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Saturday, November 15, 2014

Solving Cubic Equations of Two Unknowns, 2

Category: Algebra

"Published in Newark, California, USA"

Solve the following systems by substitution:



Solution:

Consider the given equations above  



The first equation can be factored by the difference of the two cubes as follows 



But  

Hence, the above equation becomes




The second equation can be written as



Substitute the value of y to the above equation, we have 







If you equate each factor to zero, then the values of x are 3 and -2. 

If x = 3, then the value of y is




If x = -2, then the value of y is 




Therefore, the solutions of the two equations are: