Systems of Linear Equations Solving 2 Equations. Review of an Equation & It’s Solution Algebra is primarily about solving for variables. The value or

Systems of Linear Equations

Solving 2 Equations

€

3x + 2y = 9

y = 3x − 7

Review of an Equation & It’s Solution

Algebra is primarily about solving for variables. The value or values of the variable which make the EQUATIONS true are SOLUTIONS.

SOLUTIONS make up the GRAPH of the equation.


The graph of a one variable equation is a number on the number line. (3x=21 ) x=7

70•


The graph of a one variable equation is a number on the number line. (3x=21 ) x=7

The graph of an inequality is a dot and heavy line & an arrow on a number line. (3x>21 ) x>7

70•

70


The graph of one variable equation is a number on the number line. (3x=21 ) x=7

The graph of an inequality is a dot and heavy line & arrow on a number line. (3x>21 ) x>7

The graph of a linear equation is a line.

y=x-1

70•

70

(0,-1)





y=x-1

The graph of a quadratic equation

is a parabola. y=x2-1

70•

70

(0,-1)

(0,-1)





y=x-1



SOLUTIONS ARE THE GRAPH

70•

70

(0,-1)

(0,-1)

A Linear Equation & It’s Solution


y = x-1(0,-1)



y = x-1(0,-1)

Every Point on the Line is a Solution.



y = x-1(0,-1)

Every Point on the Line is a Solution.

To determined if a point is a solution (on the line) plug in the x & y values into the equation and see if it is true.



y = x-1(0,-1)

Is (5,4) a solution of y = x-1?



y = x-1(0,-1)

Is (5,4) a solution of y = x-1?

(0,-1)

• (5,4)

y = x-1?

4 = 5 - 1 so (5,4) is a solution

Yes it is

Yes it is

(0,-1)

y = x-1?

4 = 5 - 1 so (5,4) is a solution

• (5,4)

Review of Solving aSolution of One Equation

SOLUTIONS OF a LINEAR EQUATION

Is (2,-3) a solution of y = 2x-7



1. Put (2,-3) (x,y) valuesinto the equation.

y = 2x-7-3=2(2)-7

x y




2. IF THE EQUATION ISTRUE THE POINT (2,-3)IS A SOLUTION.

If the equation is not truethe point isn’t a solution

y = 2x-7-3=2(2)-7-3=4-7-3=-3

SO (2,-3) IS ASOLUTION

x y






y = 2x-7-3=2(2)-7-3=4-7-3=-3


x y

Systems of Equations

Given 2 linear equations.

WHAT IS THE SOLUTION?

Systems of Equations Given 2 linear equations

The single point where they intersect is the solution.

•

Systems of EquationsHave 3 Possible Answers


•

ONE

(Lines Intersect)


•

ONE NONE

(Lines Intersect) (Lines are Parallel)


•

ONE NONE INFINITE

(Lines Intersect) (Lines are Parallel) (2 lines on each other)


The Solution is where the two lines meet (or intersect)

(0,0) 1 2 3

1

2

4

3

5 6€

y = x −1

m2 = 1

€

y = −x + 3

m1 = −1

-1

• 1 Solution

(0,-1)

(0,3)

(2,1)


(0,0) 1 2 3

1

2

4

3

5 6-1

How many solutions are there to parallel lines?

(0,-1)

(0,3)

(2,1)


(0,0) 1 2 3

1

2

4

3

5 6-1

Parallel Lines have No Solutions(They never meet)

(0,-1)

(0,3)

(2,1)


(0,0) 1 2 3

1

2

4

3

5 6-1

Similar LinesHave Solutions(They meet everywhere)

(0,-1)

(0,3)

(2,1)

Determine if a given point is a Solution to a Sys of Eq.

€

y = −x + 3

y = x −1Is (5,4) a solution?

1. Put (x,y) point into each equation.

2. If both equations are true the point is a solution.


€

y = −x + 3

y = x −1Is (5,4) a solution?



€

4 = −5 + 3?

4 = 5 −1?

Not True

(5,4) IS NOT A SOLUTION

True


€

2x + 3y =12

x − 4y = −5Is (3,2) a solution?



€

5x − 2y = −5

3x − 7y = −32Is (1,5) a solution?1.

