IDNLearn.com connects you with a global community of knowledgeable individuals. Get the information you need from our community of experts who provide accurate and comprehensive answers to all your questions.
Sagot :
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.
[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.
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is committed to providing accurate answers. Thanks for stopping by, and see you next time for more solutions.