# How do you factor and solve for y: 14y^2-31xy=10x^2?

Nov 21, 2016

$y = - \frac{2}{7} x$ or

$y = \frac{5}{2} x$

#### Explanation:

The equation $14 {y}^{2} - 31 x y = 10 {x}^{2}$ is just like any quadratic equation, as it can be written as

$14 {y}^{2} - 31 x y - 10 {x}^{2} = 0$ and dividing each term by ${x}^{2}$

$14 {y}^{2} / {x}^{2} - 31 \frac{x y}{x} ^ 2 - 10 = 0$

or $14 {\left(\frac{y}{x}\right)}^{2} - 31 \frac{y}{x} - 10 = 0$ and assuming $\frac{y}{x} = u$ this is

$14 {u}^{2} - 31 u - 10 = 0$

Note that in such cases we have $f \left(x , y\right)$ and degree of each term is $2$. Hence, it is called as homogeneous equation.

Let us now factorize it in normal way and for this we have to identify two numbers whose sum is $- 31$ and product is $14 x \left(- 10\right) = - 140$. These are $- 35$ and $4$.

$14 {y}^{2} - 31 x y - 10 {x}^{2} = 0$

or $14 {y}^{2} - 35 x y + 4 x y - 10 {x}^{2} = 0$

or $7 y \left(2 y - 5 x\right) + 2 x \left(2 y - 5 x\right) = 0$

or $\left(7 y + 2 x\right) \left(2 y - 5 x\right) = 0$

Therefore either $7 y + 2 x = 0$ i.e. $7 y = - 2 x$ i.e. $y = - \frac{2}{7} x$

or $2 y - 5 x = 0$ i.e. $2 y = 5 x$ i.e. $y = \frac{5}{2} x$