€

x = −1

x − y = −2Is (-1,1) a solution?2.

€

y =1

4x

3x − y = 33

Is (12,3) a solution?

3.

AM 182&3SOLVING Systems of Equations GRAPHICALLY

€

y = −x + 3

y = x −1Solve


€

y = −x + 3

y = x −1Solve

1. Get each equation in y=mx+b form

Both equations are already solved for y.

(0,0) 1 2 3

1

2

4

3

5 6

€

y = −x + 3

m1 = −1

-1(0,-1)

(0,3)


€

y = −x + 3

y = x −1Solve


2. Graph 1st Line

(0,0) 1 2 3

1

2

4

3

5 6€

y = x −1

m2 = 1

€

y = −x + 3

m1 = −1

-1(0,-1)

(0,3)


€

y = −x + 3

y = x −1Solve


2. Graph 1st Line3. Graph 2nd Line

(0,0) 1 2 3

1

2

4

3

5 6€

y = x −1

m2 = 1

€

y = −x + 3

m1 = −1

-1

• Solution

(0,-1)

(0,3)

(2,1)


€

y = −x + 3

y = −x −1Solve


2. Graph 1st Line3. Graph 2nd Line4. Sol. Is

Intersection

AM 182&3 Solve the system graphically

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

Find Solution

1. Get each equation in y = mx + b form.

2. Graph each equation

3. Solution is the intersection

Note: Parallel lines have no solutions & if the lines are the same they have infinite solutions.

–x –x

y = –x +5

x + y = 5 +2x

+2x

y = 2x -4

-2x + y = -4

the solution is (3, 2)


€

y = −2x + 3

y = 2x −1Solve

AM 184&5 SOLVING Systems of Equations BY SUBSTITUTION

€

y = −x + 3

y = x −1


€

y = −x + 3

y = x −1

1. Solve one equation for one variable.(Already done in this case)


€

y = −x + 3

y = x −1


2. Substitute results into the other equation.

€

−x + 3 = x −1


+1 +1+x+x€

y = −x + 3

y = x −1



3. Solve to get the value of one of the variables.

€

4 = 2x

€

−x + 3 = x −1

€

x = 2


+1 +1+x+x€

y = −x + 3

y = x −1




4. Substitute this into either of theoriginal equations and solve forother variable.

€

4 = 2x

€

−x + 3 = x −1

€

x = 2

€

y = 2 −1

y =1


+1 +1+x+x€

y = −x + 3

y = x −1





5. Answer is this (x,y) Point

€

4 = 2x

€

−x + 3 = x −1

€

x = 2

€

y = 2 −1

y =1

(2,1)


+1 +1+x+x€

y = −x + 3

y = x −1





5. Answer is this (x,y) Point

6. Check by putting (x,y) back into originalequations.

€

4 = 2x

€

−x + 3 = x −1

€

x = 2

€

y = 2 −1

y =1

(2,1)

€

1= −2 + 3

1= 2 −1

SOLVING Systems of Equations BY SUBSTITUTION & GRAPHING

+1 +1+x+x€

y = −x + 3

y = x −1

€

4 = 2x

€

−x + 3 = x −1

€

x = 2

€

y = 2 −1

y =1

(2,1)(0,0) 1 2 3

1

2

4

3

€

y = x − 1

€

y = − x + 3

-1

• Solution(2,1)

