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To determine whether the substitution method or the elimination method is better for solving the following system of equations, we'll need to analyze the structure of the equations:
[tex]\[ \begin{array}{l} -2x + 2y = 9 \\ 3x + 6y = -12 \end{array} \][/tex]
### Analysis for Substitution Method
The substitution method involves solving one equation for one of the variables and then substituting this expression into the other equation. Let’s see if this is straightforward:
1. Solving for one variable:
From the first equation [tex]\(-2x + 2y = 9\)[/tex], to solve for [tex]\(y\)[/tex]:
[tex]\[ 2y = 2x + 9 \][/tex]
[tex]\[ y = x + \frac{9}{2} \][/tex]
This is relatively straightforward.
2. Substituting into the second equation:
Substitute [tex]\(y = x + \frac{9}{2}\)[/tex] into the second equation:
[tex]\[ 3x + 6\left(x + \frac{9}{2}\right) = -12 \][/tex]
[tex]\[ 3x + 6x + 27 = -12 \][/tex]
[tex]\[ 9x + 27 = -12 \][/tex]
[tex]\[ 9x = -39 \][/tex]
[tex]\[ x = -\frac{39}{9} = -\frac{13}{3} \][/tex]
Substitute [tex]\(x = -\frac{13}{3}\)[/tex] back into [tex]\(y = x + \frac{9}{2}\)[/tex]:
[tex]\[ y = -\frac{13}{3} + \frac{9}{2} \][/tex]
[tex]\[ y = -\frac{26}{6} + \frac{27}{6} \][/tex]
[tex]\[ y = \frac{1}{6} \][/tex]
### Analysis for Elimination Method
The elimination method involves adding or subtracting the equations to eliminate one of the variables:
1. Align coefficients:
The two equations are:
[tex]\[ -2x + 2y = 9 \][/tex]
[tex]\[ 3x + 6y = -12 \][/tex]
To align the coefficients of [tex]\(y\)[/tex], let's multiply the first equation by 3:
[tex]\[ -6x + 6y = 27 \][/tex]
Now we have:
[tex]\[ -6x + 6y = 27 \][/tex]
[tex]\[ 3x + 6y = -12 \][/tex]
2. Subtract the second equation from the first:
[tex]\[ (-6x + 6y) - (3x + 6y) = 27 - (-12) \][/tex]
[tex]\[ -6x + 6y - 3x - 6y = 39 \][/tex]
[tex]\[ -9x = 39 \][/tex]
[tex]\[ x = -\frac{39}{9} = -\frac{13}{3} \][/tex]
Substitute [tex]\(x = -\frac{13}{3}\)[/tex] back into either original equation, let's use the first one:
[tex]\[ -2\left(-\frac{13}{3}\right) + 2y = 9 \][/tex]
[tex]\[ \frac{26}{3} + 2y = 9 \][/tex]
[tex]\[ 2y = 9 - \frac{26}{3} \][/tex]
[tex]\[ 2y = \frac{27}{3} - \frac{26}{3} \][/tex]
[tex]\[ 2y = \frac{1}{3} \][/tex]
[tex]\[ y = \frac{1}{6} \][/tex]
### Conclusion
Both methods lead to the solution [tex]\(x = -\frac{13}{3}\)[/tex] and [tex]\(y = \frac{1}{6}\)[/tex]. However, the elimination method leverages the coefficient alignment, making it straightforward to remove one variable. Therefore, the elimination method is indeed more effective here because:
1. Both equations are already in standard form.
2. It takes advantage of the same coefficients for easier variable elimination.
Thus, the elimination method is better because both equations are written in standard form and two variables have the same coefficient.
[tex]\[ \begin{array}{l} -2x + 2y = 9 \\ 3x + 6y = -12 \end{array} \][/tex]
### Analysis for Substitution Method
The substitution method involves solving one equation for one of the variables and then substituting this expression into the other equation. Let’s see if this is straightforward:
1. Solving for one variable:
From the first equation [tex]\(-2x + 2y = 9\)[/tex], to solve for [tex]\(y\)[/tex]:
[tex]\[ 2y = 2x + 9 \][/tex]
[tex]\[ y = x + \frac{9}{2} \][/tex]
This is relatively straightforward.
2. Substituting into the second equation:
Substitute [tex]\(y = x + \frac{9}{2}\)[/tex] into the second equation:
[tex]\[ 3x + 6\left(x + \frac{9}{2}\right) = -12 \][/tex]
[tex]\[ 3x + 6x + 27 = -12 \][/tex]
[tex]\[ 9x + 27 = -12 \][/tex]
[tex]\[ 9x = -39 \][/tex]
[tex]\[ x = -\frac{39}{9} = -\frac{13}{3} \][/tex]
Substitute [tex]\(x = -\frac{13}{3}\)[/tex] back into [tex]\(y = x + \frac{9}{2}\)[/tex]:
[tex]\[ y = -\frac{13}{3} + \frac{9}{2} \][/tex]
[tex]\[ y = -\frac{26}{6} + \frac{27}{6} \][/tex]
[tex]\[ y = \frac{1}{6} \][/tex]
### Analysis for Elimination Method
The elimination method involves adding or subtracting the equations to eliminate one of the variables:
1. Align coefficients:
The two equations are:
[tex]\[ -2x + 2y = 9 \][/tex]
[tex]\[ 3x + 6y = -12 \][/tex]
To align the coefficients of [tex]\(y\)[/tex], let's multiply the first equation by 3:
[tex]\[ -6x + 6y = 27 \][/tex]
Now we have:
[tex]\[ -6x + 6y = 27 \][/tex]
[tex]\[ 3x + 6y = -12 \][/tex]
2. Subtract the second equation from the first:
[tex]\[ (-6x + 6y) - (3x + 6y) = 27 - (-12) \][/tex]
[tex]\[ -6x + 6y - 3x - 6y = 39 \][/tex]
[tex]\[ -9x = 39 \][/tex]
[tex]\[ x = -\frac{39}{9} = -\frac{13}{3} \][/tex]
Substitute [tex]\(x = -\frac{13}{3}\)[/tex] back into either original equation, let's use the first one:
[tex]\[ -2\left(-\frac{13}{3}\right) + 2y = 9 \][/tex]
[tex]\[ \frac{26}{3} + 2y = 9 \][/tex]
[tex]\[ 2y = 9 - \frac{26}{3} \][/tex]
[tex]\[ 2y = \frac{27}{3} - \frac{26}{3} \][/tex]
[tex]\[ 2y = \frac{1}{3} \][/tex]
[tex]\[ y = \frac{1}{6} \][/tex]
### Conclusion
Both methods lead to the solution [tex]\(x = -\frac{13}{3}\)[/tex] and [tex]\(y = \frac{1}{6}\)[/tex]. However, the elimination method leverages the coefficient alignment, making it straightforward to remove one variable. Therefore, the elimination method is indeed more effective here because:
1. Both equations are already in standard form.
2. It takes advantage of the same coefficients for easier variable elimination.
Thus, the elimination method is better because both equations are written in standard form and two variables have the same coefficient.
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