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:
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 sum 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
By using completing the square method, the value of x is
If 
, then y equals
If 
, then y equals
Since the solutions of the two equations have imaginary numbers or complex numbers, then the two equations have no solution.
