5.3 Solving Systems of Linear Equations by the Addition Method

Solving Using Addition Method

1. Write both eqns. in standard form (Ax + By = C).2. Get opposite coefficients for one of the variables.

You may need to mult. one or both eqns. by a nonzero number to do this.

3. Add the eqns., vertically.4. Solve the remaining eqn. 5. Substitute the value for the variable from step 4

into one of the original eqns. and solve for the other variable.

6. Check soln. in BOTH eqns., if necessary.

Ex. Solve by the addition method: x + y = 3 x – y = 5

1. Done2. x + y = 3

x – y = 53. x + y = 3

x – y = 5 4. 2x + 0 = 8 2x = 8

2x = 8 2 2

x = 4

5. x + y = 3 4 + y = 3 sub 4 for x y + 4 – 4 = 3 – 4

y = -1

Soln: {(4, -1)}

6. Check: x + y = 3 x

– y = 54 + (-1) = 3 4 – (-1) = 3

3 = 3 4 + 1 = 5

5 = 5

add

Ex. Solve by the addition method: x + y = 9 -x + y = -3

1. Done2. x + y = 9 -x + y = -33. x + y = 9

-x + y = -34. 0 + 2y = 6 2y = 6

2y = 6 2 2

y = 3

5. x + y = 9 x + 3 = 9 sub 3 for y x + 3 – 3 = 9 – 3

x = 6

Soln: {(6, 3)}

6. Check: x + y = 9 -x

+ y = -3 6 + 3 = 9 -6 + 3 = -3

9 = 9 -3 = -3

add

Ex. Solve by the addition method: -5x + 2y = -6 10x + 7y = 34

1. Done2. -5x + 2y = -6 2(-5x + 2y)=2(-6) -10x + 4y = -12 10x + 7y = 34 10x + 7y = 34 10x + 7y = 343. -10x + 4y = -12

10x + 7y = 344. 0 + 11y = 22 11y = 22

11y = 22 11 11

y = 2

add

6. Check: -5x + 2y = -6 10x + 7y = 34-5(2) + 2(2) = -6 10(2) + 7(2) = 34 -10 + 4 = -6 20 + 14 = 34 -6 = -6 34 = 34

5. 10x + 7y = 34 10x + 7(2) = 34 sub 2 for y

10x + 14 = 34 10x + 14 – 14 = 34 – 14

10x = 20 10x = 20

10 10 x = 2

Soln: {(2, 2)}

Ex. Solve by the addition method: 3x + 2y = -1 -7y = -2x – 9

1. Rewrite 2nd eqn. in standard form (Ax + By = C)-7y = -2x – 9 -7y + 2x = -2x – 9 + 2x2x – 7y = -9

2. 3x + 2y = -1 7(3x + 2y) =7(-1) 21x + 14y = -7 2x – 7y = -9 2(2x – 7y) = 2(-9) 4x – 14y = -183. 21x + 14y = -7 4x – 14y = -184. 25x + 0 = -25 25x = -25

25x = -25 25 25

x = -1

add

5. 3x + 2y = -1 3(-1) + 2y = -1 sub -1 for x

-3 + 2y = -1 -3 + 2y + 3 = -1 + 3

2y = 2 2y = 2 2 2

y = 1

Soln: {(-1, 1)}

Ex. Solve by the addition method: -2x = 4y + 1 2x + 4y = -1

1. Rewrite 1st eqn. in standard form (Ax + By = C)-2x = 4y + 1-2x – 4y = 4y + 1 – 4y -2x – 4y = 1

2. -2x – 4y = 1 2x + 4y = -1

3. -2x – 4y = 1 2x + 4y = -14. 0 + 0 = 0

add

No variables remain and a TRUE stmt. lines coincideinfinite number of solns.dependent eqns.

Soln: {(x, y)|2x + 4y = -1}

Ex. Solve by the addition method: -3x – 6y = 4 3(x + 2y + 7) = 0

1. Rewrite 2nd eqn. in standard form (Ax + By = C)3(x + 2y + 7) = 0 3x + 6y + 21 = 0 3x + 6y + 21 – 21 = 0 – 21

3x + 6y = -212. -3x – 6y = 4

3x + 6y = -21 3. -3x – 6y = 4 3x + 6y = -214. 0 + 0 = -17

add

No variables remain and a FALSE stmt. lines are parallelno solutioninconsistent system

Answer: no soln. or ø (empty set)

Groups

Page 315 – 316: 27, 41, 59

Groups or class discussion27 -> answer has fractions41-> clear fractions first59-> distribute first