Algebraic Fractions Equality Explained

The problem presents us with a set of equal algebraic fractions: $ \frac{x}{x+y+z} = \frac{y}{x+y+z} = \frac{z}{x+y+z} = \frac{2k}{3} $ We are also given that the variables $x, y, z$ are positive numbers ($x > 0, y > 0, z > 0$). This condition is important because it ensures that the denominator, $x+y+z$, is a positive value and therefore not equal to zero.

Equating Numerators from Equal Fractions

A fundamental property of fractions is that if they have the same non-zero denominator, then for the fractions to be equal, their numerators must also be equal. In this specific problem, the denominator $x+y+z$ is common to the first three fractions. Since these fractions are stated to be equal, we can deduce that their numerators must be equal:

Simplifying Algebraic Fractions

To proceed, let's represent the common value of $x, y,$ and $z$ with a single variable, say $a$. Since $x, y,$ and $z$ are all positive, $a$ must also be positive ($a > 0$). Now, we substitute $a$ for $x, y,$ and $z$ in the fractions:

Given that $a > 0$, we can simplify this expression by canceling $a$ from both the numerator and the denominator:

This tells us that each of the algebraic fractions $\frac{x}{x+y+z}$, $\frac{y}{x+y+z}$, and $\frac{z}{x+y+z}$ simplifies to the value $\frac{1}{3}$.

Solving for k

The problem states that these fractions are also equal to $\frac{2k}{3}$. Since we've determined that the fractions equal $\frac{1}{3}$, we can now set up an equation involving $k$:

To find the value of $k$, we can first eliminate the denominator by multiplying both sides of the equation by 3:

This simplifies to:

Now, we isolate $k$ by dividing both sides by 2:

Calculating 6k Value

The final step is to calculate the value of $6k$. Using the value of $k$ we just found ($k = \frac{1}{2}$), we perform the multiplication:

Therefore, the value of $6k$ is 3.