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# 7.17.2 - PowerPoint PPT Presentation

Systems of Linear Equations and Their Solutions.

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### Systems of Linear Equations and Their Solutions

We have seen that all equations in the form Ax + By = C are straight lines when graphed. Two such equations, such as those listed below, are called a system of linear equations. A solution to a system of linear equations is an ordered pair that satisfies all equations in the system. For example, (3, 4) satisfies the system

x + y = 7 (3 + 4 is, indeed, 7.)

x – y = -1 (3 – 4 is indeed, -1.)

Thus, (3, 4) satisfies both equations and is a solution of the system. The solution can be described by saying that x = 3 and y = 4. The solution can also be described using set notation. The solution set to the system is {(3, 4)} - that is, the set consisting of the ordered pair (3, 4).

Solution Because 4 is the x-coordinate and -1 is the y-coordinate of (4, -1),

we replace x by 4 and y by -1.

x + 2y = 2 x – 2y = 6

4 + 2(-1) = 2 4 – 2(-1) = 6

4 + (-2) = 2 4 – (-2) = 6

2 = 2 true 4 + 2 = 6

6 = 6 true

The pair (4, -1) satisfies both equations: It makes each equation true. Thus, the

pair is a solution of the system. The solution set to the system is {(4, -1)}.

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### Example: Determining Whether an Ordered Pair Is a Solution of a System

Determine whether (4, -1) is a solution of the system

x + 2y = 2

x – 2y = 6.

y

y

x

x

x

Exactly one solution

No Solution (parallel lines)

Infinitely many solutions

(lines coincide)

### The Number of Solutions to a System of Two Linear Equations

The number of solutions to a system of two linear equations in two variables is given by one of the following.

Number of SolutionsWhat This Means Graphically

Exactly one ordered-pair solution The two lines intersect at one point.

No solution The two lines are parallel.

Infinitely many solutions The two lines are identical.

One way to determine the type of solution you expect to get is by looking at the coefficients of each variable in the two equations. Consider the general systems:

Compare the corresponding coefficients:

Same line:

Parallel lines:

Unique Solution:

### Solving Linear Systems by Substitution

• Solve either of the equations for one variable in terms of the other. (If one of the equations is already in this form, you can skip this step.)

• Substitute the expression found in step 1 into the other equation. This will result in an equation in one variable.

• Solve the equation obtained in step 2.

• Back-substitute the value found in step 3 into the equation from step 1. Simplify and find the value of the remaining variable.

• Check the proposed solution in both of the system's given equations.

Step 2Substitute the expression from step 1 into the other equation. We substitute 2y - 3 for x in the first equation.

x = 2y – 3 5 x – 4y = 9

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### Example: Solving a System by Substitution

Solve by the substitution method:

5x – 4y = 9

x – 2y = -3.

Solution

Step 1Solve either of the equations for one variable in terms of the other. We begin by isolating one of the variables in either of the equations. By solving for x in the second equation, which has a coefficient of 1, we can avoid fractions.

x - 2y = -3 This is the second equation in the given system.

x = 2y - 3 Solve for x by adding 2y to both sides.

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### Example: Solving a System by Substitution

Solve by the substitution method:

5x – 4y = 9

x – 2y = -3.

Solution

This gives us an equation in one variable, namely

5(2y - 3) - 4y = 9.

The variable x has been eliminated.

Step 3Solve the resulting equation containing one variable.

5(2y – 3) – 4y = 9 This is the equation containing one variable.

10y – 15 – 4y = 9 Apply the distributive property.

6y – 15 = 9 Combine like terms.

6y = 24 Add 15 to both sides.

y = 4 Divide both sides by 6.

### Example: Solving a System by Substitution

Solve by the substitution method:

5x – 4y = 9

x – 2y = -3.

Solution

Step 4Back-substitute the obtained value into the equation from step 1. Now that we have the y-coordinate of the solution, we back-substitute 4 for y in the equation x = 2y – 3.

x = 2y – 3 Use the equation obtained in step 1.

x = 2 (4) – 3 Substitute 4 for y.

x = 8 – 3 Multiply.

x = 5 Subtract.

With x = 5 and y = 4, the proposed solution is (5, 4).

Step 5Check. Take a moment to show that (5, 4) satisfies both given equations. The solution set is {(5, 4)}.

### Solving Linear Systems by Addition

• If necessary, rewrite both equations in the form Ax + By = C.

• If necessary, multiply either equation or both equations by appropriate nonzero numbers so that the sum of the x-coefficients or the sum of the y-coefficients is 0.

• Add the equations in step 2. The sum is an equation in one variable.

• Solve the equation from step 3.

• Back-substitute the value obtained in step 4 into either of the given equations and solve for the other variable.

• Check the solution in both of the original equations.

Step 1Rewrite both equations in the form Ax + By = C. We first arrange the system so that variable terms appear on the left and constants appear on the right. We obtain

2x - 7y = -17

3x + 5y = 17

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### Example: Solving a System by the Addition Method

2x = 7y - 17

5y = 17 - 3x.

Step 2 If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x-coefficients or the sum of the y-coefficients is 0. We can eliminate x or y. Let's eliminate x by multiplying the first equation by 3 and the second equation by -2.

Multiply by -2.

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Solution

Steps 3 and 4 Add the equations and solve for the remaining variable.

Divide both sides by -31.

Simplify.

Step 5Back-substitute and find the value for the other variable. Back-substitution of 85/31 for y into either of the given equations results in cumbersome arithmetic. Instead, let's use the addition method on the given system in the form Ax + By = C to find the value for x. Thus, we eliminate y by multiplying the first equation by 5 and the second equation by 7.

Multiply by 7.

Solution