Solving For Y: A System Of Equations Challenge

Hey guys! Today, we're diving deep into a super cool math problem that'll test your algebra skills. We're tackling a system of equations, and our mission, should we choose to accept it, is to find the $y$-coordinate of the solution. It sounds fancy, but trust me, with a bit of focus and some handy algebraic tricks, we'll nail this. We've got two equations here: a linear one and a quadratic one. This means they'll look a bit different when graphed – one's a straight line, and the other's a curve (a parabola, to be precise!). Our goal is to find where these two graphs intersect, because that intersection point is our solution. And out of that point, we're specifically hunting for the $y$-value. So, grab your notebooks, get comfy, and let's break down how to find that elusive $y$-coordinate. We'll go step-by-step, making sure we understand each move. Remember, practice makes perfect, and solving these kinds of problems is a fantastic way to boost your mathematical prowess. Let's get started on this algebraic adventure!

Understanding the Equations: Line Meets Parabola

Alright team, let's get a good look at the two equations we're working with. We have:

Equation 1: $y = 8x + 9$ Equation 2: $y = x^2 + 4x + 13$

Notice how both equations are already solved for $y$? This is a huge advantage, guys! It means we can use a method called substitution really easily. Since both $8x + 9$ and $x^2 + 4x + 13$ are equal to $y$, they must be equal to each other. Think of it like this: if A = C and B = C, then A must equal B. So, we can set the right-hand sides of both equations equal to each other. This is the key step that allows us to eliminate $y$ temporarily and solve for $x$. Once we have the value(s) of $x$ that satisfy both equations simultaneously, we can then plug those $x$ values back into either of the original equations to find the corresponding $y$ value. Since we're asked for the $y$-coordinate of the solution, this process is exactly what we need. It's like solving a puzzle where each piece fits together perfectly. The linear equation represents a straight path, and the quadratic equation represents a curved path. Where do these paths cross? That crossing point is our solution, and we need the vertical position ($y$-value) of that crossing.

Setting Up the Substitution

The first equation, $y = 8x + 9$, is our linear equation. It describes a straight line with a slope of 8 and a $y$-intercept of 9. The second equation, $y = x^2 + 4x + 13$, is our quadratic equation. It describes a parabola, which is a U-shaped curve. Finding the solution to a system of equations means finding the point(s) $(x, y)$ that satisfy both equations at the same time. Graphically, this is where the line and the parabola intersect. Because both equations are already set equal to $y$, we can use the substitution method. This means we can take the expression for $y$ from the first equation ($8x + 9$) and substitute it into the second equation wherever we see $y$. Or, we could substitute the expression from the second equation into the first. It doesn't matter which way you go, but often it's easier to substitute the simpler expression (the linear one) into the more complex one (the quadratic). So, we'll set $8x + 9$ equal to $x^2 + 4x + 13$. This gives us a single equation with only one variable, $x$. This is our pathway to finding the $x$-coordinates of the intersection points. Once we find the $x$'s, we're one step closer to our ultimate goal: the $y$-coordinate!

The Equation to Solve for x

So, as we just discussed, since both $8x + 9$ and $x^2 + 4x + 13$ are equal to $y$, we can set them equal to each other. This gives us the following equation:

$8x + 9 = x^2 + 4x + 13$

This equation now only contains the variable $x$. Our next step is to rearrange this equation into the standard form of a quadratic equation, which is $ax^2 + bx + c = 0$. To do this, we need to move all the terms to one side of the equation, usually the side where $x^2$ is positive, to make it easier to solve. Let's subtract $8x$ and $9$ from both sides of the equation:

$0 = x^2 + 4x - 8x + 13 - 9$

Now, let's combine the like terms:

$0 = x^2 + (4x - 8x) + (13 - 9)$ $0 = x^2 - 4x + 4$

So, our quadratic equation is $x^2 - 4x + 4 = 0$. This is the equation we need to solve for $x$. Finding the values of $x$ that satisfy this equation will tell us the $x$-coordinates of the points where our line and parabola intersect. Once we have these $x$-values, we can plug them back into either of our original equations to find the corresponding $y$-values. This equation, $x^2 - 4x + 4 = 0$, is crucial because it holds the key to finding our solution. Keep your eyes peeled, because solving this quadratic equation is our next big step!

Solving the Quadratic Equation for x

We've arrived at the quadratic equation $x^2 - 4x + 4 = 0$. Now, we need to find the value(s) of $x$ that make this equation true. There are a few ways to solve quadratic equations: factoring, using the quadratic formula, or completing the square. Let's see if this one is easy to factor. We're looking for two numbers that multiply to give us 4 (the constant term) and add up to give us -4 (the coefficient of the $x$ term). Can you think of any numbers? If you think about it, $-2$ and $-2$ fit the bill! $(-2) imes (-2) = 4$ and $(-2) + (-2) = -4$. This means our quadratic equation can be factored as:

$(x - 2)(x - 2) = 0$

Or, more simply:

$(x - 2)^2 = 0$

To solve for $x$, we set the factor equal to zero:

$x - 2 = 0$

Adding 2 to both sides gives us:

$x = 2$

What's interesting here, guys, is that we only got one value for $x$. This means that the line and the parabola don't intersect at two distinct points; they actually touch at exactly one point. This is called a tangent point. The line is tangent to the parabola. This is a special case, and it happens when the discriminant of the quadratic equation is zero, which our equation $x^2 - 4x + 4 = 0$ confirms, as $b^2 - 4ac = (-4)^2 - 4(1)(4) = 16 - 16 = 0$. So, our solution for $x$ is $x=2$. This is the $x$-coordinate of our single intersection point.

