Certainly! Let's solve the given system of linear equations step-by-step. The system is:
[tex]\[
\begin{array}{l}
-2x - 3y = -23 \\
4x + 6y = 46
\end{array}
\][/tex]
First, let's look at the second equation:
[tex]\[
4x + 6y = 46
\][/tex]
We notice that the second equation can be simplified. Observe that dividing all terms by 2 simplifies it:
[tex]\[
2x + 3y = 23
\][/tex]
So, we now have the two equations:
1. [tex]\(-2x - 3y = -23\)[/tex]
2. [tex]\(2x + 3y = 23\)[/tex]
Notice that these two are actually additive inverses of each other. If we add the equations directly, we observe:
[tex]\[
(-2x - 3y) + (2x + 3y) = -23 + 23
\][/tex]
This results in:
[tex]\[
0 = 0
\][/tex]
The left-hand side cancels out completely, leading to an identity. This indicates that the equations are dependent on each other. This means there are infinitely many solutions which satisfy these two equations.
To express the solution more clearly, let’s solve one of the equations for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]. We'll use the simplified form of the second equation for convenience:
[tex]\[
2x + 3y = 23 \implies 2x = 23 - 3y \implies x = \frac{23 - 3y}{2}
\][/tex]
So the solutions can be expressed parametrically where [tex]\(y\)[/tex] can take any real value, and [tex]\(x\)[/tex] is given by:
[tex]\[
x = \frac{23}{2} - \frac{3}{2}y
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
\boxed{x = \frac{23}{2} - \frac{3}{2}y}
\][/tex]
Where [tex]\(y\)[/tex] is a free parameter that can take any real number.
So, for any real value of [tex]\(y\)[/tex], we can find a corresponding [tex]\(x\)[/tex] using the expression above, satisfying both of the original equations.
