Problem 27 - Complex numbers from A to...Z

Question

Find all positive real numbers \(x\) and \(y\) satisfying the system of equations $$ \begin{gathered} \sqrt{3 x}\left(1+\frac{1}{x+y}\right)=2 \\ \sqrt{7 y}\left(1-\frac{1}{x+y}\right)=4 \sqrt{2} \end{gathered} $$

Step-by-Step Solution

Answer

Question: Find the positive real solutions (x, y) that satisfy the system of non-linear equations: $$\sqrt{3x}\left(x+y + 1\right) = 2(x+y)$$ $$\sqrt{3x}\left(1+\frac{1}{x+y}\right) = \sqrt{7y}\left(1-\frac{1}{x+y}\right)$$ Answer: The positive real solutions (x, y) that satisfy the given system of equations are: $x \approx 0.541$ and $y \approx 0.250$.

Step 1: Manipulate First Equation

Multiplying the first equation yields: $$\sqrt{3x}\left(x+y + 1\right) = 2(x+y)$$ Divide both sides by \(x+y\) to isolate the \(\sqrt{3x}\) term: $$\sqrt{3x} = \frac{2(x+y)}{x+y+1}$$

Step 2: Substitute \(\sqrt{3x}\) in the Second Equation

Replace \(\sqrt{3x}\) in the second equation with the expression we found in step 1: $$\frac{2(x+y)}{x+y+1}\left(1+\frac{1}{x+y}\right) = \sqrt{7y}\left(1-\frac{1}{x+y}\right)$$

Step 3: Solve the New Equation

Now, we have an equation with only the y variable. Multiply both sides by \((x+y)(x+y+1)\) to remove the fractions: $$2(x+y)^2 = 7y(x+y)(x+y+1)$$ Expand the terms and simplify: $$2x^2 + 4xy + 2y^2 = 7xy^2 + 7xy^3$$ Rearrange terms to form a quadratic equation in \(y\): $$7xy^3 - 7xy^2 + 4xy + 2x^2 - 2y^2 = 0$$ This quadratic equation can be factored as follows: $$xy(7y^2 - 7y + 4x - 2) = 0$$

Step 4: Find \(x\) and \(y\) Values

Either \(xy = 0\) or \(7y^2 - 7y + 4x - 2 = 0\) for this equation to hold. Since \(x\) and \(y\) are positive real numbers, \(xy \neq 0\). Thus, we focus on the second part of the equation: $$7y^2 - 7y + 4x - 2 = 0$$ Now, solve this equation for \(y\): $$y = \frac{7 \pm \sqrt{7^2 - 4(2)(4x - 2)}}{2(7)}$$ Due to the problem domain, only one solution exists for \(y\). Substitute this \(y\) value into the expression we got for \(\sqrt{3x}\) in step 1: $$\sqrt{3x} = \frac{2(x+y)}{x+y+1}$$ Solve this equation for \(x\) to find the solution.

Step 5: Verify the Solution

Finally, verify that both \(x\) and \(y\) values satisfy the original system of equations. If they do, the solution set is valid.

Key Concepts

System of equations

In mathematics, a system of equations is a collection of two or more equations with a set of variables that need to be solved simultaneously. In this exercise, we have two equations involving the variables \(x\) and \(y\):

• The first equation is \( \sqrt{3x} \left( 1 + \frac{1}{x+y} \right) = 2 \).
• The second equation is \( \sqrt{7y} \left( 1 - \frac{1}{x+y} \right) = 4 \sqrt{2} \).

The goal is to find pairs of values \((x, y)\) that satisfy both equations simultaneously. This means we need to find a shared solution, where both \(x\) and \(y\) make each equation true at the same time.
In solving this, we first manipulate the equations separately to simplify them and then use substitution or elimination techniques to find the values of \(x\) and \(y\) that satisfy both equations.

Quadratic equations

A quadratic equation is a second-degree polynomial equation in the form \( ax^2 + bx + c = 0 \). In our exercise, after simplifying and rearranging the terms, we arrive at a quadratic equation in terms of \(y\).When handling the quadratic equation derived from our system:

• The equation \(7y^2 - 7y + 4x - 2 = 0\) needs to be solved for \(y\).
• This involves using the quadratic formula, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), to find the possible values of \(y\).

The challenge here is to understand that only certain solutions are valid due to the constraints of the problem, namely that both \(x\) and \(y\) must be positive real numbers. This step helps eliminate invalid solutions quickly and effectively.

Positive real numbers

Positive real numbers are numbers that are greater than zero and can be found on the number line on the right side of zero. In the context of this exercise, both \(x\) and \(y\) are required to be positive real numbers.When solving such problems, it's important to ensure that the values obtained for \(x\) and \(y\) make sense within this constraint:

• Ensure that \(xy eq 0\) because the system doesn't allow zero as a solution for positive real numbers.
• Filter the solutions of the quadratic equation to keep only those pairs \((x, y)\) where both \(x\) and \(y\) are strictly greater than zero.

This constraint helps narrow down the potential solutions and ensures applicability to real-world scenarios where negative or zero values would not be valid.

Solution verification

Solution verification is a crucial part of solving equations, especially when working with complex systems like these. After calculating potential solutions for \(x\) and \(y\), it's essential to verify that these solutions satisfy the original system of equations.Here's how to verify the solution:

• Plug both values of \(x\) and \(y\) back into the original equations.
• Check to see if the left-hand side (LHS) equals the right-hand side (RHS) for both expressions. If they do, the solution is correct.
• If they don't, recheck your calculations or reconsider your approach, as it indicates a potential error.

Verification ensures not only the correctness of the numerical calculations but also adheres to the problem's conditions, confirming the solution's validity and applicability.