n = 13m - 2n = -1

D (1, 1)

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A graphing

B substitution

C elimination

y = -2xy = -0.5x + 3

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39 Now solve it!

A (-6, 12)

B (2, -4)y = -2xy = -0.5x + 3

C (-2, 4)

D (1, -2)

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A graphing

B substitution

C elimination

2x - y = 4x + 3y = 16

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41 Now solve it!

A (6, 5)

B (-4, 7)

C (-4, 4)

2x - y = 4x + 3y = 16

D (4, 4)

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A graphing

B substitution

C elimination

u = 4v3u - 3v = 7

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43 Now solve it!

A ( , )B ( , )C (28, 7)

u = 4v3u - 3v = 7

D (7, ) 7 4

28 9

28 9

7 9

7 9

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