Let's solve the given system of equations:
[tex]\[
\begin{array}{l}
3x - 4y = 6 \quad \text{(Equation 1)} \\
x = \frac{4}{3} y + 2 \quad \text{(Equation 2)}
\end{array}
\][/tex]
### Step 1: Substitution
Since Equation 2 directly provides a value for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex], we can substitute this expression into Equation 1.
From Equation 2:
[tex]\[
x = \frac{4}{3} y + 2
\][/tex]
Substitute this into Equation 1:
[tex]\[
3\left(\frac{4}{3} y + 2\right) - 4y = 6
\][/tex]
### Step 2: Simplifying the Equation
First, distribute the 3:
[tex]\[
3 \left(\frac{4}{3} y\right) + 3 \cdot 2 - 4y = 6
\][/tex]
[tex]\[
4y + 6 - 4y = 6
\][/tex]
Now, simplify the equation:
[tex]\[
4y - 4y + 6 = 6
\][/tex]
[tex]\[
6 = 6
\][/tex]
### Step 3: Analyzing the Result
The result [tex]\( 6 = 6 \)[/tex] indicates that the equation is always true and does not depend on the value of [tex]\( y \)[/tex]. This means that there are infinitely many solutions, and the relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] can be described as:
### Step 4: Expressing the Relationship
From Equation 2, we have:
[tex]\[
x = \frac{4}{3} y + 2
\][/tex]
This equation indicates a linear relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex]. For any value of [tex]\( y \)[/tex], we can compute the corresponding value of [tex]\( x \)[/tex] using this relationship.
### Final Solution
The system of equations can be represented by the equation:
[tex]\[
x = \frac{4}{3} y + 2
\][/tex]
Or, equivalently using decimal values:
[tex]\[
x = 1.33333333333333y + 2.0
\][/tex]
This expression provides the relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in this system, representing an infinite number of solutions along the line described by this equation.
