Understanding the System of Linear Equations

We are given a system of two linear equations with two variables, x and y:

1. \(x + 2y = 13\)
2. \(3x + 6y = 9\)

To determine the number of solutions for a system of linear equations in the form \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\), we can compare the ratios of the coefficients and the constant terms.

For the given equations:

• Equation 1: \(a_1 = 1\), \(b_1 = 2\), \(c_1 = 13\)
• Equation 2: \(a_2 = 3\), \(b_2 = 6\), \(c_2 = 9\)

Now, let's calculate the ratios:

We compare these ratios to determine the type of solution(s):

• If \(a_1/a_2 \neq b_1/b_2\), there is a unique solution. The lines intersect at one point.
• If \(a_1/a_2 = b_1/b_2 = c_1/c_2\), there are infinitely many solutions. The lines are coincident (the same line).
• If \(a_1/a_2 = b_1/b_2 \neq c_1/c_2\), there is no solution. The lines are parallel and distinct.

In our case, we have:

\(a_1/a_2 = 1/3\)
\(b_1/b_2 = 1/3\)
\(c_1/c_2 = 13/9\)

Comparing these values, we see that \(a_1/a_2 = b_1/b_2\), but \(c_1/c_2\) is different. Specifically, \(1/3 = 1/3 \neq 13/9\).

This condition, \(a_1/a_2 = b_1/b_2 \neq c_1/c_2\), indicates that the system of equations has no solution.

Geometrically, the two equations represent two distinct parallel lines that never intersect.

Let's check this by trying to manipulate the equations. If we multiply the first equation (\(x + 2y = 13\)) by 3, we get \(3(x + 2y) = 3(13)\), which simplifies to \(3x + 6y = 39\).

Now we compare this with the second equation (\(3x + 6y = 9\)). We have two equations that state the same expression (\(3x + 6y\)) is equal to two different values (39 and 9). This is a contradiction, \(39 \neq 9\), which confirms that there is no pair of values (x, y) that can satisfy both equations simultaneously.

Therefore, the system of equations has no solution.