To solve the system of linear equations 3 x + 5 y = 2 and 6 x + 10 y = 8 by using the linear combination method, Laura correctly multiplied the first equation by –2 to get Negative 6 x minus 10 y = negative 4 and then correctly added the equations Negative 6 x minus 10 y = negative 4 and 6 x + 10 y = 8 to get 0 = 4. How many solutions are there to the system of equations?

Question

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Answer 1

Let's analyze the system of equations given:

1. \( 3x + 5y = 2 \)
2. \( 6x + 10y = 8 \)

First, we can notice that the second equation is actually a multiple of the first equation. Specifically, if we multiply the first equation by 2, we get:

\[ 2(3x + 5y) = 2 \cdot 2 \implies 6x + 10y = 4 \]

Now, we can compare the two equations:

• The first equation can be rewritten as \( 6x + 10y = 4 \).
• The second equation is \( 6x + 10y = 8 \).

Thus, we have:

1. \( 6x + 10y = 4 \)
2. \( 6x + 10y = 8 \)

When Laura multiplied the first equation by -2, she obtained the equation:

\[ -6x - 10y = -4 \]

Then she added this to the second equation \( 6x + 10y = 8 \):

\[ (-6x - 10y) + (6x + 10y) = -4 + 8 \]

This simplifies to:

\[ 0 = 4 \]

This statement \( 0 = 4 \) is a contradiction, indicating that the system of equations is inconsistent.

Since the two equations represent parallel lines that never intersect, there are no solutions to this system of equations.

Therefore, the answer is:

0 solutions.