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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.

Sagot :

GinnyAnswer
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]
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