Find the real values of x and y for which the following equation is satisfied ((1+i)x - 2i)/(3+i) +

Q: Why can't we just cross-multiply the fractions directly without finding a common denominator?

You technically could, but it would lead to a much more complicated expression with complex numbers in the denominators on both sides. Finding the common denominator $(3+i)(3-i) = 10$ is a much more elegant and efficient strategy. It immediately transforms the problem into a standard linear system, which is much easier to solve.

Q: Wait, why do we equate the real and imaginary parts separately? Can't we just solve for $x$ in terms of $y$?

Because $x$ and $y$ are real numbers, the equation $A + iB = C + iD$ implies $A=C$ and $B=D$ because $1$ and $i$ are linearly independent over the real numbers. This is a powerful property of complex numbers that allows us to split one complex equation into two real linear equations, making the system solvable.

Q: I keep getting a sign error when expanding $(1+i)(3-i)$. How do I avoid this?

The best way to avoid sign errors is to write out the expansion step-by-step using the FOIL method: $(1)(3) + (1)(-i) + (i)(3) + (i)(-i)$. Then, explicitly write $i^2 = -1$ and substitute it back in. For example, $(i)(-i) = -i^2 = -(-1) = 1$. Taking that extra second to write out the substitution prevents the most common mistakes in complex algebra.

Question

Find the real values of $x$ and $y$ for which the following equation is satisfied $$\frac{(1+i)x - 2i}{3+i} + \frac{(2-3i)y + i}{3-i} = i$$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are going to demystify a problem that might look like a tangled mess of imaginary units and fractions, but is, in reality, a beautifully structured puzzle.

We are tasked with finding the real values of $x$ and $y$ that satisfy the equation:

$$ \frac{(1+i)x - 2i}{3+i} + \frac{(2-3i)y + i}{3-i} = i $$

At first glance, the presence of $i$ in the denominators might feel intimidating. However, the secret to this problem is not brute force, but rather the symmetry of complex conjugates.

### The Conjugate Strategy

The first thing we notice is that our denominators, $(3+i)$ and $(3-i)$, are complex conjugates. When we multiply these two, we are essentially calculating the square of the modulus of the complex number $3+i$.

The math unfolds as:

$$ (3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 10 $$

By multiplying the entire equation by this common denominator of $10$, we clear the fractions entirely. This transforms our equation into a much more manageable form:

$$ ((1+i)x - 2i)(3-i) + ((2-3i)y + i)(3+i) = 10i $$

### The Algebraic Expansion

Now, we must be diligent. We need to expand these terms without letting a single sign slip through our fingers.

Let us take the first term: $((1+i)x - 2i)(3-i)$. Distributing the terms, we get $(1+i)(3-i)x - 2i(3-i)$.

Expanding $(1+i)(3-i)$ gives us $3 - i + 3i - i^2$, which simplifies to $4+2i$. Similarly, $-2i(3-i)$ becomes $-6i + 2i^2$, which is $-6i - 2$. Combining these, the first part of our equation becomes $(4+2i)x - 2 - 6i$.

We apply the same careful logic to the second term: $((2-3i)y + i)(3+i)$. This expands to $(2-3i)(3+i)y + i(3+i)$.

The product $(2-3i)(3+i)$ yields $6 + 2i - 9i - 3i^2$, which simplifies to $9-7i$. Adding $i(3+i) = 3i + i^2 = 3i - 1$, we get $(9-7i)y + 3i - 1$.

### The Separation of Worlds

With our expansions complete, we bring everything together:

$$ (4+2i)x - 2 - 6i + (9-7i)y + 3i - 1 = 10i $$

Now, we group the real parts and the imaginary parts. The real parts are $4x + 9y - 3$, and the imaginary parts are $i(2x - 7y - 3)$.

Setting this equal to $10i$, we have:

$$ (4x + 9y - 3) + i(2x - 7y - 3) = 0 + 10i $$

Because $1$ and $i$ are linearly independent, we can equate the real and imaginary parts separately. This gives us our system of equations:

$$ 4x + 9y - 3 = 0 $$
$$ 2x - 7y - 3 = 10 $$

### The Final Intersection

We are left with two linear equations: $4x + 9y = 3$ and $2x - 7y = 13$. To solve this, we multiply the second equation by $2$ to get $4x - 14y = 26$.

Subtracting this from the first equation, we find:

$$ 23y = -23 \Rightarrow y = -1 $$

Substituting $y = -1$ back into $2x - 7y = 13$, we get $2x + 7 = 13$, so $2x = 6$, and finally $x = 3$.

We have arrived at our destination: **$x = 3$ and $y = -1$**. You have successfully navigated the complex plane and turned a daunting equation into a simple linear system.

Animated Solution for Mathematics - Complex Numbers: Find the real values of $x$ and $y$ for which the following equation is satisfied $\frac{( 1 + i ) x - 2 i}{3 + i} + \frac{( 2 - 3 i ) y + i}{3 - i} = i$ .

Visualized Solution

Analyze the Equation

Find Common Denominator

Eliminate Fractions

Expand the First Term

Expand the Second Term

Combine and Group Terms

Equate Real and Imaginary Parts

Form Equation 1

Form Equation 2

Strategy for Solving

Solve for $x$ and $y$

Summary and Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Conjugate Strategy

The Algebraic Expansion

The Separation of Worlds

The Final Intersection

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the real values of x and y for which the following equation is satisfied 3+i(1+i)x−2i​+3−i(2−3i)y+i​=i.