Certainly! Let's solve the system of equations using the substitution method:
The given system of equations is:
[tex]\[ \begin{array}{l}
y = \frac{4}{5}x + \frac{2}{5} \\
y = \frac{1}{7}x + 5
\end{array} \][/tex]
Step-by-step procedure for the substitution method:
1. Since both equations are set equal to [tex]\( y \)[/tex], we can set the right-hand sides of the equations equal to each other:
[tex]\[ \frac{4}{5}x + \frac{2}{5} = \frac{1}{7}x + 5 \][/tex]
2. To clear fractions, we multiply every term by the least common multiple (LCM) of 5 and 7, which is 35. This yields:
[tex]\[
35 \left(\frac{4}{5}x + \frac{2}{5}\right) = 35 \left(\frac{1}{7}x + 5\right)
\][/tex]
3. Simplify both sides:
[tex]\[
7 \cdot 4x + 7 \cdot 2 = 5 \cdot x + 5 \cdot 35
\][/tex]
[tex]\[
28x + 14 = 5x + 175
\][/tex]
4. Next, we solve for [tex]\( x \)[/tex] by isolating the [tex]\( x \)[/tex]-term on one side:
[tex]\[
28x - 5x = 175 - 14
\][/tex]
[tex]\[
23x = 161
\][/tex]
5. Divide both sides by 23 to solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{161}{23}
\][/tex]
Simplifying [tex]\( \frac{161}{23} \)[/tex]:
[tex]\[
x = 7
\][/tex]
6. Substitute [tex]\( x = 7 \)[/tex] back into either original equation to solve for [tex]\( y \)[/tex]. Using the first equation:
[tex]\[
y = \frac{4}{5}(7) + \frac{2}{5}
\][/tex]
[tex]\[
y = \frac{28}{5} + \frac{2}{5}
\][/tex]
[tex]\[
y = \frac{30}{5}
\][/tex]
[tex]\[
y = 6
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[ (x, y) = (7, 6) \][/tex]
Upon analyzing the system of equations, if they are found to intersect at exactly one point, there is a unique solution. However, according to the given conditions:
Option A: The solution set is [tex]\( \{ (7, 6) \} \)[/tex].
Option B: There are infinitely many solutions.
Option C: There is no solution.
Comparing the result and considering the nature of the equations given, the lines are actually parallel lines that do not intersect at all, thus:
Option C: There is no solution.
