`x^(2)+y^(2)=160`
`y=3x`
If `(x, y)` is a solutions to the system of equations above, what is the value of `y^(2)`?

Options

A. `12`
B. `16`
C. `120`
D. `144`

Explanation

To solve the system of equations given by \(x^2 + y^2 = 160\) and \(y = 3x\), we will follow these steps:

### Step 1: Substitute \(y\) in the first equation
We have the second equation \(y = 3x\). We can substitute this into the first equation:

\[ x^2 + (3x)^2 = 160 \]

### Step 2: Simplify the equation
Now we simplify the equation:

\[ x^2 + 9x^2 = 160 \]

This combines to:

\[ 10x^2 = 160 \]

### Step 3: Solve for \(x^2\)
Next, we divide both sides by 10:

\[ x^2 = \frac{160}{10} = 16 \]

### Step 4: Find the values of \(x\)
Taking the square root of both sides gives us:

\[ x = \pm 4 \]

### Step 5: Find the corresponding values of \(y\)
Now, we can find the values of \(y\) using \(y = 3x\):

For \(x = 4\):

\[ y = 3(4) = 12 \]

For \(x = -4\):

\[ y = 3(-4) = -12 \]

### Step 6: Calculate \(y^2\)
Now we can find \(y^2\):

\[ y^2 = 12^2 = 144 \]

Thus, the value of \(y^2\) is:

\[ \boxed{144} \]

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