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To determine the number of solutions for the given system of linear equations:
[tex]\[ \begin{aligned} y &= \frac{2}{3}x + 2 \\ 6x - 4y &= -10 \end{aligned} \][/tex]
we need to follow these steps.
1. Convert the first equation to standard form:
The first equation is given in slope-intercept form [tex]\( y = \frac{2}{3}x + 2 \)[/tex]. To convert this to the standard form [tex]\( Ax + By = C \)[/tex], we can move all terms involving [tex]\(x\)[/tex] and [tex]\(y\)[/tex] to one side:
[tex]\[ y - \frac{2}{3}x = 2 \implies \frac{2}{3}x - y = -2 \quad \text{or} \quad 2x - 3y = -6 \quad \text{(multiplying through by 3)} \][/tex]
2. Write the system of equations in standard form:
Now, the system of equations can be written as:
[tex]\[ \begin{aligned} 2x - 3y &= -6 \quad \text{(1)} \\ 6x - 4y &= -10 \quad \text{(2)} \end{aligned} \][/tex]
3. Setting up the determinant:
We use the determinant method (Cramer's Rule) to determine if the system has a unique solution, no solution, or infinitely many solutions. The determinant ([tex]\(\Delta\)[/tex]) of the coefficients is calculated as follows:
[tex]\[ \Delta = \left| \begin{array}{cc} 2 & -3 \\ 6 & -4 \\ \end{array} \right| = (2)(-4) - (6)(-3) = -8 + 18 = 10 \][/tex]
4. Check the determinant ([tex]\(\Delta\)[/tex]):
Since [tex]\(\Delta \neq 0\)[/tex], the system has a unique solution.
5. Find the unique solution ([tex]\(x\)[/tex] and [tex]\(y\)[/tex]) using Cramer's Rule:
[tex]\[ x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta} \][/tex]
where [tex]\(\Delta_x\)[/tex] and [tex]\(\Delta_y\)[/tex] are the determinants of matrices obtained by replacing the coefficients of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] with the constants from the right-hand side.
[tex]\[ \Delta_x = \left| \begin{array}{cc} -6 & -3 \\ -10 & -4 \\ \end{array} \right| = (-6)(-4) - (-10)(-3) = 24 - 30 = -6 \][/tex]
[tex]\[ \Delta_y = \left| \begin{array}{cc} 2 & -6 \\ 6 & -10 \\ \end{array} \right| = (2)(-10) - (6)(-6) = -20 + 36 = 16 \][/tex]
With [tex]\(\Delta = 10\)[/tex], we have:
[tex]\[ x = \frac{-6}{10} = -0.6, \quad y = \frac{16}{10} = -1.6 \][/tex]
6. Conclusion:
The system has a unique solution, which is [tex]\((-0.6, -1.6)\)[/tex].
Therefore:
[tex]\[ \text{One solution:} \ (-0.6, -1.6) \][/tex]
[tex]\[ \begin{aligned} y &= \frac{2}{3}x + 2 \\ 6x - 4y &= -10 \end{aligned} \][/tex]
we need to follow these steps.
1. Convert the first equation to standard form:
The first equation is given in slope-intercept form [tex]\( y = \frac{2}{3}x + 2 \)[/tex]. To convert this to the standard form [tex]\( Ax + By = C \)[/tex], we can move all terms involving [tex]\(x\)[/tex] and [tex]\(y\)[/tex] to one side:
[tex]\[ y - \frac{2}{3}x = 2 \implies \frac{2}{3}x - y = -2 \quad \text{or} \quad 2x - 3y = -6 \quad \text{(multiplying through by 3)} \][/tex]
2. Write the system of equations in standard form:
Now, the system of equations can be written as:
[tex]\[ \begin{aligned} 2x - 3y &= -6 \quad \text{(1)} \\ 6x - 4y &= -10 \quad \text{(2)} \end{aligned} \][/tex]
3. Setting up the determinant:
We use the determinant method (Cramer's Rule) to determine if the system has a unique solution, no solution, or infinitely many solutions. The determinant ([tex]\(\Delta\)[/tex]) of the coefficients is calculated as follows:
[tex]\[ \Delta = \left| \begin{array}{cc} 2 & -3 \\ 6 & -4 \\ \end{array} \right| = (2)(-4) - (6)(-3) = -8 + 18 = 10 \][/tex]
4. Check the determinant ([tex]\(\Delta\)[/tex]):
Since [tex]\(\Delta \neq 0\)[/tex], the system has a unique solution.
5. Find the unique solution ([tex]\(x\)[/tex] and [tex]\(y\)[/tex]) using Cramer's Rule:
[tex]\[ x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta} \][/tex]
where [tex]\(\Delta_x\)[/tex] and [tex]\(\Delta_y\)[/tex] are the determinants of matrices obtained by replacing the coefficients of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] with the constants from the right-hand side.
[tex]\[ \Delta_x = \left| \begin{array}{cc} -6 & -3 \\ -10 & -4 \\ \end{array} \right| = (-6)(-4) - (-10)(-3) = 24 - 30 = -6 \][/tex]
[tex]\[ \Delta_y = \left| \begin{array}{cc} 2 & -6 \\ 6 & -10 \\ \end{array} \right| = (2)(-10) - (6)(-6) = -20 + 36 = 16 \][/tex]
With [tex]\(\Delta = 10\)[/tex], we have:
[tex]\[ x = \frac{-6}{10} = -0.6, \quad y = \frac{16}{10} = -1.6 \][/tex]
6. Conclusion:
The system has a unique solution, which is [tex]\((-0.6, -1.6)\)[/tex].
Therefore:
[tex]\[ \text{One solution:} \ (-0.6, -1.6) \][/tex]
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