Tyler and Han are trying to solve this system by substitution:

[tex]\[
\left\{
\begin{array}{l}
x + 3y = -5 \\
9x + 3y = 3
\end{array}
\right.
\][/tex]

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.



Answer :

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]