Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
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]
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
