To solve the system of equations using the substitution method, we start with the two equations:
- \( 2x + 3y = 4 \) (Equation 1)
- \( -4x - 6y = -8 \) (Equation 2)
First, let's solve one of the equations for one variable. We can solve Equation 1 for \( y \):
\[ 3y = 4 - 2x \] \[ y = \frac{4 - 2x}{3} \]
Now that we have \( y \) expressed in terms of \( x \), we can substitute this expression into Equation 2:
Substituting \( y = \frac{4 - 2x}{3} \) into Equation 2:
\[ -4x - 6\left(\frac{4 - 2x}{3}\right) = -8 \]
Simplifying the left side:
\[ -4x - \frac{24 - 12x}{3} = -8 \] \[ -4x - 8 + 4x = -8 \]
The \( -4x \) and \( +4x \) cancel out:
\[ -8 = -8 \]
This is a true statement, which indicates that the two equations are dependent — they represent the same line. Thus, every point on the line is a solution.
To express the solution, we rewrite \( y \) in terms of \( x \):
\[ y = \frac{4 - 2x}{3} \]
So the solution set is:
\[ {(x, \frac{4 - 2x}{3}) \mid x \in \mathbb{R}} \]
This indicates that there are infinitely many solutions that lie along the line represented by \( 2x + 3y = 4 \).
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