Let's solve the given system of equations step-by-step to determine their solution set and potential dependency:
1. System of Equations:
[tex]\[
\begin{array}{r}
\frac{7}{4} x - \frac{7}{3} y = 0 \\
y = \frac{3}{4} x
\end{array}
\][/tex]
2. Rewriting the Second Equation:
The second equation is already solved for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{3}{4} x
\][/tex]
3. Form of the First Equation:
The first equation can be rewritten as:
[tex]\[
\frac{7}{4} x - \frac{7}{3} y = 0
\][/tex]
4. Both equations are linear and can be compared:
To compare whether these equations are dependent, let's express the second equation in a similar format to the first one:
[tex]\[
y = \frac{3}{4} x \implies \frac{3}{4} x - y = 0
\][/tex]
Note that it can be rewritten as:
[tex]\[
\frac{3}{4} x - 1 \cdot y = 0
\][/tex]
5. Comparing Coefficient Ratios:
Now, let’s compare the coefficients:
First Equation: [tex]\(\frac{7}{4} x - \frac{7}{3} y = 0\)[/tex]
- [tex]\(a_1 = \frac{7}{4}\)[/tex]
- [tex]\(b_1 = -\frac{7}{3}\)[/tex]
Second Equation: [tex]\(\frac{3}{4} x - y = 0\)[/tex]
- [tex]\(a_2 = \frac{3}{4}\)[/tex]
- [tex]\(b_2 = -1\)[/tex]
Calculate the ratios:
[tex]\[
\frac{a_1}{a_2} = \frac{\frac{7}{4}}{\frac{3}{4}} = \frac{7}{3}
\][/tex]
[tex]\[
\frac{b_1}{b_2} = \frac{-\frac{7}{3}}{-1} = \frac{7}{3}
\][/tex]
Since the ratios [tex]\(\frac{a_1}{a_2}\)[/tex] and [tex]\(\frac{b_1}{b_2}\)[/tex] are equal, the equations are dependent.
6. Type of Solution:
Since the equations are dependent, they represent the same line in a two-dimensional plane. As such, there are infinitely many solutions to this system.
7. Solution Set:
The solution set [tex]\( \{(x, y)\} \)[/tex] represents all points [tex]\((x, y)\)[/tex] that satisfy either of the given equations.
To conclude:
- The equations are dependent.
- The system has infinitely many solutions.
- The solution set is [tex]\(\{(x, y)\}\)[/tex].
