Solve For Y: Equation Analysis

Hey guys! Today, we're diving into a classic math problem that often trips people up if they're not careful. We've got this equation:

$$\frac{4}{y^2-3 y}=\frac{1}{y^2}$$

And we need to figure out which of the following statements about the value of 'y' is true. The options are:

A. $y=4$ B. $y=0,4$ C. $y=0,-1$ D. $y=0$

Let's break this down step-by-step and make sure we don't fall into any common traps. This is all about understanding algebraic manipulation and being aware of restrictions on variables, especially when they appear in denominators. So, grab your calculators (or just your brains!) and let's get to it!

Understanding the Equation and Potential Pitfalls

When we look at the equation $\frac{4}{y^2-3 y}=\frac{1}{y^2}$, the first thing that should jump out at you is that we have variables in the denominators. This is super important because, in mathematics, we cannot divide by zero. This means we need to identify any values of 'y' that would make either of the denominators equal to zero. If we find such values, they are not valid solutions to the equation.

Let's look at the denominators:

1. $y^2 - 3y$: To find the values that make this zero, we set it equal to zero: $y^2 - 3y = 0$. We can factor out a 'y' here: $y(y-3) = 0$. This equation is true if $y=0$ or if $y-3=0$, which means $y=3$. So, $y$ cannot be 0 or 3.
2. $y^2$: This denominator is zero only when $y=0$. So, $y$ cannot be 0.

Combining these restrictions, we know right off the bat that $y$ cannot be 0 or 3. This immediately tells us that any answer choice including $y=0$ as a valid solution is likely incorrect, unless there's a specific reason derived from the solving process that cancels it out or we're looking for extraneous solutions. But generally, these are the values we need to exclude from our potential solutions.

Now, let's move on to solving the equation itself. We'll use cross-multiplication, a handy technique for equations involving fractions. Remember, this technique works best when you have a single fraction on each side of the equals sign. We already do!

So, cross-multiplying gives us:

$4 \times y^2 = 1 \times (y^2 - 3y)$

This simplifies to:

$4y^2 = y^2 - 3y$

See? It's starting to look much cleaner already. The next step is to get all the terms on one side of the equation to set it equal to zero. This is a standard approach for solving polynomial equations, especially quadratic ones like this appears to be.

Let's subtract $y^2$ from both sides:

$4y^2 - y^2 = -3y$

$3y^2 = -3y$

Now, let's move the $-3y$ term to the left side by adding $3y$ to both sides:

$3y^2 + 3y = 0$

We now have a quadratic equation in standard form ($ax^2 + bx + c = 0$, although here $c=0$). The best way to solve this is by factoring. We can see that both terms have a common factor of $3y$. Let's factor that out:

$3y(y + 1) = 0$

For this product to be zero, at least one of the factors must be zero. So, we set each factor equal to zero:

1. $3y = 0 Rightarrow y = 0$
2. $y + 1 = 0 Rightarrow y = -1$

So, our potential solutions are $y=0$ and $y=-1$.

Evaluating the Potential Solutions Against Restrictions

Okay, guys, this is where things get really important. We found two potential solutions: $y=0$ and $y=-1$. But remember our earlier discussion about the denominators? We established that $y$ cannot be 0 or 3 because those values would make the original equation undefined (division by zero!).

Let's look at our potential solutions:

• $y = 0$: If we plug $y=0$ back into the original equation, the denominators become $0^2 - 3(0) = 0$ and $0^2 = 0$. Since we cannot divide by zero, $y=0$ is an extraneous solution. It's a solution that arises from the algebraic manipulation but doesn't satisfy the original equation due to the domain restrictions.
• $y = -1$: Let's check this one.
  • Left side: $\frac{4}{(-1)^2 - 3(-1)} = \frac{4}{1 + 3} = \frac{4}{4} = 1$
  • Right side: $\frac{1}{(-1)^2} = \frac{1}{1} = 1$ Since the left side equals the right side (1 = 1), $y=-1$ is a valid solution.

So, after considering the restrictions, the only true solution to the equation is $y=-1$. Now, let's compare this to our answer choices:

A. $y=4$ B. $y=0,4$ C. $y=0,-1$ D. $y=0$

Looking at our valid solution, $y=-1$, none of the options directly state only $y=-1$. This means we need to re-examine the question and options carefully. The question asks which of the following statements is true. It doesn't necessarily mean all solutions must be listed or that only the true solutions are listed. It's asking which statement itself accurately reflects the situation based on the equation.

Let's analyze each option in light of our findings:

• A. $y=4$: We found $y=-1$ is the solution. Is $y=4$ a solution? Let's check:

  • Left side: $\frac{4}{4^2 - 3(4)} = \frac{4}{16 - 12} = \frac{4}{4} = 1$
  • Right side: $\frac{1}{4^2} = \frac{1}{16}$ $1 \neq \frac{1}{16}$, so $y=4$ is not a solution. Statement A is false.

• B. $y=0,4$: We already know $y=0$ is not a valid solution, and $y=4$ is not a solution either. Statement B is false.

• C. $y=0,-1$: We found that $y=-1$ is a solution, but $y=0$ is not a valid solution because it makes the denominators zero. So, this statement lists both a valid solution ($y=-1$) and an invalid one ($y=0$). However, the way the question is phrased,