Question

Show and explain how replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions as the one shown.

8x + 7y = 39

4x – 14y = –68

Answer 1

To demonstrate how replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions, we'll take the two equations given:

1. \( 8x + 7y = 39 \) (Equation 1)
2. \( 4x - 14y = -68 \) (Equation 2)

### Step 1: Multiply Equation 2 by a Suitable Constant

We want to manipulate these equations. Let's multiply Equation 2 by \( 2 \):

\[
2(4x - 14y) = 2(-68)
\]

This gives us:

\[
8x - 28y = -136 \quad \text{(Equation 3)}
\]

### Step 2: Add Equation 1 to Equation 3

Now, we will add Equation 1 to Equation 3:

\[
(8x + 7y) + (8x - 28y) = 39 - 136
\]

This simplifies to:

\[
8x + 8x + 7y - 28y = 39 - 136
\]

Combine like terms:

\[
16x - 21y = -97 \quad \text{(Equation 4)}
\]

### Step 3: Show Equivalent Systems

Now we have transformed our system from \( (1) \) and \( (2) \) to \( (1) \) and \( (4) \):

1. \( 8x + 7y = 39 \)
2. \( 16x - 21y = -97 \)

The two systems are equivalent, which means they have the same set of solutions.

### Key Concept

The process we used is based on the property that adding or subtracting a multiple of one equation from another does not change the solution set of the system. This is because any solution that satisfies original equations must satisfy the new equation formed from the manipulation, resulting in an equivalent system.

In summary, we replaced Equation 2 with a new equation (where we used a multiple of Equation 2 and added it to Equation 1), and the new equation still represents the same solutions to the original system.