To solve the given system of equations:
[tex]\[ \begin{cases}
3x + y = 2 \\
7x - 4y = 30
\end{cases} \][/tex]
we can use the method of substitution or elimination. Here, I'll demonstrate using the elimination method.
### Step 1: Align the equations
First, we write the equations clearly one under the other:
[tex]\[
\begin{array}{l}
3x + y = 2 \quad \text{(Equation 1)} \\
7x - 4y = 30 \quad \text{(Equation 2)} \\
\end{array}
\][/tex]
### Step 2: Make coefficients of [tex]\( y \)[/tex] identical
To eliminate [tex]\( y \)[/tex], we can manipulate the equations so that the coefficients of [tex]\( y \)[/tex] in both equations are the same in absolute value. We can achieve this by multiplying Equation 1 by 4:
[tex]\[
\begin{array}{l}
4(3x + y) = 4(2) \\
\Rightarrow 12x + 4y = 8 \quad \text{(Equation 3)}
\end{array}
\][/tex]
Now we have:
[tex]\[
\begin{array}{l}
12x + 4y = 8 \quad \text{(Equation 3)} \\
7x - 4y = 30 \quad \text{(Equation 2)} \\
\end{array}
\][/tex]
### Step 3: Add the equations to eliminate [tex]\( y \)[/tex]
Add Equation 3 and Equation 2 to eliminate [tex]\( y \)[/tex]:
[tex]\[
(12x + 4y) + (7x - 4y) = 8 + 30
\][/tex]
Simplify the left-hand side and the right-hand side:
[tex]\[
12x + 4y + 7x - 4y = 38
\][/tex]
[tex]\[
19x = 38
\][/tex]
### Step 4: Solve for [tex]\( x \)[/tex]
Divide both sides by 19:
[tex]\[
x = \frac{38}{19}
\][/tex]
[tex]\[
x = 2
\][/tex]
### Step 5: Substitute [tex]\( x \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]
Use Equation 1 to find [tex]\( y \)[/tex]:
[tex]\[
3x + y = 2
\][/tex]
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
3(2) + y = 2
\][/tex]
[tex]\[
6 + y = 2
\][/tex]
Subtract 6 from both sides:
[tex]\[
y = 2 - 6
\][/tex]
[tex]\[
y = -4
\][/tex]
### Conclusion
The solution to the system of equations is:
[tex]\[
(x, y) = (2, -4)
\][/tex]
So, the correct answer is [tex]\( (2, -4) \)[/tex].