Factoring vs. Quadratic Formula

We chose to factor the quadratic equation $x^2 - 4x + 4 = 0$ because it was a perfect square trinomial, making it super straightforward. However, not all quadratics are this easy to factor. If you encounter one that isn't, the quadratic formula is your best friend. The quadratic formula solves for $x$ in any equation of the form $ax^2 + bx + c = 0$ and is given by:

x = rac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In our case, $a=1$, $b=-4$, and $c=4$. Plugging these values into the formula would give us:

x = rac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4)}}{2(1)} x = rac{4 \pm \sqrt{16 - 16}}{2} x = rac{4 \pm \sqrt{0}}{2} x = rac{4}{2} $x = 2$

As you can see, the quadratic formula also yields $x=2$. This consistency is reassuring! It confirms that our factoring was correct and that there is indeed only one solution for $x$. This single $x$-value is the $x$-coordinate of the unique point where our line and parabola meet. Mastering both factoring and the quadratic formula gives you the flexibility to tackle any quadratic equation you come across in your math journey. They are powerful tools in your algebraic arsenal!

The Solution for x

After applying either factoring or the quadratic formula to the equation $x^2 - 4x + 4 = 0$, we found a single, definitive solution for $x$:

$x = 2$

This value, $x=2$, represents the $x$-coordinate of the point where the line $y = 8x + 9$ and the parabola $y = x^2 + 4x + 13$ intersect. The fact that we found only one value for $x$ tells us that the line is tangent to the parabola; they meet at exactly one point. This is an important piece of information. If we had found two distinct values for $x$, it would mean the line crossed the parabola at two separate points. If we had found no real values for $x$ (which would happen if the value under the square root in the quadratic formula was negative), it would mean the line and the parabola do not intersect at all. But in our case, $x=2$ is our confirmed $x$-coordinate. Now, we're ready for the final, and arguably most important, step: finding the $y$-coordinate of this intersection point. We're almost there, guys!

Finding the y-Coordinate

We've successfully found the $x$-coordinate of the solution, which is $x=2$. Now, our ultimate goal is to find the $y$-coordinate of this solution point. Remember, the solution is a point $(x, y)$ that lies on both the line and the parabola. This means that if we plug our value of $x=2$ into either of the original equations, we should get the same $y$-value. This is a great way to check our work! Let's use the first equation, the linear one, because it's simpler:

$y = 8x + 9$

Substitute $x=2$ into this equation:

$y = 8(2) + 9$ $y = 16 + 9$ $y = 25$

So, based on the first equation, the $y$-coordinate is 25. Now, to be absolutely sure, let's plug $x=2$ into the second equation, the quadratic one:

$y = x^2 + 4x + 13$

Substitute $x=2$ into this equation:

$y = (2)^2 + 4(2) + 13$ $y = 4 + 8 + 13$ $y = 12 + 13$ $y = 25$

Wow, it matches! Both equations give us $y=25$ when $x=2$. This confirms that our solution point is $(2, 25)$, and the $y$-coordinate of this solution is indeed 25. This is fantastic news, guys! We've navigated through substitution, solved a quadratic equation, and finally found the $y$-coordinate. High fives all around!

Using Both Equations for Verification

It's always a smart move in mathematics to verify your answer. We did this by plugging our $x$-value ($x=2$) into both of the original equations. Let's recap why this is so important. The solution to a system of equations is a point that satisfies all equations in the system. If we find an $x$-value and then plug it into one equation to get a $y$-value, we've only shown that the point lies on that one equation. By plugging that same $x$-value into the other equation and getting the exact same $y$-value, we've proven that the point lies on both equations simultaneously. This makes us incredibly confident in our answer. In our case, plugging $x=2$ into $y = 8x + 9$ gave us $y = 25$. Then, plugging $x=2$ into $y = x^2 + 4x + 13$ also gave us $y = 25$. The perfect match confirms that the point $(2, 25)$ is the unique solution to this system of equations. This verification step is crucial for catching any calculation errors and ensuring the accuracy of your final answer. It's your safety net in the world of algebra!

The Final Answer: The y-Coordinate

After all our hard work and careful calculations, we have reached the final answer. We started with a system of two equations: a linear equation and a quadratic equation. By using the substitution method, we set the expressions for $y$ equal to each other, which led us to a quadratic equation in terms of $x$: $x^2 - 4x + 4 = 0$. We solved this quadratic equation and found a single solution for $x$, which was $x=2$. This single value indicated that the line and the parabola were tangent at one point. Finally, we substituted this $x$-value back into both of the original equations to find the corresponding $y$-value. Both equations yielded $y=25$. Therefore, the solution to the system of equations is the point $(2, 25)$. The question specifically asked for the $y$-coordinate of the solution. And that, my friends, is 25.

Conclusion: A Job Well Done!

So there you have it, team! We successfully tackled a system of equations involving a line and a parabola. We used the powerful technique of substitution to eliminate one variable and form a single quadratic equation. Then, we solved that quadratic equation to find the $x$-coordinate of the intersection point. Crucially, we verified our $x$-value by plugging it back into both original equations, ensuring that we obtained the same $y$-value for both. This meticulous process led us to the unique solution point $(2, 25)$. And the $y$-coordinate, which is what the question asked for, is a solid 25. Problems like these are fantastic for building your algebraic muscles. They teach you about the relationships between different types of functions and how to find common ground between them. Keep practicing, keep exploring, and don't be afraid to dive into more complex problems. You've got this, and I'm here to help you every step of the way. Great job, everyone!