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CLASS NOTES: §9 – 1 thru §9 – 5 Systems of Equations §9 – 1: The Graphing Method What is a LINEAR EQUATION?

What is a SYSTEM OF LINEAR EQUATIONS?

What does it mean to SOLVE A SYSTEM OF LINEAR EQUATIONS?

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x − 2y = 54x + 3y = 9

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HOW DO I SOLVE a system of linear equations?

There are THREE different ways to do this. You can use whichever method you prefer since they will all give you the same answer. The THREE ways (called “methods”) are:

• Drawing a GRAPH (§9 – 1)

• SUBSTITUTION (§9 – 2)

• ELIMINATION (also called “ADDITION”) (§9 – 4 and §9 – 5)

In this case, the point of intersection is

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3 , −1( ). This is the SOLUTION for

this system.

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EX 1 Solve the system by graphing.

2x − y = 8x + y = 1

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EX 2 Solve the system by graphing.

x − 2y = −6x − 2y = 2

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EX 3 Solve the system by graphing.

2x + 3y = 64x + 6y = 12

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§9 – 2: The Substitution Method

EX 1 x + y = 6x − y = 4

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First, solve one equation for y.

Substitute this expression for y into the other equation and solve for x.

Substitute the value of x in the equation in step 1 and solve for y.

The Graphing Method – To solve a system of linear equations in two variables, draw the graph of each linear equation in the same coordinate plane.

4. If the lines intersect, there is only one solution which is the point of intersection.

5. If the lines are parallel, there is no solution.

6. If the lines coincide, there are infinitely many solutions.

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EX 2 2x − 3y = 4x + 4y = −9

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First, solve one equation for x.

Substitute this expression for x into the other equation and solve for y.

Substitute the value of y in the equation in step 1 and solve for x.

EX 3 2x − 8y = 6x − 4y = 8

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This false statement means that there is no point of intersection of the two lines.

• This means the system has no solution.

• This also means that the two

lines are parallel.

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EX 4 y2

= 2 − x

6x + 3y = 12

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Multiply both sides by 2 to get y by itself.

This true statement means that there are many points of intersection between the two lines.

• This means the system has infinitely many solutions.

• This also means that the two

equations represent the same line.

The Substitution Method – To solve a system of linear equations in two variables:

1. Solve one equation for one of the variables.

2. Substitute this expression in the other equation and solve for the other variable.

3. Substitute this value in the equation in step 1 and solve for the other variable.

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§9 – 4: The Addition or Subtraction Method (Elimination)

EX 1 Solve 5x − y = 123x + y = 4

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Add the similar terms of the two equations. Solve the resulting equation. Substitute the x value into either of the two original equations to find y.

EX 2 Solve 6c + 7d = −156c − 2c = 12

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Multiply one of the equations by -1. Then add the similar terms of the equations. Solve the resulting equation. Substitute the value for d into either of the two original equations to find c.

Notice that the y’s are eliminated.

If you add these two equations, c will not eliminate because they are both positive. One needs to be negative!

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EX 3 Solve x + y = 14x − y = 4

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EX 4 Solve 3r + 2s = 23r + s = 7

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The Addition (Elimination) Method – To solve a system of linear equations in two variables:

1. Add the equations to eliminate one variable. If the two equations have the same sign, multiply one by -1 so they have opposite signs.

2. Solve the resulting equation for the remaining variable.

3. Substitute the known variable into either of the two original equations and solve for the other variable.

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§9 – 5: Multiplication with the Addition Method (Elimination)

EX 1 Solve 4x − 5y = 233x + 10y = 31

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Multiply the first equation by 2. Then add the two equations. The y terms will eliminate.

Solve the resulting equation.

Substitute the x value into either of the two original equations to find y.

EX 2 Solve 6x + y = 63x + 2y = 9

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Notice that no variable can be eliminated by adding these equations. However, if we multiply the first equation by 2, then it will have -10y and the y’s will eliminate!

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EX 3 Solve 3a + 4b = 25a + 9b = 1

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Multiply the first equation by 5. Multiply the second equation by -3.

EX 4 Solve 4s − 5t = 33s + 2t = −15

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Transform BOTH equations by multiplication so that the a terms are opposites of the same number.