9x + 10y = 18; this is an equation having 'x' and 'y' variables with 9 and 10 as coefficients respectively.
Place 'y' term on left side and 'x' term on the right side so we get,
10y = 18 – 9x;
10y = 9 (2 -x) ;
y = (9/10) (2-x) ; this is the required value of 'y' in terms of 'x'.
For quadratic equation, the equations are in the form of second degree due to which there is a possibility of Square root at the end of the result. Let us see an example of a quadratic equation and solve it for 'y' in terms of 'x':
x = uy2 + vy + w;
This is a quadratic equation with two variables 'x' and 'y', where u, v, w are the constant or we can say variable coefficients.
Divide all terms by the coefficient of 'y' then we get,
x/u = y2 + vy/u + w/u;
Let us assume that, w/u = a + b, a = (v/2u)2,
Then the equation will be:
x/u = y2 + vy/u + a + b;
x/u – b = y2 + vy/u + a;
x/u – b = (y + v/2u)2, where a = (v/2u)2,
Now interchange the equation from right to left then,
y + v/2u = √(x/u – b);
y = √(x/u -b) – v/2u, this is the required result for 'y' in terms of 'x'.