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By Root 304 0

y = form, but you can use it anytime you can conveniently put one of the equations into that form. If trying to solve for x or for y in terms of the other variable is too difficult or gives you a messy substitution, combinations may be a better method.

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EXERCISE 5.3

Solve each system of equations by substitution.

1. y = x

x + y = 10

2. x + y = 12

y = 2x

3. x − y = −6

y = 3x

4. x + y = 57

x = y + 3

5. x + 2y = 65

y = x + 4

6. x − y = 46

x = 7y − 2

7. 3x − 4y = 5

y = 2x − 5

8. 5x + 7y = 73

x − 2y = 1

9. 5x − 3y = 49

x − 4y = 3

10. y = 2x + 7

y = 3x + 8

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Solving systems of equations by combination

If both equations are in standard form(or can easily be transformed to standard form), then the method of linear combinations, or elimination, allows you to add or subtract the equations in a way that will make one variable drop out. Make sure that both equations are in standard form and that like terms are aligned under one another before you begin.

Addition

If the coefficients of one variable are opposites, add the equations to eliminate that variable.

Solve for the remaining variable and substitute back into either equation to find the value of the second variable.

2(−2) − 3y = −10

−4 − 3y = −10

−3y = −6

y = 2

Subtraction

If the coefficients of one variable are identical, you can subtract the equations to eliminate that variable, or you can multiply one of the equations by −1 and then add.

Solve for the remaining variable and substitute back into either equation to find the value of the second variable.

4(5) + 2y = 12

20 + 2y = 12

2y = −8

y = −4

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EXERCISE 5.4

Solve each system of equations.

1. x + y = 8

x − y = 4

2. x + y = 17

x − y = 3

3. 2x + y = 9

x − y = 3

4. 3x + 2y = 23

x − 2y = 5

5. 5x + 12y = 16

2x − 12y = −2

6. 8x + y = 19

2x + y = 7

7. 7x + 5y = 21

7x + 9y = 21

8. x + 9y = 93

x + 4y = 43

9. x + y = 1

2x + y = 9

10. x + 2y = 6

3x − 2y = 6

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Combinations with multiplication

If the coefficients of the variable are neither identical nor opposite, you can still use combinations, but you need to do a little work first. Multiply one or both equations by constants to create equivalent equations with matching or opposite coefficients.

If you need to solve the system , you can multiply the top equation by 2 and the bottom equation by 3. That will turn the system into . This system has the same solution as the original, but in this version, you can eliminate y by adding the equations. Once the coefficients are identical or opposite, eliminate a variable by adding or subtracting, and then substitute back into one of the original equations.

Be careful to multiply through the entire equation by the constant. Multiplying the variable terms but not the constant is a common error. Checking your solution in both of the original equations will help you catch your errors.

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EXERCISE 5.5

Solve each system of equations.

1. 3a + 2b = 15

2a + b = 8

2. 3x + 2y = 23

x − y = −9

3. 8x + 2y = 16

2x − y = 7

4. 5x − 2y = 29

3x + 4y = 33

5. 4x − 3y = 14

3x + 2y = 19

6. 2x + 3y = 4

3x − 8y = −9

7. 3x − 7y = 30

5x + y = 12

8. 7x − 3y = 1

2x − y = 1

9. 2x + 4y = 5

4x + 5y = 6

10. 2x + 3y = 21

4x + y = 9

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Dependent and inconsistent systems

Not every system of equations has a unique solution. Since parallel lines never intersect, if the two equations produce parallel lines when graphed, the system has no solution. This occurs when the equations have the same slope but different y-intercepts. A system that has no solution is inconsistent. A system that has a unique solution is consistent.

If a system is made up of two equations that produce the same graph, every point on the line is a point of intersection, and so the system has infinitely many solutions. Such a system is dependent. You can recognize a dependent system because one equation will be a constant multiple of the other.

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EXERCISE 5.6

Label each system as consistent, inconsistent, or dependent.

1.