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Problem 10

# Determine whether each system of linear equations has (a) one and only one solution, (b) infinitely many solutions, or (c) no solution. Find all solutions whenever they exist. $$\frac{5}{4} x-\frac{2}{3} y=3$$ $$\frac{1}{4} x+\frac{5}{3} y=6$$

Expert verified
The system of linear equations has one unique solution, which is $$x=4$$ and $$y=2$$.
See the step by step solution

## Step 1: Calculate the determinant

To determine the number of solutions possible, we first need to calculate the determinant of the coefficient matrix. The coefficient matrix for the given system of equations is $A = \begin{bmatrix} \frac{5}{4} & -\frac{2}{3} \\ \frac{1}{4} & \frac{5}{3} \end{bmatrix}$ To calculate the determinant, we use the formula $$D = ad - bc$$, where $$a = \frac{5}{4}, \: b = -\frac{2}{3}, \: c = \frac{1}{4}, \: d = \frac{5}{3}$$ So, $D = \frac{5}{4} \cdot \frac{5}{3} - \left(-\frac{2}{3}\right) \cdot \frac{1}{4} = \frac{25}{12} + \frac{1}{6} = \frac{27}{12}$

## Step 2: Analyzing the determinant

Since the determinant $$D = \frac{27}{12}$$ is not zero, the given system of linear equations has a unique solution. Now we will use Cramer's rule to find the solution.

## Step 3: Find the solution using Cramer's rule

Cramer's rule allows us to find the values of $$x$$ and $$y$$ in a unique solution using determinants. The solution for $$x$$ and $$y$$ can be calculated using the formulas: $$x = \frac{D_x}{D}$$ $$y = \frac{D_y}{D}$$ where $$D_x$$ is the determinant obtained by substituting the constant term of each equation for its corresponding $$x$$'s coefficient in matrix $$A$$: $D_x = \begin{vmatrix} 3 & -\frac{2}{3} \\ 6 & \frac{5}{3} \end{vmatrix}$ $$D_y$$ is the determinant obtained by substituting the constant term of each equation for its corresponding $$y$$'s coefficient in matrix $$A$$: $D_y = \begin{vmatrix} \frac{5}{4} & 3 \\ \frac{1}{4} & 6 \end{vmatrix}$ Now, let's calculate $$D_x$$ and $$D_y$$: $$D_x = 3 \cdot \frac{5}{3} - 6 \cdot \left(-\frac{2}{3}\right) = 5 + 4 = 9$$ $$D_y = 6 \cdot \frac{5}{4} - 3 \cdot \frac{1}{4} = \frac{27}{4} - \frac{3}{4} = \frac{24}{4}$$ Now, we can find the values of $$x$$ and $$y$$: $$x = \frac{9}{\frac{27}{12}} = \frac{9}{\frac{27}{12}} \cdot \frac{12}{1} = \frac{108}{27} = 4$$ $$y = \frac{\frac{24}{4}}{\frac{27}{12}} = \frac{6}{\frac{9}{4}} \cdot \frac{4}{1} = 6 \cdot \frac{4}{3} = 2 \cdot 4 = 2$$

## Step 4: Verify the solution

Now that we have found the values of $$x=4$$ and $$y=2$$, let's verify if they satisfy both given equations: 1. $$\frac{5}{4} \cdot 4 - \frac{2}{3} \cdot 2 = 5 - \frac{4}{3} = 5 - \frac{4}{3} = 3$$: True 2. $$\frac{1}{4} \cdot 4 + \frac{5}{3} \cdot 2 = 1 + \frac{10}{3} = 1 + 3\frac{1}{3} = 6$$: True Since both equations are satisfied by the values of $$x=4$$ and $$y=2$$, we can conclude that the given system of linear equations has one unique solution $$(4, 2)$$.

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