Question

Solve the given system using elimination.

Answer 1

r = 2, s = - 1

Explanation:

Given the 2 equations

2r + 8s = - 4 → (1)

7r = - 6s + 8 ( add 6s to both sides )

7r + 6s = 8 → (2)

Multiplying (1) by 7 and (2) by - 2, then adding will eliminate the r- term

14r + 56s = - 28 → (3)

- 14r - 12s = - 16 → (4)

Add (3) and (4) term by term to eliminate r , that is

44s = - 44 ( divide both sides by 44 )

s = - 1

Substitute s = - 1 into either of the 2 equations and solve for r

Substituting into (1)

2r + 8(- 1) = - 4

2r - 8 = - 4 ( add 8 to both sides )

2r = 4 ( divide both sides by 2 )

r = 2

solution is r = 2 , s = - 1

Answer 2

s = -1

r = 2

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Equality Properties

Algebra I

• Solving systems of equations using substitution/elimination

Explanation:

Step 1: Define Systems

2r + 8s = -4

7r = -6s + 8

Step 2: Rewrite Systems

2r + 8s = -4

1. Factor: 2(r + 4s) = -4
2. Divide 2 on both sides: r + 4s = -2
3. Subtract 4s on both sides: r = -4s - 2
4. Multiply -7 on both sides: -7r = 28s + 14

Step 3: Redefine Systems

-7r = 28s + 14

7r = -6s + 8

Step 4: Solve for s

Elimination

1. Combine equations: 0 = 22s + 22
2. Isolate s term: -22 = 22s
3. Isolate s: -1 = s
4. Rewrite: s = -1

Step 5: Solve for r

1. Define equations: 2r + 8s = -4
2. Substitute in s: 2r + 8(-1) = -4
3. Multiply: 2r - 8 = -4
4. Isolate r term: 2r = 4
5. Isolate r: r = 2