Question

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

Answer 1

`y=-2/7x` or

`y=5/2x`

Explanation:

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

`14y^2-31xy-10x^2=0` and dividing each term by `x^2`

`14y^2/x^2-31(xy)/x^2-10=0`

or `14(y/x)^2-31y/x-10=0` and assuming `y/x=u` this is

`14u^2-31u-10=0`

Note that in such cases we have `f(x,y)` 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 `14x(-10)=-140`. These are `-35` and `4`.

`14y^2-31xy-10x^2=0`

or `14y^2-35xy+4xy-10x^2=0`

or `7y(2y-5x)+2x(2y-5x)=0`

or `(7y+2x)(2y-5x)=0`

Therefore either `7y+2x=0` i.e. `7y=-2x` i.e. `y=-2/7x`

or `2y-5x=0` i.e. `2y=5x` i.e. `y=5/2x`