To solve the given system of equations by substitution, we want to isolate one variable from one of the equations. Let's consider the options.
Given system of equations:
[tex]\[
\begin{array}{l}
2x + 8y = 12 \\
3x - 8y = 11
\end{array}
\][/tex]
Option A: Solve for [tex]\( y \)[/tex] in the first equation
1. Start with the first equation: [tex]\( 2x + 8y = 12 \)[/tex].
2. Isolate [tex]\( y \)[/tex]: [tex]\( 8y = 12 - 2x \)[/tex].
3. Divide by 8: [tex]\( y = \frac{12 - 2x}{8} \)[/tex].
This step isolates [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex].
Option B: Solve for [tex]\( x \)[/tex] in the first equation
1. Start with the first equation: [tex]\( 2x + 8y = 12 \)[/tex].
2. Isolate [tex]\( x \)[/tex]: [tex]\( 2x = 12 - 8y \)[/tex].
3. Divide by 2: [tex]\( x = \frac{12 - 8y}{2} \)[/tex].
This step isolates [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex].
Option C: Solve for [tex]\( x \)[/tex] in the second equation
1. Start with the second equation: [tex]\( 3x - 8y = 11 \)[/tex].
2. Isolate [tex]\( x \)[/tex]: [tex]\( 3x = 11 + 8y \)[/tex].
3. Divide by 3: [tex]\( x = \frac{11 + 8y}{3} \)[/tex].
This step isolates [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex].
Option D: Solve for [tex]\( y \)[/tex] in the second equation
1. Start with the second equation: [tex]\( 3x - 8y = 11 \)[/tex].
2. Isolate [tex]\( y \)[/tex]: [tex]\( -8y = 11 - 3x \)[/tex].
3. Divide by -8: [tex]\( y = \frac{3x - 11}{8} \)[/tex].
This step isolates [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex].
To determine which is the best variable to solve for, we look for the easiest and most direct approach. Solving for [tex]\( y \)[/tex] in the first equation involves straightforward arithmetic and no negative coefficients in the divisor, which is a simple and direct way to start. Thus, the best variable to solve for is [tex]\( y \)[/tex] in the first equation.
Therefore, the correct answer is:
A. [tex]\( y \)[/tex], in the first equation.
