To solve the system of equations by substitution, we need to determine which variable would be the easiest to isolate from one of the equations. Let's analyze our options step by step.
The given system of equations is:
[tex]\[
\begin{array}{l}
3x + 6y = 9 \quad \text{(Equation 1)} \\
2x - 10y = 13 \quad \text{(Equation 2)}
\end{array}
\][/tex]
1. Option A: Solve for \( x \) in the second equation
- Equation 2: \( 2x - 10y = 13 \)
- Solving for \( x \):
[tex]\[
2x = 13 + 10y \\
x = \frac{13 + 10y}{2}
\][/tex]
This is a valid option, but involves dealing with a fraction which can be more complex to substitute later.
2. Option B: Solve for \( y \) in the first equation
- Equation 1: \( 3x + 6y = 9 \)
- Solving for \( y \):
[tex]\[
6y = 9 - 3x \\
y = \frac{9 - 3x}{6} \\
y = \frac{3 - x}{2}
\][/tex]
This introduces a fraction as well, but it's a possible path.
3. Option C: Solve for \( y \) in the second equation
- Equation 2: \( 2x - 10y = 13 \)
- Solving for \( y \):
[tex]\[
-10y = 13 - 2x \\
y = \frac{2x - 13}{10}
\][/tex]
This also includes a fraction, which adds complexity.
4. Option D: Solve for \( x \) in the first equation
- Equation 1: \( 3x + 6y = 9 \)
- Solving for \( x \):
[tex]\[
3x = 9 - 6y \\
x = 3 - 2y
\][/tex]
This approach results in a simple expression for \( x \) which is straightforward to substitute into the other equation.
Given these steps, the best variable to solve for is \( x \) in the first equation because it results in a relatively simple expression without fractions. Thus, the correct answer is:
D. [tex]\( x \)[/tex], in the first equation