Solve by Graphing & Substitution

€

x + 2y = 7

x = y − 2

€

y = −2x − 9

x = −5

Prob 1 Prob 2

Prob 3 Prob 4

SOLVING Systems of Equations BY ELIMINATION/ADDITION

ONE VARIABLE IS ELIMINATED BY ADDING TWO EQUATIONS TOGETHER

SOLVING Systems of Equations BY ELIMINATION/ADDITION

€

5 + 3 = 8

3 + 2 = 5

8 + 5 = 13Equations can easily be added and the new equation is true

AM 186&7 SOLVING Systems of Eq. By ELIM/ADD

€

y = −x + 3

y = x −1Solve

€

y = −x + 3

y = +x −1€

y = −x + 3

y = x −1

1. Line up equation variablesand #.

SolveAM 186&7 SOLVING Systems of Eq. By ELIMINATION/ADD

€

y = −x + 3

y = +x −1€

y = −x + 3

y = x −1


2. Add Combining Like Terms

€

2y = 0 + 2

AM 186&7 SOLVING Systems of Eq. By ELIMINATION/ADD

€

y = −x + 3

y = +x −1€

y = −x + 3

y = x −1


2. Add Combining Like Terms

3. Solve for 1 variable.

€

2y = 0 + 22 2

€

y = +1


€

y = −x + 3

y = +x −1€

y = −x + 3

y = x −1


2. Combine Like Terms

3. Solve for 1 variable.

4. Put answer into either equation and solve for the other variable.

€

2y = 0 + 22 2

€

y = +1

€

1= x −1

€

2 = x

+1+1

(2,1)

Answer is the Point where lines cross


€

y = −x + 3

y = x −1


2. Combine Like Terms3. Solve for 1 variable.4. Put answer into either

equation and solve for the other variable.

5. CHECK ANS. BY PUTTINGANS. BACK INTO EACH EQ.

(2,1)

€

y = −x + 3

1= −2 + 3

1=1

y = x −1

1= 2 −1

1=1


(0,0) 1 2 3

1

2

4

3

€

y = x −1

€

y = −x + 3

-1

• Solution(2,1)

Comparison of SOLVING Systems of Equations BY GRAPHING & ELIM/ADD

€

y = −x + 3

y = x −1

€

2y = 2

y = 1

€

y = x −1

1= x −1

2 = x (2,1)

€

y = −x + 3

y = x −1

SOLVE FOR (x,y) BY ADDITION

€

y = −x + 3

y = x + 25

€

2y = −x + 7

−y = x + 3

€

y = −x + 3

−y = −x −1€

y = −x + 3

y = x −1

SOLUTIONS

€

y = −x + 3

y = x + 25

2y = 28

y = 14

€

14 = −x + 3

x = −11

-14+x +x-14

(-11,14)

€

2y = −x + 7

−y = x + 3

y = 10

€

2(10) = −x + 7

20 = −x + 7

−13 = x

-7+x -7

(-13,10)

€

y = −x + 3

−y = −x −1

0 = −2x + 2

2x = 2

x = 1

€

y = −(1) + 3

y = 2

-1+x +x-1

(1,2)

€

y = −x + 3

y = x −1

2y = +2

y = 1 (2,1)

€

1 = −x + 3

x = 2

What if just adding 2 linear equations do not eliminate one of the variables?

Multiply one or both equations to eliminate one variable by adding.

€

4y = −2x + 2

y = x + 5

Solve: Find (x,y)Point where theseTwo lines cross.

AM 186&7 SOLVE Systems of Eq. By ELIM./ADD

€

4y = −2x + 2

y = x + 5

1. Multiply the second equation by 2 to eliminate a variable when adding equations.

x2

€

4y = −2x + 2

2y = 2x +10


€

4y = −2x + 2

y = x + 5


€

4y = −2x + 2

y = x + 5


2. Add equations

x2

€

4y = −2x + 2

2y = 2x +10


€

4y = −2x + 2

y = x + 5

€

6y =12


€

4y = −2x + 2

y = x + 5


2. Add equations

3. Solve for one variable

x2

€

4y = −2x + 2

2y = 2x +10


€

4y = −2x + 2

y = x + 5

€

6y =12

y = 2


€

4y = −2x + 2

y = x + 5


2. Add equations


4. Sub. found variable into either eq. to find other variable.

x2

€

4y = −2x + 2

2y = 2x +10


€

4y = −2x + 2

y = x + 5

€

6y =12

y = 2

€

2 = x + 5

−3 = x

-5-5

(-3,2) ANSWER


€

4y = −2x + 2

y = x + 5


2. Add equations


4. Sub. found variable into either eq. to find other variable.

x2

€

4y = −2x + 2

2y = 2x +10


€

4y = −2x + 2

y = x + 5

€

6y =12

y = 2

€

2 = x + 5

−3 = x

-5-5

(-3,2) ANSWER


REVIEW Multiply to Set-Up

Solving by Addition

€

4y = −2x + 3

y = x + 5

Multiplying the second equation by 2 allows us to eliminate a variable by adding equations.

x2

€

4y = −2x + 3

2y = 2x +10

(Remember to multiply both sides of the equation by 2)

Sometimes you have to multiply both equations

€

4y = −3x + 3

3y = 5x + 5

Sometimes you have to multiply both equations

€

4y = −3x + 3

3y = 5x + 5 x(-4)

€

12y = −9x + 9

−12y = 20x + 20

x3

Now y can be eliminated by

Adding equations.

