Solving For Y In Terms Of X A Step By Step Guide

Hey guys! Today, we're diving into a super important topic in algebra: solving systems of equations. Specifically, we're going to focus on how to solve for y and express it in terms of x. This is a fundamental skill that you'll use in many areas of math, so let's get right to it!

Understanding Systems of Equations

Before we jump into the nitty-gritty, let's make sure we're all on the same page about what a system of equations actually is. Simply put, a system of equations is a set of two or more equations that contain the same variables. Our goal is to find values for those variables that make all of the equations true simultaneously. It's like finding the sweet spot where all the equations agree!

In this particular case, we're dealing with a system of two linear equations, which means each equation represents a straight line when graphed. The solution to the system is the point where these lines intersect. This point gives us the values of x and y that satisfy both equations. But, we're not just looking for any solution; we want to express y in terms of x. This means we want an equation that looks like y = something times x plus a constant. This form is incredibly useful because it tells us exactly how y changes as x changes. Expressing y in terms of x is a core concept in algebra and is super useful for understanding the relationship between variables. It allows us to easily see how the value of y depends on the value of x, and this is crucial in many real-world applications. Whether you are graphing lines, analyzing data, or solving word problems, mastering this skill will give you a solid foundation. So, buckle up and let’s dive into the methods to conquer this algebraic challenge!

The Equations We're Tackling

Okay, let's take a look at the specific system of equations we're going to solve:

$\begin{array}{l} 6x - 2y = 10 \\ 2x + 3y = 51 \end{array}$

Our mission, should we choose to accept it (and we do!), is to find an expression for y in terms of x. This means we want to rewrite the equations so that we end up with something like y = [some expression involving x]. There are a couple of popular methods we can use to achieve this: substitution and elimination. Let's explore both to see which one works best for our situation. The beauty of math is that there often isn't just one way to solve a problem! Understanding different methods gives you flexibility and helps you choose the most efficient approach. Think of these methods as tools in your algebraic toolbox. The more tools you have, the better equipped you'll be to tackle any problem that comes your way. So, as we delve into substitution and elimination, pay attention to the steps involved and consider which method resonates best with your problem-solving style. Remember, the goal isn't just to get the answer, but to understand the process and why it works. This deeper understanding will serve you well in more advanced math courses and in real-world applications where problem-solving skills are highly valued. Now, let's get started and uncover the secrets to expressing y in terms of x!

Method 1: Substitution

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This effectively eliminates one variable, leaving us with a single equation in one variable that we can solve. Let's start with the first equation:

$6x - 2y = 10$

Our goal is to isolate y. First, let's subtract 6x from both sides:

$-2y = -6x + 10$

Now, we'll divide both sides by -2 to solve for y:

$y = 3x - 5$

Awesome! We've successfully solved the first equation for y. This gives us a crucial stepping stone in our journey. The elegance of the substitution method lies in its ability to simplify complex systems by reducing the number of variables. By expressing one variable in terms of another, we create a pathway to unravel the relationships hidden within the equations. Think of it as untangling a knot – carefully loosening one strand at a time until the whole thing comes undone. This skill is particularly useful when dealing with non-linear systems or systems with multiple variables, where other methods might become cumbersome. So, as we move forward, remember that substitution is not just a trick, but a powerful technique grounded in logical reasoning. Mastering it will unlock new dimensions in your mathematical problem-solving abilities and empower you to tackle even the most intricate challenges with confidence. Now that we have y isolated, let's see how we can use this to solve the rest of the system!

Substituting into the Second Equation

Now that we have y expressed in terms of x (y = 3x - 5), we can substitute this expression into the second equation:

$2x + 3y = 51$

Replace y with (3x - 5):

$2x + 3(3x - 5) = 51$

Now, we have an equation with only x, which we can solve. First, distribute the 3:

$2x + 9x - 15 = 51$

Combine like terms:

$11x - 15 = 51$

Add 15 to both sides:

$11x = 66$

Finally, divide both sides by 11:

$x = 6$

Great! We've found the value of x. But remember, our ultimate goal is to express y in terms of x. So, we're not quite done yet. But the hardest part is over! The power of substitution shines through in this step, where we transform a complex equation into a solvable form. This highlights a crucial aspect of problem-solving: breaking down a problem into smaller, manageable steps. By carefully substituting and simplifying, we create a clear path towards the solution. This strategic approach is not just valuable in mathematics but also in various aspects of life, where tackling large tasks requires a systematic methodology. So, as you continue your mathematical journey, remember the importance of methodical substitution and simplification. These techniques are not just about crunching numbers; they are about developing a mindset that empowers you to conquer any challenge, big or small. Now, let’s use this value of x to finally unlock the expression for y!

Expressing y in Terms of x (Finally!)

Remember, we already found y in terms of x when we solved the first equation:

$y = 3x - 5$

This is exactly what we were looking for! We've successfully expressed y in terms of x. We could also substitute the value of x we found (x=6) into this equation to find the numerical value of y, but the question specifically asked for y in terms of x, so we're all set. The final step brings us full circle, emphasizing the importance of understanding what the problem is truly asking for. In this case, it wasn't just about finding a numerical solution, but about expressing the relationship between variables. This highlights the nuances of mathematical problem-solving and the need to pay close attention to the question's wording. Moreover, this final expression, y = 3x - 5, encapsulates the entire relationship between x and y in a concise and elegant form. It reveals the slope and y-intercept of the line, giving us a complete picture of the linear equation. So, take a moment to appreciate the beauty of this final expression – it's not just an answer, it's a representation of a fundamental mathematical concept! And with that, we have successfully conquered the challenge. Let's celebrate our achievement and continue our journey to explore the fascinating world of mathematics!

Solution

The solution for y in terms of x is:

$y = 3x - 5$

Method 2: Elimination (Bonus!)

Just for kicks, let's quickly look at the elimination method. This method involves manipulating the equations so that either the x or y coefficients are opposites. Then, when we add the equations together, one variable cancels out.

Our equations are:

$\begin{array}{l} 6x - 2y = 10 \\ 2x + 3y = 51 \end{array}$

Let's eliminate x. Multiply the second equation by -3:

$-6x - 9y = -153$

Now, add this modified equation to the first equation:

$\begin{array}{rcrcr} 6x & - & 2y & = & 10 \\ -6x & - & 9y & = & -153 \\ \hline 0x & - & 11y & = & -143 \end{array}$

This simplifies to:

$-11y = -143$

Divide both sides by -11:

y = rac{-143}{-11} = 13

Now, substitute this value of y back into either of the original equations to solve for x. Let's use the first equation:

$6x - 2(13) = 10$

$6x - 26 = 10$

$6x = 36$

$x = 6$

Now, to express y in terms of x, substitute x = 6 into the original equation 6x - 2y = 10 to solve for y:

$6(6) - 2y = 10$

$36 - 2y = 10$

$-2y = -26$

$y = 13$

However, to express y in terms of x, we rearrange the equation to solve for y:

$-2y = 10 - 6x$

$y = 3x - 5$

So, even using elimination, we arrive at the same expression for y in terms of x!

Key Takeaways

• Systems of equations are sets of equations with the same variables.
• Solving for y in terms of x means rewriting the equation as y = [expression involving x].
• The substitution method involves solving one equation for one variable and substituting that expression into the other equation.
• The elimination method involves manipulating equations to eliminate one variable when added together.

I hope this explanation was helpful! Remember, practice makes perfect, so keep working on these skills, and you'll become a master of systems of equations in no time. Keep up the great work, guys! You've got this!