To determine the best variable to solve for and from which equation, let's carefully analyze the given system of equations:
[tex]\[
\begin{array}{l}
3x + 6y = 9 \\
2x - 10y = 13
\end{array}
\][/tex]
### Option A: Solve for [tex]\( x \)[/tex] in the first equation
Starting with the first equation:
[tex]\[
3x + 6y = 9
\][/tex]
To solve for [tex]\( x \)[/tex]:
[tex]\[
3x = 9 - 6y
\][/tex]
[tex]\[
x = \frac{9 - 6y}{3}
\][/tex]
This simplifies to:
[tex]\[
x = 3 - 2y
\][/tex]
The solution for [tex]\( x \)[/tex] in the first equation is relatively straightforward.
### Option B: Solve for [tex]\( y \)[/tex] in the second equation
Starting with the second equation:
[tex]\[
2x - 10y = 13
\][/tex]
To solve for [tex]\( y \)[/tex]:
[tex]\[
-10y = 13 - 2x
\][/tex]
[tex]\[
y = \frac{13 - 2x}{-10}
\][/tex]
This simplifies to:
[tex]\[
y = -\frac{13}{10} + \frac{1}{5}x
\][/tex]
The solution for [tex]\( y \)[/tex] involves dealing with a negative fraction and may be considered a bit more complex.
### Option C: Solve for [tex]\( y \)[/tex] in the first equation
Starting with the first equation:
[tex]\[
3x + 6y = 9
\][/tex]
To solve for [tex]\( y \)[/tex]:
[tex]\[
6y = 9 - 3x
\][/tex]
[tex]\[
y = \frac{9 - 3x}{6}
\][/tex]
This simplifies to:
[tex]\[
y = \frac{3 - x}{2}
\][/tex]
The solution for [tex]\( y \)[/tex] in the first equation is also straightforward with simple fractions.
### Option D: Solve for [tex]\( x \)[/tex] in the second equation
Starting with the second equation:
[tex]\[
2x - 10y = 13
\][/tex]
To solve for [tex]\( x \)[/tex]:
[tex]\[
2x = 13 + 10y
\][/tex]
[tex]\[
x = \frac{13 + 10y}{2}
\][/tex]
This simplifies to:
[tex]\[
x = \frac{13}{2} + 5y
\][/tex]
Among these options, solving for [tex]\( x \)[/tex] in the first equation looks to involve fewer steps and simpler fractions compared to the other options.
Thus, the best option is:
### A. [tex]\( x \)[/tex], in the first equation.
