Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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]
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
