Solving The System Of Equations Xy=25 And 3x-y=35 A Step-by-Step Guide

Hey guys! Today, we're diving into the fascinating world of systems of equations. Specifically, we're going to tackle a problem that involves both a product and a linear equation. Don't worry if it sounds intimidating – we'll break it down step by step so it's super easy to follow. We'll explore different methods to solve this type of problem and highlight some common pitfalls to avoid. So, grab your thinking caps, and let's get started!

The Challenge: Cracking the Code of Equations

So, the million-dollar question is: What's the solution to the following system of equations?

\begin{cases}
xy = 25 \\
3x - y = 35
\end{cases}

We've got some options to choose from:

A) (5, 5) B) (10, 2.5) C) (7, 3.57) D) (8, 3.125)

This looks like a classic system of equations problem, but with a twist! We've got a non-linear equation (xy = 25) hanging out with a linear one (3x - y = 35). This means we can't just use simple elimination or substitution straight away. We'll need a clever approach to crack this code. The core challenge here is to find the values of x and y that satisfy both equations simultaneously. Think of it like a puzzle where two pieces need to fit perfectly together. Each equation represents a constraint, and the solution is the point where these constraints overlap. We need to find the pair of numbers (x, y) that makes both equations true at the same time. This involves a bit of algebraic manipulation and strategic thinking.

Method 1: The Substitution Sleuth

One of the most reliable methods for solving systems of equations is substitution. The core idea behind substitution is simple: we isolate one variable in one equation and then substitute that expression into the other equation. This effectively reduces our two-variable problem into a single-variable problem, which is much easier to solve. Let's see how it works in our case. First, let's take the second equation, 3x - y = 35, and isolate y. We can do this by adding y to both sides and subtracting 35 from both sides:

3x - y = 35
3x - 35 = y
y = 3x - 35

Now, we have an expression for y in terms of x. This is our golden ticket! We can now substitute this expression into the first equation, xy = 25:

x(3x - 35) = 25

See what we did there? We replaced y with (3x - 35), and now we have an equation with only x! This is a quadratic equation, which we know how to solve. Let's expand and rearrange the equation:

3x^2 - 35x = 25
3x^2 - 35x - 25 = 0

Now we have a standard quadratic equation in the form ax^2 + bx + c = 0. We can solve this using the quadratic formula, factoring, or completing the square. For this case, the quadratic formula is our best bet.

Cracking the Quadratic: Unleashing the Formula

The quadratic formula is a powerful tool for solving equations of the form ax^2 + bx + c = 0. It states that:

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

In our equation, 3x^2 - 35x - 25 = 0, we have a = 3, b = -35, and c = -25. Let's plug these values into the quadratic formula:

x = \frac{-(-35) \pm \sqrt{(-35)^2 - 4(3)(-25)}}{2(3)}
x = \frac{35 \pm \sqrt{1225 + 300}}{6}
x = \frac{35 \pm \sqrt{1525}}{6}

Now, let's simplify the square root. 1525 can be factored as 25 * 61, so \sqrt{1525} = \sqrt{25 * 61} = 5\sqrt{61}. This gives us:

x = \frac{35 \pm 5\sqrt{61}}{6}

We have two possible solutions for x:

x_1 = \frac{35 + 5\sqrt{61}}{6}
x_2 = \frac{35 - 5\sqrt{61}}{6}

Let's approximate these values. \sqrt{61} is approximately 7.81, so:

x_1 \approx \frac{35 + 5(7.81)}{6} \approx \frac{35 + 39.05}{6} \approx \frac{74.05}{6} \approx 12.34
x_2 \approx \frac{35 - 5(7.81)}{6} \approx \frac{35 - 39.05}{6} \approx \frac{-4.05}{6} \approx -0.675

So, we have two possible x values: approximately 12.34 and -0.675. Now, we need to find the corresponding y values.

Finding the 'y' to Complete the Pair

Remember our expression for y? We found that y = 3x - 35. Now, we'll use our two x values to find the corresponding y values. Let's start with x_1 ≈ 12.34:

y_1 = 3(12.34) - 35 \approx 37.02 - 35 \approx 2.02

So, one possible solution is approximately (12.34, 2.02). Now, let's find the y value for x_2 ≈ -0.675:

y_2 = 3(-0.675) - 35 \approx -2.025 - 35 \approx -37.025

This gives us another possible solution: approximately (-0.675, -37.025). Now, let's check these solutions against our original equations.

