Question

solve using elimination.
2x + 3y = 6
4x - 3y = 8
(give details)

Answer 1

To solve the system of equations using the elimination method, we have the following two equations:

1. \( 2x + 3y = 6 \) \ \ \ (Equation 1)
2. \( 4x - 3y = 8 \) \ \ \ (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.