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The problem presents two equations as a system to solve by substitution:

\[
\begin{aligned}
x + 3y &= -5 \\
9x + 3y &= 3
\end{aligned}
\]

### Step 1: **Tyler's approach: Isolate \(x\) in the first equation.**

Starting with:
\[
x + 3y = -5
\]

Isolate \(x\):
\[
x = -5 - 3y
\]

### Step 2: **Substitute Tyler's equation into the second equation.**

Substitute \(x = -5 - 3y\) into \(9x + 3y = 3\):

\[
9(-5 - 3y) + 3y = 3
\]

Simplify the equation:
\[
-45 - 27y + 3y = 3
\]

\[
-45 - 24y = 3
\]

Solve for \(y\):
\[
-24y = 48
\]
\[
y = -2
\]

### Step 3: **Substitute \(y = -2\) back into Tyler's equation.**

Now substitute \(y = -2\) into \(x = -5 - 3y\):

\[
x = -5 - 3(-2) = -5 + 6 = 1
\]

Thus, the solution is \(x = 1\) and \(y = -2\).

---

### Step 4: **Han's approach: Isolate \(3y\) in the first equation.**

Han isolates \(3y\) in the first equation \(x + 3y = -5\):

\[
3y = -5 - x
\]

### Step 5: **Substitute Han's equation into the second equation.**

Substitute \(3y = -5 - x\) into \(9x + 3y = 3\):

\[
9x + (-5 - x) = 3
\]

Simplify:
\[
9x - 5 - x = 3
\]

\[
8x = 8
\]

Solve for \(x\):
\[
x = 1
\]

### Step 6: **Substitute \(x = 1\) back into Han's equation.**

Now substitute \(x = 1\) into \(3y = -5 - x\):

\[
3y = -5 - 1 = -6
\]

Solve for \(y\):
\[
y = -2
\]

Thus, Han's approach also gives the solution \(x = 1\) and \(y = -2\).

---

Both Tyler's and Han's methods lead to the same solution, \(x = 1\) and \(y = -2\).

Would you like more details or have any questions?

### Here are 5 related questions to expand on the topic:
1. What would happen if you isolated \(y\) instead of \(x\) in either method?
2. How would the solution change if the system had no solutions or infinitely many solutions?
3. Can you solve the system by using the elimination method instead of substitution?
4. Why does substituting into the second equation help verify the solution?
5. How do you know if a system of equations is consistent or inconsistent?

### Tip:
Always double-check your substitution by plugging the solution back into both original equations to ensure accuracy.

Tyler and Han are trying to solve this system by substitution: { x + 3y = -5, 9x + 3y = 3 }. Tyler's first step is to isolate x in the first equation to get x = -5 - 3y. Han's first step is to isolate 3y in the first equation to get 3y = -5 - x. Show that both first steps can be used to solve the system and will yield the same solution.

Explore how to solve a system of linear equations using the substitution method with two different starting points. This problem demonstrates isolating variables in equations and verifying that both approaches yield the same solution. Suitable for students in Grades 8-10.

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