The Moment of Truth: Verifying Our Solutions

It's crucial to check our solutions to make sure they actually work in both original equations. This is especially important when dealing with non-linear equations, as we might encounter extraneous solutions (solutions that arise from our algebraic manipulations but don't actually satisfy the original equations). Let's start by checking the approximate solution (12.34, 2.02):

• Equation 1: xy = 25

  (12.34)(2.02) \approx 24.93
  

  This is close to 25, so it seems promising.

• Equation 2: 3x - y = 35

  3(12.34) - 2.02 \approx 37.02 - 2.02 \approx 35
  

  This one checks out perfectly!

Now, let's check the other approximate solution (-0.675, -37.025):

• Equation 1: xy = 25

  (-0.675)(-37.025) \approx 25.0
  

  This also looks good!

• Equation 2: 3x - y = 35

  3(-0.675) - (-37.025) \approx -2.025 + 37.025 \approx 35
  

  This one checks out as well!

So, we have two potential solutions. However, looking back at our answer choices, none of them perfectly match our calculated solutions. This suggests there might be a slight error in the answer choices or that we need to look for an exact solution.

A Closer Look at the Options: Spotting the Best Fit

Let's revisit the answer choices and see if any of them come close to satisfying our equations:

A) (5, 5) B) (10, 2.5) C) (7, 3.57) D) (8, 3.125)

Let's test each option in our equations:

• Option A: (5, 5)

  • Equation 1: (5)(5) = 25 (Checks out!)
  • Equation 2: 3(5) - 5 = 15 - 5 = 10 (Doesn't equal 35)

  So, option A is not the solution.

• Option B: (10, 2.5)

  • Equation 1: (10)(2.5) = 25 (Checks out!)
  • Equation 2: 3(10) - 2.5 = 30 - 2.5 = 27.5 (Doesn't equal 35)

  Option B is also not the solution.

• Option C: (7, 3.57)

  • Equation 1: (7)(3.57) \approx 24.99 (Close to 25)
  • Equation 2: 3(7) - 3.57 = 21 - 3.57 = 17.43 (Doesn't equal 35)

  Option C is not the solution.

• Option D: (8, 3.125)

  • Equation 1: (8)(3.125) = 25 (Checks out!)
  • Equation 2: 3(8) - 3.125 = 24 - 3.125 = 20.875 (Doesn't equal 35)

  Option D is not the solution either.

It seems like none of the provided options perfectly satisfy both equations. This could indicate an error in the options themselves, or it might mean that the exact solutions are not simple integers or decimals.

Conclusion: Unraveling the Mystery

While we've explored the substitution method and the quadratic formula, and meticulously checked the provided options, we haven't found a perfect match among the answer choices. This highlights an important aspect of problem-solving: sometimes, the answer isn't neatly presented. In this case, our calculations using the quadratic formula gave us approximate solutions, and none of the multiple-choice answers perfectly aligned. This could be due to a slight error in the provided options, or it might suggest that the exact solution involves more complex numbers or a different form. Even though we didn't find a direct match, the process of working through the problem has been invaluable. We've reinforced our understanding of systems of equations, mastered the substitution method, and confidently wielded the quadratic formula. These are skills that will serve us well in tackling future mathematical challenges! Remember, guys, the journey of problem-solving is just as important as the destination. Keep practicing, keep exploring, and never be afraid to dive into a challenge, even if the answer isn't immediately clear.

If I had to choose the closest option based on our work, I'd lean towards none of the above given the discrepancy. However, it's crucial to double-check the original problem and options for any potential errors.

Take Away

• When solving systems of equations, especially those involving non-linear equations, it's crucial to check your solutions in both original equations.
• The quadratic formula is a powerful tool for solving quadratic equations, but it can lead to complex solutions.
• Sometimes, the provided answer choices might not be perfectly accurate, and it's important to recognize this and consider the closest possible solution.
• Don't be discouraged if you don't find a perfect solution immediately. The problem-solving process itself is a valuable learning experience.