Question: In a planetary system, the distances of three exoplanets from their star are $7y + 2$, $3y - 4$, and $5y + 6$ light-years. If their average distance is 12 light-years, find $y$. - Treasure Valley Movers

Discover the Hidden Math Behind Distant Worlds
Could light-years truly reveal a story about hidden planets orbiting distant stars? Right now, scientific curiosity—and growing public fascination with exoplanets—has sparked interest in how astronomers measure vast cosmic distances. The question at the heart of this trend is: In a planetary system, the distances of three exoplanets from their star are $7y + 2$, $3y - 4$, and $5y + 6$ light-years. If their average distance is 12 light-years, can we unlock the value of $y$ through simple math? Solving this isn’t just about numbers—it reflects how data-driven storytelling connects people to the search for life beyond Earth.

Why Exoplanet Distance Puzzles Are Capturing Attention

In recent years, space science has moved from remote discovery to public engagement. With breakthroughs like the James Webb Space Telescope capturing atmospheric clues on distant worlds, audiences crave clear explanations of the fundamentals. Measuring exoplanet distances using light-years brings abstract space into tangible terms, sparking interest across U.S. astronomy communities, education platforms, and science communicators.
This particular problem—finding $y$ when the average distance is 12 light-years—mirrors how real astronomy uses algebra to decode planetary systems. It’s not just academic; it resonates with anyone interested in how science measures the unknown, making it a perfect fit for Discover’s mission to raise curiosity without sensationalism.

Breaking Down the Formula: Finding $y$ with Logic

The average distance of the three exoplanets is calculated by summing their distances and dividing by three:

$$ \frac{(7y + 2) + (3y - 4) + (5y + 6)}{3} = 12 $$

Combining like terms:

$$ \frac{15y + 4}{3} = 12 $$

Multiplying both sides by 3:

$$ 15y + 4 = 36 $$

Solving for $y$:

$$ 15y = 32 \quad \Rightarrow \quad y = \frac{32}{15} $$

This result shows the precise value needed to balance the system mathematically. While unlikely to hint at direct real