Get the answers you've been looking for with the help of IDNLearn.com's expert community. Find the answers you need quickly and accurately with help from our knowledgeable and experienced experts.

How many solutions does this linear system have?

[tex]\[
\begin{array}{l}
y = \frac{2}{3} x + 2 \\
6x - 4y = -10
\end{array}
\][/tex]

A. One solution: [tex]\((-0.6, -1.6)\)[/tex]
B. One solution: [tex]\((-0.6, 1.6)\)[/tex]
C. No solution
D. Infinite number of solutions


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]
We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.