# Solve (3x+y)/8 = (x-y)/5 = (x²-y²)/5. What is the values for x and y?

Jul 4, 2015

The two solutions are: $\left(x , y\right) = \left(0 , 0\right)$ and $\left(x , y\right) = \left(\frac{13}{6} , - \frac{7}{6}\right)$

#### Explanation:

$\frac{3 x + y}{8} = \frac{x - y}{5} = \frac{{x}^{2} - {y}^{2}}{5}$

Start with $\frac{x - y}{5} = \frac{{x}^{2} - {y}^{2}}{5}$. Multiply by $5$ and factor the right side:

$\left(x - y\right) = \left(x - y\right) \left(x + y\right)$.

Collect on one side:
$\left(x - y\right) \left(x + y\right) - \left(x - y\right) = 0$.

Factor $\left(x - y\right)$

$\left(x - y\right) \left(x + y - 1\right) = 0$.

So $x - y = 0$ or $x + y - 1 = 0$

This gives us: $y = x$ or $y = 1 - x$

Now use the first two expressions together with these solutions for $y$.

$\frac{3 x + y}{8} = \frac{x - y}{5}$
Leads to: $15 x + 5 y = 8 x - 8 y$.

So $7 x + 13 y = 0$

Solution 1
Now, when $y = x$, we get $20 x = 0$, so $x = 0$ and thus $y = 0$

Solution 2
When $y = 1 - x$, we get

$7 x + 13 \left(1 - x\right) = 0$

$7 x + 13 - 13 x = 0$

$- 6 x = - 13$

$x = \frac{13}{6}$ and
$y = 1 - x = 1 - \frac{13}{6} = - \frac{7}{6}$

Checking these solutions

$\frac{3 x + y}{8} = \frac{x - y}{5} = \frac{{x}^{2} - {y}^{2}}{5}$

For $\left(0 , 0\right)$, we get

$\frac{0}{8} = \frac{0}{5} = \frac{0}{5}$

For $\left(\frac{13}{6} , - \frac{7}{6}\right)$, we get:

$\frac{3 \left(\frac{13}{6}\right) + \left(- \frac{7}{6}\right)}{8} = \frac{39 - 7}{48} = \frac{32}{48} = \frac{2}{3}$
$\frac{\left(\frac{13}{6}\right) - \left(- \frac{7}{6}\right)}{5} = \frac{20}{30} = \frac{2}{3}$
$\frac{{\left(\frac{13}{6}\right)}^{2} - {\left(- \frac{7}{6}\right)}^{2}}{5} = \frac{169 - 49}{36 \cdot 5} = \frac{120}{36 \cdot 5} = \frac{20}{6 \cdot 5} = \frac{2}{3}$