To classify the system of equations, we can follow a detailed process step-by-step. Let's begin by rewriting the given equations in standard form [tex]\(Ax + By = C\)[/tex].
The given system of equations is:
[tex]\[
\begin{aligned}
2x + y - 4 &= 0 \quad \text{(Equation 1)} \\
\frac{1}{3}x + y + 5 &= 0 \quad \text{(Equation 2)}
\end{aligned}
\][/tex]
First, we need to rewrite Equation 2 to eliminate the fraction:
[tex]\[
\begin{aligned}
\frac{1}{3}x + y + 5 &= 0 \\
\Rightarrow 3 \left( \frac{1}{3}x + y + 5 \right) &= 3 \cdot 0 \\
\Rightarrow x + 3y + 15 &= 0 \quad \text{(Equation 2 in standard form)}
\end{aligned}
\][/tex]
Now we have the system of equations:
[tex]\[
\begin{aligned}
2x + y &= 4 \quad \text{(Equation 1 in standard form)} \\
x + 3y &= -15 \quad \text{(Equation 2 in standard form)}
\end{aligned}
\][/tex]
To determine the type of the system, we examine the coefficients of the variables in both equations. This involves setting up the coefficient matrix and checking its determinant. The coefficient matrix [tex]\(A\)[/tex] and the constant matrix [tex]\(B\)[/tex] are:
[tex]\[
A = \begin{pmatrix}
2 & 1 \\
1 & 3
\end{pmatrix}, \quad
B = \begin{pmatrix}
4 \\
-15
\end{pmatrix}
\][/tex]
Next, we calculate the determinant of [tex]\(A\)[/tex]:
[tex]\[
\det(A) = (2 \cdot 3) - (1 \cdot 1) = 6 - 1 = 5
\][/tex]
The determinant of the coefficient matrix [tex]\(A\)[/tex] is 5. Since the determinant is non-zero ([tex]\(\det(A) = 5\)[/tex]), the system of equations is classified as “intersecting”. This means that the two lines represented by these equations intersect at a single unique point.
Thus, the correct answer is:
intersecting
