In mathematics, a change of variables is a basic technique used to simplify problems in which the original variables are replaced with functions of other variables. The intent is that when expressed in new variables, the problem may become simpler, or equivalent to a better understood problem. Change of variables is an operation that is related to substitution. However these are different operations, as can be seen when considering differentiation (chain rule) or integration (integration by substitution). A very simple example of a useful variable change can be seen in the problem of finding the roots of the sixth-degree polynomial:
x 6 − 9 x 3 + 8 = 0. {\displaystyle x^{6}-9x^{3}+8=0.}
Sixth-degree polynomial equations are generally impossible to solve in terms of radicals (see Abel–Ruffini theorem). This particular equation, however, may be written
( x 3 ) 2 − 9 ( x 3 ) + 8 = 0 {\displaystyle (x^{3})^{2}-9(x^{3})+8=0}
(this is a simple case of a polynomial decomposition). Thus the equation may be simplified by defining a new variable u = x 3 {\displaystyle u=x^{3}} . Substituting x by u 3 {\displaystyle {\sqrt[{3}]{u}}} into the polynomial gives
u 2 − 9 u + 8 = 0 , {\displaystyle u^{2}-9u+8=0,}
which is just a quadratic equation with the two solutions:
u = 1 and u = 8. {\displaystyle u=1\quad {\text{and}}\quad u=8.}
The solutions in terms of the original variable are obtained by substituting x3 back in for u, which gives
x 3 = 1 and x 3 = 8. {\displaystyle x^{3}=1\quad {\text{and}}\quad x^{3}=8.}
Then, assuming that one is interested only in real solutions, the solutions of the original equation are
x = ( 1 ) 1 / 3 = 1 and x = ( 8 ) 1 / 3 = 2. {\displaystyle x=(1)^{1/3}=1\quad {\text{and}}\quad x=(8)^{1/3}=2.}
Simple example Consider the system of equations
x y + x + y = 71 {\displaystyle xy+x+y=71}
x 2 y + x y 2 = 880 {\displaystyle x^{2}y+xy^{2}=880}
where x {\displaystyle x} and y {\displaystyle y} are positive integers with x > y {\displaystyle x>y} . (Source: 1991 AIME) Solving this normally is not very difficult, but it may get a little tedious. However, we can rewrite the second equation as x y ( x + y ) = 880 {\displaystyle xy(x+y)=880} . Making the substitutions s = x + y {\displaystyle s=x+y} and t = x y {\displaystyle t=xy} reduces the system to s + t = 71 , s t = 880 {\displaystyle s+t=71,st=880} . Solving this gives ( s , t ) = ( 16 , 55 ) {\displaystyle (s,t)=(16,55)} and ( s , t ) = ( 55 , 16 ) {\displaystyle (s,t)=(55,16)} . Back-substituting the first ordered pair gives us x + y = 16 , x y = 55 , x > y {\displaystyle x+y=16,xy=55,x>y} , which gives the solution ( x , y ) = ( 11 , 5 ) . {\displaystyle (x,y)=(11,5).} Back-substituting the second ordered pair gives us x + y = 55 , x y = 16 , x > y {\displaystyle x+y=55,xy=16,x>y} , which gives no solutions. Hence the solution that solves the system is ( x , y ) = ( 11 , 5 ) {\displaystyle (x,y)=(11,5)} .
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