# Solving of Equations with Method of Introducing New Variable

It is easier to explain essence of the method of introducing new variable on example.

Example 1. Solve equation (x^2-3x)^2+3(x^2-3x)-28=0.

Let x^2-3x=y, then initial equation can be rewritten as y^2+3y-28=0. This equation has two roots: y_1=-7,y_2=4.

Thus, we obtained set of two equations: x^2-3x=-7,x^2-3x=4, i.e. x^2-3x+7=0,x^2-3x-4=0.

First equation doesn't have roots because its discriminant D=(-3)^2-4*1*7=-19<0. Second equation has roots x=4 and x=1. These roots are also roots of initial equation.

Example 2. Solve equation 24/(x^2+2x-8)-15/(x^2+2x-3)=2.

Let x^2+2x=y, then equation can be rewritten as (24)/(y-8)-(15)/(y-3)=2.

Common denominator is (y-8)(y-3). So, (24(y-3))/((y-8)(y-3))-(15(y-8))/((y-3)(y-8))=(2(y-3)(y-8))/((y-3)(y-8)). This can be rewritten as (24(y-3)-15(y-8)-2(y-3)(y-8))/((y-8)(y-3))=0.

Domain of this equation is all y, except y=3 and y=8.

Now, fraction equals zero, when numerator equals zero: 24(y-3)-15(y-8)-2(y-3)(y-8)=0 or y(2y-31)=0. Therefore, either y=0 or y=31/2. Both of these roots are in domain of the equation.

But y=x^2+2x, so we obtain set of equations: x^2+2x=0,x^2+2x=31/2.

First equation has solutions 0 and -2. Second equation has solutions x=-1+-sqrt(33/2).

All these solutions are solutions of initial equation: 0,-2,-1+sqrt(33/2),-1-sqrt(33/2).