Solve For Y: 9(y^2+w)-6=9R

Hey math whizzes and algebra adventurers! Today, we're diving deep into a problem that might look a little daunting at first glance, but trust me, guys, it's all about breaking it down step-by-step. We're going to tackle the equation $9 (y^2+w)-6=9 R$ and figure out how to solve for y. Remember, the key here is to isolate 'y' on one side of the equation, and we'll do this using the trusty rules of algebra. We're assuming all other variables – that's 'w' and 'R' – are non-zero, which simplifies things a bit. So, grab your calculators, your notebooks, and let's get this done!

The Starting Point: Understanding the Equation

Alright, let's look at our equation: $9 (y^2+w)-6=9 R$. Our ultimate goal is to get 'y' all by itself. Right now, 'y' is tucked inside parentheses, being squared, and then multiplied by 9, and then we subtract 6. That's a lot going on! But don't worry, we'll peel back these layers one by one. Think of it like unwrapping a present – each step reveals a bit more. We're given that 'w' and 'R' are non-zero, which is a good hint that we don't need to worry about dividing by zero or any weird edge cases involving those variables. This is a standard algebraic manipulation problem, perfect for flexing those problem-solving muscles. The question is straightforward: solve for y. This means we need to rearrange the equation until we have an expression that looks like '$y = ext{[some expression involving R and w]}$'. This is a common task in mathematics, especially when you're working with formulas and need to express one variable in terms of others. So, let's get into the nitty-gritty of how we achieve this isolation.

Step 1: Isolate the Term with 'y'

Our first move is to get the part of the equation that contains 'y' – which is $9 (y^2+w)$ – by itself on one side. Right now, we have a '-6' chilling on the left side. To get rid of it, we do the opposite: we add 6 to both sides of the equation. This is crucial for maintaining balance. Whatever you do to one side, you must do to the other. So, let's add 6:

$9 (y^2+w)-6 + 6 = 9 R + 6$

This simplifies to:

$9 (y^2+w) = 9 R + 6$

See? We're one step closer. Now, the term with 'y' is almost isolated. The next obstacle is the '9' that's multiplying the entire parenthesis $(y^2+w)$. To undo multiplication, we perform division. So, we'll divide both sides of the equation by 9:

rac{9 (y^2+w)}{9} = rac{9 R + 6}{9}

This leaves us with:

(y^2+w) = rac{9 R + 6}{9}

We can simplify the right side a little further by dividing each term in the numerator by 9:

(y^2+w) = rac{9 R}{9} + rac{6}{9}

(y^2+w) = R + rac{2}{3}

Awesome! We've successfully isolated the term $y^2+w$. This involved two key algebraic operations: addition to move a constant term and division to remove a multiplier. These are fundamental steps in solving equations, and we've applied them perfectly here. The right side, R + rac{2}{3}, now represents the value of $y^2+w$. It's important to note that while we simplified rac{6}{9} to rac{2}{3}, you could also leave it as rac{9R+6}{9} if that felt clearer, but simplification is usually good practice. This intermediate result shows us how the sum of $y^2$ and $w$ relates to the other variables in the equation.

Step 2: Isolate $y^2$

We're getting warmer! Our equation is now (y^2+w) = R + rac{2}{3}. Our target is still 'y', but right now we have $y^2$. To get $y^2$ by itself, we need to move the '+w' to the other side. Just like before, we perform the opposite operation. Since we're adding 'w', we subtract 'w' from both sides:

y^2 + w - w = R + rac{2}{3} - w

This simplifies to:

y^2 = R + rac{2}{3} - w

Boom! We've now isolated $y^2$. This is a significant step because $y^2$ is directly related to 'y' by a simple square root. The expression on the right side, R + rac{2}{3} - w, is the value that $y^2$ must equal. It's a combination of the original variable 'R' and the variable 'w', adjusted by a constant fraction. This shows how 'y' is dependent on the values of 'R' and 'w'. Remember our assumption that 'w' and 'R' are non-zero? That's still in play, and it doesn't complicate this step. We're just isolating terms. The key takeaway here is that we successfully separated $y^2$ from any other additive or subtractive terms on its side of the equation, bringing us one step closer to finding the value of 'y' itself. The algebraic move here is subtraction, the inverse of addition, used to move the 'w' term across the equals sign.

Step 3: Solve for 'y'

We've reached the final frontier, guys! We have y^2 = R + rac{2}{3} - w. To get 'y' from $y^2$, we need to take the square root of both sides. This is the inverse operation of squaring. When we take the square root, we have to remember that a number squared can result in a positive value, so there are often two possible solutions: a positive one and a negative one. Therefore, we must include the 'plus or minus' symbol (±).

oxed{y = oxed{oldsymbol{ ormalsize oldsymbol{ pmath{oldsymbol{ pmath{y}}}}}

So, there you have it! We've successfully solved for y in the equation $9 (y^2+w)-6=9 R$. The solution is y = oldsymbol{ ormalsize oldsymbol{ pmath{oldsymbol{ pmath{y}}}}}. This process involved a series of standard algebraic manipulations: adding a constant, dividing by a coefficient, subtracting a variable, and finally, taking a square root while accounting for both positive and negative possibilities. It's a great example of how we can systematically unpack complex equations to find the value of a specific variable. Keep practicing these steps, and you'll be an algebra master in no time! Remember, math is all about practice and understanding the fundamental rules. Happy solving!