To solve the system of equations using the elimination method, we have the following two equations:
- \( 2x + 3y = 6 \) \ \ \ (Equation 1)
- \( 4x - 3y = 8 \) \ \ \ (Equation 2)
Step 1: Align the Equations
We will align the equations for clarity:
\[ \begin{align*} 2x + 3y & = 6 \quad (1) \ 4x - 3y & = 8 \quad (2) \end{align*} \]
Step 2: Eliminate \(y\)
To eliminate \(y\), we can add the two equations. Notice that the coefficient of \(y\) in Equation 1 is positive (+3) and in Equation 2 is negative (-3). When we add both equations, the \(y\) terms will cancel out.
First, let's set the equations so we can add them directly:
\[ \begin{align*} 2x + 3y & = 6 \
- (4x - 3y & = 8) \end{align*} \]
Now, perform the addition:
\[ (2x + 4x) + (3y - 3y) = 6 + 8 \]
This simplifies to:
\[ 6x + 0y = 14 \]
So we are left with:
\[ 6x = 14 \]
Step 3: Solve for \(x\)
Now we can solve for \(x\):
\[ x = \frac{14}{6} = \frac{7}{3} \]
Step 4: Substitute Back to Find \(y\)
Next, we substitute \(x = \frac{7}{3}\) back into one of the original equations to find \(y\). We will use Equation 1:
\[ 2x + 3y = 6 \]
Substituting for \(x\):
\[ 2\left(\frac{7}{3}\right) + 3y = 6 \]
This simplifies to:
\[ \frac{14}{3} + 3y = 6 \]
To isolate \(3y\), we first subtract \(\frac{14}{3}\) from both sides. We convert 6 into a fraction with a denominator of 3:
\[ 6 = \frac{18}{3} \]
So now we have:
\[ 3y = \frac{18}{3} - \frac{14}{3} = \frac{4}{3} \]
Now, divide by 3:
\[ y = \frac{4}{3} \div 3 = \frac{4}{9} \]
Final Solution
The solution to the system of equations is:
\[ \boxed{\left(\frac{7}{3}, \frac{4}{9}\right)} \]
This means the values \(x = \frac{7}{3}\) and \(y = \frac{4}{9}\) satisfy both equations in the original system.