4.3 Solving Systems of Linear
Equations by Elimination (Addition)
Solving Using Elimination 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 (eliminating variable).
4. Solve the remaining equation.
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. Done
2. x + y = 3
x–y=5
3. 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
add
Soln: (4, -1)
6. Check:
x+y=3
4 + (-1) = 3
3=3
x–y=5
4 – (-1) = 3
4+1=5
5=5
Ex. Solve by the addition method: x + y = 9
-x + y = -3
1. Done
2. x + y = 9
-x + y = -3
3. x + y = 9
-x + y = -3
4. 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
add
Soln: (6, 3)
6. Check:
x+y=9
6+3=9
9=9
-x + y = -3
-6 + 3 = -3
-3 = -3
Ex. Solve by the addition method: -5x + 2y = -6
10x + 7y = 34
1. Done
2. -5x + 2y = -6  2(-5x + 2y)=2(-6)  -10x + 4y = -12
10x + 7y = 34  10x + 7y = 34  10x + 7y = 34
3. -10x + 4y = -12
add
10x + 7y = 34
4.
0 + 11y = 22
11y = 22
11y = 22
11 11
y=2
5.
10x + 7y = 34
6. Check:
10x + 7(2) = 34 sub 2 for y
-5x + 2y = -6
10x + 7y = 34
10x + 14 = 34
-5(2) + 2(2) = -6 10(2) + 7(2) = 34
10x + 14 – 14 = 34 – 14
-10 + 4 = -6
20 + 14 = 34
10x = 20
-6 = -6
34 = 34
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 + 2x
2x – 7y = -9
2. 3x + 2y = -1  7(3x + 2y) =7(-1)  21x + 14y = -7
2x – 7y = -9  2(2x – 7y) = 2(-9)  4x – 14y = -18
3. 21x + 14y = -7 add
4x – 14y = -18
4. 25x + 0 = -25
25x = -25
25x = -25
25
25
x = -1
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
No variables remain and a TRUE stmt.
2. -2x – 4y = 1
lines coincide
2x + 4y = -1
infinite number of solns.
3. -2x – 4y = 1
dependent eqns.
add
2x + 4y = -1
Soln: Infinitely many solutions
4. 0 + 0 = 0
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 = -21
No variables remain and a FALSE stmt.
2. -3x – 6y = 4
lines are parallel
3x + 6y = -21
no solution
3. -3x – 6y = 4
inconsistent system
add
3x + 6y = -21
Answer: no solution
4.
0 + 0 = -17