What would you have to do to set these equations up for

addition?

€

2y = −5x + 6

7y = 3x + 5

1.


addition?

€

2y = −5x + 6

7y = 3x + 5 x(-2)

€

14y = −35x + 42

−14y = −6x +10

x(7)

Now y can be eliminated by

Adding equations.

1.


addition?

€

2y = −5x + 6

7y = 3x + 5 x(5)

€

6y = −15x +18

35y = 15x + 25

x(3)

Now x can be eliminated by

Adding equations.

Or

1.

What would you have to do to set these equations up for addition?

€

3x = −5y + 6

y = 3x − 22.

2.

2. Add equations to solve

1. Line up Like Terms €

3x = −5y + 6

y = 3x − 2

€

3x = −5y + 6

y = 3x − 2-3x-3x -y-y

€

−3x = −y − 2



€

2y + x = −4

7 = 3x + 5y

3.

€

2y + x = −4

7 = 3x + 5y

3.

-5y-5y

€

−3x − 5y = −72. Add equations to solve

€

10y + 5x = −20

-7-3x -7-3x •5

1. Line up Like Terms

€

2y + x = −4

7 = 3x + 5y



€

8 = −x + 9y

4y = 3x + 6

4.

4.

2. Add equations to solve

1. Line up Like Terms

€

8 = −x + 9y

4y = 3x + 6

€

8 = −x + 9y

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

€

0 = 3x − 4 y + 6

€

24 = −3x + 27y

•3



The graph of one variable equation is a number onthe number line. (3x=21 ) x=7

The graph of an inequality is a dot and heavy line &arrow on a number line. (3x>21 ) x>7


y=x-1



70•

70

(0,-1)

(0,-1)

SOLUTIONS ARE THE GRAPH

SOLUTIONS OF aLINEAR EQUATION





y = 2x-7-3=2(2)-7-3=4-7-3=-3


x y

€

y = − x + 3

y = x − 1


2. Combine Like Terms3. Solve for 1 variable.4. Put answer into either

equation and solve for the other variable.

5. CHECK ANS. BY PUTTINGANS. BACK INTO EACH EQ.

(2,1)

€

y = − x + 3

1 = − 2 + 3

1 = 1

y = x − 1

1 = 2 − 1

1 = 1

AM 186&7 SOLVING Systemsof Eq. By ELIMINATION/ADD

Determine if a given pointis a Solution to a Sys of Eq.

€

y = − x + 3

y = x − 1

(5,4) Is a?solution

1. ( , ) Put x y point into .each equation

2. If both equations are true the point is

.a solution€

4 = − 5 + 3 ?

4 = 5 − 1 ?

Not True

(5,4) IS NOT A SOLUTION

True

Systems of EquationsGiven 2 linear equations

The single point where theyintersect is the solution.

•

€

4 y = − 2 x + 2

y = x + 5

1. Multiply the secondequation by 2 toeliminate a variablewhen addingequations.

2. Add equations


4. Sub. found variableinto either eq. to findother variable.

x2

€

4 y = − 2 x + 2

2 y = 2 x + 10


€

4 y = − 2 x + 2

y = x + 5

€

6 y = 12

y = 2

€

2 = x + 5

− 3 = x

-5-5

(-3,2) ANSWER

AM 186&7 SOLVE Systemsof Eq. By ELIM./ADD


•

ONE NONE INFINITE

(Lines Intersect) (Lines are Parallel) (2 lines on each other)

SOLVING Systems of Equations BYSUBSTITUTION & GRAPHING

+1 +1+x+x€

y = − x + 3

y = x − 1

€

4 = 2 x

€

− x + 3 = x − 1

€

x = 2

€

y = 2 − 1

y = 1

(2,1) (0,0) 1 2 3

1

2

4

3

€

y = x − 1

€

y = − x + 3

-1

• Solution(2,1)

Documents

Systems of Linear Equations Solving 2 Equations. Review of an Equation & It’s Solution Algebra is primarily about solving for variables. The value or