Sure! Let's work through the problem step by step.
We are given two relationships concerning the ages of Ele's car ([tex]\(x\)[/tex] years old) and Kat's car ([tex]\(y\)[/tex] years old):
1. Kat's car is 3 times as old as Ele's car. This can be written as the equation:
[tex]\[
y = 3x
\][/tex]
2. Kat's car is also 6 years older than Ele's car. This can be written as the equation:
[tex]\[
y = x + 6
\][/tex]
Now, let's consider the given multiple-choice options:
### Option (A)
[tex]\[
x = 3y \\
x - y = 6
\][/tex]
Analyzing Option (A):
- The equation [tex]\(x = 3y\)[/tex] would imply that Ele's car is 3 times as old as Kat's car, which contradicts the problem statement.
- Therefore, Option (A) is incorrect.
### Option (B)
[tex]\[
x = 3y \\
y - x = 6
\][/tex]
Analyzing Option (B):
- Similar to Option (A), the equation [tex]\(x = 3y\)[/tex] suggests that Ele's car is 3 times as old as Kat's car, which contradicts the problem statement.
- Therefore, Option (B) is incorrect.
### Option (C)
[tex]\[
y = 3x
\][/tex]
Analyzing Option (C):
- The equation [tex]\(y = 3x\)[/tex] correctly represents the statement that Kat's car is 3 times as old as Ele's car.
- We also derived [tex]\(y = x + 6\)[/tex], and both conditions are simultaneously met when the equations are considered together.
Let's verify by solving the equations together:
1. [tex]\(y = 3x\)[/tex]
2. [tex]\(y = x + 6\)[/tex]
Since both equations equal [tex]\(y\)[/tex], we can set them equal to each other:
[tex]\[
3x = x + 6
\][/tex]
Solving for [tex]\(x\)[/tex], we subtract [tex]\(x\)[/tex] from both sides:
[tex]\[
2x = 6
\][/tex]
Dividing both sides by 2:
[tex]\[
x = 3
\][/tex]
Substituting [tex]\(x = 3\)[/tex] back into [tex]\(y = 3x\)[/tex]:
[tex]\[
y = 3(3) = 9
\][/tex]
So, Ele's car is 3 years old, and Kat's car is 9 years old. The equation [tex]\(y = x + 6\)[/tex] also holds true because [tex]\(9 = 3 + 6\)[/tex].
Therefore, the system of equations that best models the situation is:
[tex]\[
\boxed{y = 3x}
\][/tex]
So, the correct answer is (C).
