Certainly! Let's solve the system of equations using the steps provided by both Tyler and Han to show that both methods yield the same solution.
The system of equations is:
[tex]\[
\begin{cases}
x + 3y = -5 \\
9x + 3y = 3
\end{cases}
\][/tex]
### Solution via Tyler's Method
Step 1: Isolate \( x \) in the first equation.
[tex]\[
x + 3y = -5 \\
x = -5 - 3y
\][/tex]
Step 2: Substitute \( x = -5 - 3y \) into the second equation.
[tex]\[
9(-5 - 3y) + 3y = 3 \\
-45 - 27y + 3y = 3 \\
-45 - 24y = 3 \\
-24y = 48 \\
y = -2
\][/tex]
Step 3: Substitute \( y = -2 \) back into \( x = -5 - 3y \) to find \( x \).
[tex]\[
x = -5 - 3(-2) \\
x = -5 + 6 \\
x = 1
\][/tex]
So, the solution using Tyler's method is \( (x, y) = (1, -2) \).
### Solution via Han's Method
Step 1: Isolate \( 3y \) in the first equation.
[tex]\[
x + 3y = -5 \\
3y = -5 - x
\][/tex]
Step 2: Express \( y \) in terms of \( x \).
[tex]\[
3y = -5 - x \\
y = \frac{-5 - x}{3}
\][/tex]
Step 3: Substitute \( y = \frac{-5 - x}{3} \) into the second equation.
[tex]\[
9x + 3\left(\frac{-5 - x}{3}\right) = 3 \\
9x - 5 - x = 3 \\
8x - 5 = 3 \\
8x = 8 \\
x = 1
\][/tex]
Step 4: Substitute \( x = 1 \) back into \( y = \frac{-5 - x}{3} \) to find \( y \).
[tex]\[
y = \frac{-5 - 1}{3} \\
y = \frac{-6}{3} \\
y = -2
\][/tex]
So, the solution using Han's method is \( (x, y) = (1, -2) \).
### Conclusion:
Both Tyler's and Han's methods yield the same solution for the system of equations:
[tex]\[
\boxed{(x, y) = (1, -2)}
\][/tex]
